Tag: complex numbers

Reminder about i-arrows

Nearly two weeks ago, I first wrote about the i-arrow visualisation of the points in the complex plane. Here’s a reminder of how they work:

Every point with complex coordinates is represented as an arrow (which I call an “i-arrow”) from one place to another on top of the Cartesian plane.

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

I want to know where the line joining two complex points is

In an earlier blog post, I investigated the complex points on various kinds of lines. I came to a complete understanding of the complex points on a real line with an equation like \(ax+by=c\) for \(a\), \(b\), \(c\) real. (They are the i-arrows drawn from one point on the real line to the other.) I also came to a complete understanding of the complex points on an unreal line with real slope with an equation like \(ax+by=C\) for \(a\), \(b\) real and \(C\) unreal. (They are the i-arrows drawn between two parallel real lines.) But I was still not sure about a lot of things to do with a line of unreal slope with an equation like \(Ax+By=C\) for at least one of \(A\), \(B\) unreal and \(C\) any complex number.

The thing that really bothered me was this: given any two points in the complex plane, it should be possible to find the unique line that joins them. That is, I want to be able to take two points, and find the other points on the line that joins them. That’s how lines should work. I know I could find the equation of the line and use that equation to find new points, but that is really unsatisfying to me. I don’t just want to fiddle around with algebra, I want a geometrical way to find the other points. I want to take a pair of i-arrows, or an i-arrow and a real point, and be able to find all the other i-arrows on the complex line that joins them using some sort of construction.

Pairs of points on a line of real slope

In that first blog post about lines and the one after it, I had made some small headway on this problem. For some very special pairs of points, I can see all of the other complex points on the line joining them. They are the situations where the two points create a line of real slope, because lines of real slope already have very simple constructions for the i-arrows of their complex points.

A real point and an i-arrow aligned with it

The real line aligned with an i-arrow is the unique real line it is on, so if an i-arrow is pointing at or pointing away from a real point, then the two are both on that unique real line the i-arrow is on.

Two i-arrows pointing along the same line

If two i-arrows both lie along the same real line, then this is the only real line either of them is on and so it is the line that joins them.

Two i-arrows sharing a base

If two i-arrows share a base, then they are on an unreal line of real slope. Draw the line joining their arrowheads, and a line parallel to that through the shared base, and the i-arrows you want are all the arrows from the base line to the arrowhead line.

Two i-arrows sharing an arrowhead

If two i-arrows share an arrowhead, they are also on an unreal line of real slope. Draw the line joining the bases, and a line parallel to that through the shared arrowhead, and the i-arrows you want are all the arrows from the base line to the arrowhead line.

Two i-arrows where the line joining the bases is parallel to the line joining the arrowheads

The above two situations are actually special cases of a more general situation where the line joining the bases of the two i-arrows is parallel to the line joining their arrowheads. In that case, they must be on an unreal line with that slope. All the other points on that line are the i-arrows drawn from the base line to the arrowhead line.

A new way to work with complex points

What I really hope for is some sort of construction that tells me the complex points on the line of unreal slope joining any two points. Up until now, I’ve been able to take either two i-arrows, or a real point and an i-arrow, and use them to find some of the other points on the line joining them. (These were described in the second blog post about unreal lines.) This week, I suddenly realised that the reason I was only able to make some of the points was that I was only combining the points together in some of the possible ways, and this led to me trying a new way to work with the complex points.

At the top, I talk about the complex point \((p+si,q+ti)\) having an i-arrow based at \((p,q)\) and extending along the vector \((s,t)\). There’s nothing stopping me, if I’m using complex number arithmetic, from literally writing the point using vector notation as \((p,q)+(s,t)i\). And if I do that, I could give those two parts names of their own. I’ll call \((p,q)\) the point \(P\) and \((s,t)\) the vector \(\mathbf{a}\) so that the complex point \((p,q)+(s,t)i\) becomes \(P+\mathbf{a}i\). This allows me to do algebra in a new way.

In the regular real Cartesian plane, all the points on the line joining point \(P\) and point \(Q\) are of the form \(P+m(Q-P)\) for real numbers \(m\). In this calculation, \(Q-P\) is the vector from \(P\) to \(Q\), and \(m(Q-P)\) is the same vector but stretched or shrunk by the factor \(m\), so that \(P+m(Q-P)\) is the point found by starting at \(P\) and moving some multiple of the journey from \(P\) to \(Q\), to arrive at some point on the line defined by \(P\) and \(Q\).

There is nothing stopping us doing the exact same thing with complex points.

Consider the complex points \(P+\mathbf{a}i\) and \(Q+\mathbf{b}i\). Then all the complex points on the line joining them are given by \(P+\mathbf{a}i + (m+\mu i)[(Q+\mathbf{b}i) – (P+\mathbf{a}i)]\) for some complex number \(m+\mu i\). Doing some algebra on this point…

\[\begin{aligned}& P+\mathbf{a}i + (m+\mu i)[(Q+\mathbf{b}i)-(P+\mathbf{a}i)]\\
&=P+\mathbf{a}i + (m+\mu i)[Q+\mathbf{b}i-P-\mathbf{a}i]\\
&=P+\mathbf{a}i + (m+\mu i)[Q- P + \mathbf{b}i-\mathbf{a}i]\\
&=P+\mathbf{a}i + (m+\mu i)[(Q- P) + (\mathbf{b}-\mathbf{a})i]\\
&=P+\mathbf{a}i + m(Q- P) + m(\mathbf{b}-\mathbf{a})i +\mu (Q- P)i- \mu (\mathbf{b}-\mathbf{a})\\
&= P + m(Q-P)-\mu (\mathbf{b}-\mathbf{a}) + [\mathbf{a} +\mu (Q- P) + m(\mathbf{b}- \mathbf{a})]i\end{aligned}\]

The i-arrow for this point has base at \(P +m(Q-P)– \mu(\mathbf{b}-\mathbf{a})\)
arrow \(\mathbf{a} +μ(Q- P) + m(\mathbf{b}-\mathbf{a})\)
and arrowhead at \(P + m(Q-P) – \mu(\mathbf{b}-\mathbf{a}) + \mathbf{a} +\mu(Q- P) + m(\mathbf{b}-\mathbf{a})\).

Notice how the new base and arrow are the old base and arrow for \(P+\mathbf{a}i\) adjusted by some multiples of the vectors \(Q-P\) and \(\mathbf{b}-\mathbf{a}\). This is already hinting at a geometrical way of figuring out new points on the line from the original two.

An i-arrow and a real point create parallel i-arrows

Suppose we have a real point \(R\) and an unreal point \(Q+\mathbf{b}i\). (That is, the point \(P+\mathbf{a}i\) in the formula above is actually the real point \(R\), so that \(P=R\) and \(\mathbf{a}=\mathbf{0}\).)

Then the other points on the line become
\(R + m(Q-P)-\mu b + [\mu(Q- R) + m\mathbf{b}]i\) which has i-arrow with
base \(R + m(Q-R)-\mu\mathbf{b}\),
arrow \(\mu(Q- R) + m\mathbf{b}\),
and arrowhead \(P + m(Q-R)-\mu\mathbf{b} + \mu(Q- R) + m\mathbf{b}\).

If \(\mu=0\), then this becomes
\(R + m(Q-R) + m\mathbf{b}i\), which has i-arrow with
base \(P + m(Q-R)\),
arrow \(m\mathbf{b}\),
and arrowhead \(R + m(Q-R)+ m\mathbf{b}\).

The point \(R+m(Q-R)\) is a point somewhere on the line joining \(R\) and \(Q\).
The vector \(m\mathbf{b}\) is parallel to \(\mathbf{b}\).
The point \(R+m(Q-R)+m\mathbf{b}= R+ m(Q+\mathbf{b}-R)\) is a point somewhere on the line joining \(R\) and \(Q+\mathbf{b}\).
So to find a new point on this line, draw the line joining \(R\) and the base \(Q\) and choose a point on this line for the base of the i-arrow, then move parallel to \(\mathbf{b}\) until you reach the line joining \(R\) and the arrowhead \(Q+\mathbf{b}\) and this is your i-arrow.

This neatly confirms the previous work I did, but as I said before, they are only some of the complex points on this complex line. In fact, they are the ones formed by adding a real multiple of the journey from \(R\) to \(Q+\mathbf{b}i\) onto \(R\). What if we don’t use a real multiple?

An i-arrow and a real point create spirals and spiderwebs

In the situation above with real point \(R\) and unreal point \(Q+\mathbf{b}i\), all the complex points on the line joining them were given by
\(R + m(Q-R)-\mu \mathbf{b} + [\mu(Q- R) + m\mathbf{b}]i\)
which has i-arrow with
base \(R + m(Q-R)-\mu\mathbf{b}\),
arrow \(\mu(Q- R) + m\mathbf{b}\),
and arrowhead \(R + m(Q-R)-\mu\mathbf{b} + \mu(Q- R) + m\mathbf{b}\).

Look what happens when \(\mu=-1\) and \(m=1\), so that the number \(m+\mu i = 1-i\). The complex point is:
\(R + (Q-R) +\mathbf{b} + [-(Q- R) + \mathbf{b}]i\)
\(= Q + \mathbf{b} + [R-Q + \mathbf{b}]i\),
which has i-arrow with
base \(Q + \mathbf{b}\),
arrow \(R-Q + \mathbf{b}\),
and arrowhead \(R + 2\mathbf{b}\).

How ridiculously simple is THAT?!

If I take a real point \(R\) and a complex point \(Q+\mathbf{b}i\), then I can create a new complex point on this line whose base is at the arrowhead of the i-arrow I already have. And to figure out where the new arrowhead is, I just go twice the length of the i-arrow I already have starting at the real point.

But wait. I could repeat this process as long as I like to create spirals like those ones I made at the end of investigating lines in a finite plane. Only this time, there’s no cumbersome fiddling around with algebra. It’s a clean and simple geometric process. So gorgeous!

I can even do this process backwards to make the spiral go in the other direction. The line from \(R\) to the existing arrowhead is twice the previous arrow, so halve it and put that pointing at the base of the existing i-arrow. And so I can make a spiral going inwards.

At first, I thought this would give me all different directions for arrows, but actually it doesn’t seem like it does. It looks like every fourth arrow points in the same direction. I wonder if I can prove that?

The first arrow goes from \(Q\) to \(Q+\mathbf{b}\) and is the vector \{b}.
The next arrow goes from \(Q+\mathbf{b}\) to \(R+2\mathbf{b}\) and is the vector \((R+2\mathbf{b})-(Q+\mathbf{b}) = (R-Q) + \mathbf{b}\).
The next arrow goes from \(R+2\mathbf{b}\) to \(R+2((R-Q) + \mathbf{b}) = R + 2(R-Q) + 2\mathbf{b}\), and is the vector \((R+2(R-Q) + 2\mathbf{b})-(R+2\mathbf{b}) = 2(R-Q)\).
The next arrow goes from \(R + 2(R-Q) + 2\mathbf{b}\) to \(R + 2(2(R-Q)) = R + 4(R-Q)\), and is the vector \((R + 4(R-Q))-(R + 2(R-Q) + 2\mathbf{b}) = 2(R-Q)-2\mathbf{b}\).
The next arrow goes from \(R + 4(R-Q)\) to \(R + 2(2(R- Q)-2\mathbf{b}) = R + 4(R-Q)-4\mathbf{b}\), and is the vector \((R + 4(R-Q)-4\mathbf{b})-(R + 4(R-Q)) = 4\mathbf{b}\).

So yes, every fourth arrow is in the same direction, and not only that, is exactly four times the length. That was completely unexpected. How weird is it that every possible complex line does this?!

Anyway, I can combine the spirals with the propagation of parallel i-arrows to create a spiderweb starting with any i-arrow and a real point.

It’s still not every i-arrow on this complex line, but it sure is pretty!

But wait! Making the spiderweb is even simpler than making the spiral. According to the calculations I did earlier, given \(R\) and \(Q+\mathbf{b}i\), I can construct the complex point \(R+2\mathbf{b}+2(R-Q)i\). Even better, I can follow parallel i-arrows inwards to get the complex point \(R+\mathbf{b}+(R-Q)i\). To be absolutely sure this works, look at the original algebra for new points on the line joining \(R\) and \(Q+\mathbf{b}i\):

\(R + m(Q-R)-\mu \mathbf{b} + [\mu(Q- R) + m\mathbf{b}]i\)
which has i-arrow with
base \(R + m(Q-R)-\mu\mathbf{b}\),
arrow \(\mu(Q- R) + m\mathbf{b}\),
and arrowhead \(R + m(Q-R)-\mu\mathbf{b} + \mu(Q- R) + m\mathbf{b}\).

If \(m=0\) and \(\mu=-1\) (corresponding to the number \(m+\mu i\) being \(-i\)), then the point is
\(R +\mathbf{b}-(Q- R)i = R+\mathbf{b}+(R-Q)i\)
which has i-arrow with
base \(R+\mathbf{b}\),
arrow \(R-Q\),
and arrowhead \(R+\mathbf{b}+(R-Q)\).

This is a remarkably simple construction to produce that spiderweb from real point \(R\) and complex point \(Q+\mathbf{b}i\).

1. Copy vector \(\mathbf{b}\) and place two of them one after the other starting at \(R\). Draw the i-arrow from \(Q+\mathbf{b}\) to this new point \(R+2\mathbf{b}\). This is a second i-arrow on the line.
2. Copy vector \(R-Q\) and place it at \(R+\mathbf{b}\). This is a third i-arrow on the line.
3. Copy vector \(R-Q\) and place two of them one after the other starting at \(R\). Draw the i-arrow from \(R+\mathbf{b}+(R-Q)\) to this new point \(R+2(R-Q)\). This is a fourth i-arrow on the line.
4. The lines through \(R\) and the bases of these four i-arrows define eight regions of the plane where the i-arrows are parallel to these four original i-arrows and have base on one line and arrowhead on the next.

A real point helps to find i-arrows based along an i-arrow

Even though the spiderweb of i-arrows shown above doesn’t tell you where every i-arrow is, we have at least covered the entire of the Cartesian plane with i-arrows. Given any point in the plane, even though we don’t have the i-arrow based at that point, we do have an i-arrow passing through it. I think this is enough information to find the i-arrow based there.

Let \(R\) be a real point and \(Q+\mathbf{b}i\) be an unreal point. As calculated before, all the complex points on the line joining them are given by
\(R + m(Q-R)-\mu\mathbf{b}+ [\mu(Q- R) + mb]i\)
which has i-arrow with
base \(R + m(Q-R)-\mu\mathbf{b}\),
arrow \(\mu(Q- R) + m\mathbf{b}\),
and arrowhead \(R + m(Q-R)-\mu\mathbf{b} + \mu(Q- R) + m\mathbf{b}\).

If I want the base \(P\) to be on the actual line segment that is the drawing of the i-arrow, that means the base has to be \(Q+\rho\mathbf{b}\) for some number \(\rho\) between 0 and 1.
That is, \(R + m(Q-R)-\mu\mathbf{b}= Q+\rho\mathbf{b}\).
I can make that happen by setting \(m=1\) and \(\mu=-\rho\).
This makes the arrow \(-\rho(Q- R) + 1\mathbf{b} = \mathbf{b} + \rho(R-Q)\),
and the arrowhead \(Q+\rho\mathbf{b} + \mathbf{b} + \rho(R-Q) = Q+\mathbf{b} + \rho(R-Q+\mathbf{b})\).

The vector \(R-Q+\mathbf{b}\) is the vector for the i-arrow with base \(Q+\mathbf{b}\)! That is, the next i-arrow outwards in the spiral. Which means that when the base of an i-arrow is on another i-arrow, then the arrowhead is on the next i-arrow in the spiral. Even more, if the base is proportion \(\rho\) along the first i-arrow, then the arrowhead is proportion \(\rho\) along the second.

The arrowhead itself being \(\mathbf{b} + \rho(R-Q)\) gives me a geometrical construction for finding the arrowhead. The vector \(\mathbf{b}\) is the first i-arrow, and the vector \(R-Q\) is the vector from \(Q\) to \(R\). So, to find the arrowhead (\(A\) in the above diagram), I go another journey of vector \(\mathbf{b}\) from \(P\), and then go parallel to the vector from \(Q\) to \(R\) until I meet the next i-arrow in the spiral.

I think this is really cool, because it also gives an alternative way to find the next i-arrow in the spiral from any i-arrow \(Q+\mathbf{b}i\) and the real point \(R\). Go another copy of the vector \(\mathbf{b}\) from \(Q+\mathbf{b}\), then go a copy of the vector \(Q-R\) from there, and this is the arrowhead of the next arrow in the spiral.

Oh my. I just realised this makes a whole new way to construct the starting arrows for the spiderweb using \(R\) and \(Q+\mathbf{b}i\).

1. Start at \(Q+\mathbf{b}\) and draw the i-arrow that is the vector \(\mathbf{b}+(R-Q)\).
2. From the arrowhead of this i-arrow, go \(–\mathbf{b}\) from there (which is halfway to \(R\)). This is the base of the next i-arrow.
3. Draw the i-arrow that is the vector \(R-Q\).
4. From the arrowhead of this i-arrow, draw the i-arrow that is the vector \((R-Q)-\mathbf{b}\).

I’m not sure if this is any better than the previous one, but it certainly has a different feel for me, and it might feel nicer in the moment when I need it.

What all of this means is that given a real point \(R\) and an unreal point \(Q+\mathbf{b}i\), I can now find the i-arrow for any complex point on the line joining these two points based at any point \(P\) in the Cartesian plane.

1. Construct the four starting i-arrows and spoke lines of the spiderweb using \(R\) and \(Q+\mathbf{b}i\).
2. Draw an i-arrow from the spiderweb through \(P\), parallel to the appropriate base i-arrow.
3. Draw the next i-arrow in the spiral from this one just drawn.
4. Construct the i-arrow based at \(P\) with arrowhead on the i-arrow just drawn.

It might not be the most simple of constructions, but I think it has a certain charm. I actually love making parts of the spiderweb until you get to the point you want.

So now I have a geometrical process that can find every other complex point on the line joining a real point and a complex point. I am so so happy.

The line joining two i-arrows arranged head to tail

The last part in this journey is to find the line joining two general complex points. If I can find the real point on this line, then the construction above will find all the other points too.

At the very least I can find the real point on a line joining two i-arrows arranged head to tail.

If \(Q+\mathbf{b}i\) and \(P+\mathbf{a}i\) are arranged so that the arrowhead of \(Q+\mathbf{b}i\)’s i-arrow is the base of \(P+\mathbf{a}i\)’s i-arrow, then they are two steps on the same spiral and it ought to be true that the real point on the line joining them is \(-2\mathbf{b}\) away from the arrowhead of \(P+\mathbf{a}i\)’s i-arrow.

Just to be sure, let me do the calculations…

The points on the line joining \(P+\mathbf{a}i\) and \(Q+\mathbf{b}i\) are of the form:
\(P + m(Q-P)- \mu(\mathbf{b}-\mathbf{a}) + [\mathbf{a} + \mu(Q- P) + m(\mathbf{b}-\mathbf{a})]i\).
This point’s i-arrow has
base at \(P +m(Q-P)-\mu(\mathbf{b}-\mathbf{a})\),
arrow \(\mathbf{a}+ \mu(Q- P) + m(\mathbf{b}-\mathbf{a})\),
and arrowhead at \(P + m(Q-P)-\mu(\mathbf{b}-\mathbf{a}) + \mathbf{a} + \mu(Q- P) + m(\mathbf{b}-\mathbf{a})\).

If the base of \(P+\mathbf{a}i\) is the arrowhead of \(Q+\mathbf{b}i\), then that means \(P = Q+\mathbf{b}\) and \(\mathbf{b}=P-Q\). So the point becomes:
\(P + m(-\mathbf{b})-\mu(\mathbf{b}-\mathbf{a}) + [\mathbf{a} + \mu(-\mathbf{b}) + m(\mathbf{b}-\mathbf{a})]i \)
\(= P -m\mathbf{b}-\mu(\mathbf{b}-\mathbf{a}) + [\mathbf{a}-\mu\mathbf{b} + m(\mathbf{b}-\mathbf{a})]i\)
whose i-arrow has
base at \(P- m\mathbf{b}-\mu(\mathbf{b}-\mathbf{a})\),
arrow \(\mathbf{a} -\mu\mathbf{b} + m(\mathbf{b}-\mathbf{a})\),
and arrowhead at \(P – m\mathbf{b} -\mu(\mathbf{b}-\mathbf{a}) + \mathbf{a} -\mu\mathbf{b} + m(\mathbf{b}-\mathbf{a})\).

If this point is real, then
\[\begin{aligned}\mathbf{a}-\mu\mathbf{b}+ m(\mathbf{b}-\mathbf{a}) &= 0\\
\mathbf{a}-\mu\mathbf{b} + m\mathbf{b}- m\mathbf{a} &= 0\\
(1-m)\mathbf{a} +(m-\mu)\mathbf{b} &= 0\end{aligned}\]

If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, then they are on a real line together, so I assume they’re not parallel, which means they’re independent vectors, so the only way to produce the zero vector as a linear combination is for both coefficients to be zero.

Therefore \(m=1\) and \(\mu=m= 1\).
So the base (which is now the real point’s location) is
\(P-\mathbf{b}-(\mathbf{b}-\mathbf{a}) = P+\mathbf{a}-2\mathbf{b}\), exactly where it should be. Nice.

But what if the two i-arrows aren’t arranged head to tail?

Two i-arrows create two i-arrows arranged head to tail

I am now left with the most general possible arrangement of two i-arrows. Two i-arrows not arranged head to tail, and not in such a way that the line joining their arrowheads is parallel to the line joining their bases. I have actually discovered in an earlier blog post that when I join these two complex points to make a line, then the i-arrows on the line whose bases are on the line joining their bases also have their arrowheads on the line joining their arrowheads. Let me confirm that with this new calculation style…

The points on the line joining \(P+\mathbf{a}i\) and \(Q+\mathbf{b}i\) are of the form:
\(P + m(Q-P)-\mu(\mathbf{b}-\mathbf{a}) + [a + \mu(Q- P) + m(\mathbf{b}-\mathbf{a})]i\).
This point’s i-arrow has
base at \(P +m(Q-P)-\mu(\mathbf{b}-\mathbf{a})\),
arrow \(\mathbf{a} + \mu(Q- P) + m(\mathbf{b}-\mathbf{a})\),
and arrowhead at \(P + m(Q-P)-\mu(\mathbf{b}-\mathbf{a}) + \mathbf{a} + \mu(Q- P) + m(\mathbf{b}-\mathbf{a})\).

If this point has its base on the line joining \(P\) and \(Q\), then the base has to be \(P+\rho(Q-P)\) for some real number \(\rho\). The base is actually \(P +m(Q-P)-\mu(\mathbf{b}-\mathbf{a})\), so we need \(\mu=0\). That means the complex point is now
\(P + m(Q-P) + [a + m(\mathbf{b}-\mathbf{a})]i\),
which has an i-arrow with
base at \(P +m(Q-P)\),
arrow \(\mathbf{a} + m(\mathbf{b}-\mathbf{a})\),
and arrowhead at \(P + m(Q-P) + \mathbf{a} + m(\mathbf{b}-\mathbf{a})\)
\(= P+\mathbf{a}+m(Q-P+\mathbf{b}-\mathbf{a})\)
\(= P+\mathbf{a}+m((Q-\mathbf{b}) -(P+\mathbf{a}))\).

This is a point on the line joining \(P+\mathbf{a}\) and \(Q+\mathbf{a}\), which are the arrowheads of the original two points. Even more, if you move the base a multiple \(m\) of the journey from \(P\) to \(Q\), then you also move the arrowhead the same multiple of the journey from \(P+\mathbf{a}\) to \(Q+\mathbf{b}\), which is exactly what I found last time.

The formula for the arrow suggests a construction to find the arrowhead given the base point too. The arrow is \(\mathbf{a} + m(\mathbf{b}-\mathbf{a})\), which is the vector \(\mathbf{a}\), plus a multiple of \((\mathbf{b}-\mathbf{a})\). If I draw the vectors \(\mathbf{a}\) and \(\mathbf{b}\) at the base point, then the vector \((\mathbf{b}-\mathbf{a})\) is the vector from the end of \(\mathbf{a}\) to the end of \(\mathbf{b}\) . Following this vector will bring me to the arrowhead of the i-arrow. But I know where that arrowhead lies: on the line joining the arrowheads of the original two i-arrows! So I don’t need to calculate how long to go, I only need to follow that arrow to the arrowhead line.

I like this construction. I like how it uses the two original arrows sitting at the base point we want to find the arrowhead we want. That seems to me to be how it really ought to work.

It should also be possible to work backwards from a specific arrowhead on the line joining the original two arrowheads to find where the matching base point is. I have to follow the vector \(\mathbf{a} + m(\mathbf{b}-\mathbf{a})\) backwards, which means following the vector \(-\mathbf{a} – m(\mathbf{b}-\mathbf{a})=-\mathbf{a} + m(-\mathbf{b}-(-\mathbf{a}))\). But this vector can be found by placing the vectors \(–\mathbf{a}\) and \(–\mathbf{b}\) at the starting point and following the line joining their vector endpoints. I know the base point is on the line joining the original two base points, so again I don’t have to calculate anything, only follow the line to where it meets the base point line.

But wait. The intersection of the base line and arrowhead line is capable of being both a base point and an arrowhead, which means I can draw a pair of i-arrows there arranged head to tail! And if I can do that, then I can find the real point on this complex line!

So, given two i-arrows for complex points \(P+\mathbf{a}i\) and \(Q+\mathbf{b}i\), I can find another pair of i-arrows on their line arranged head to tail like this:

1. Join the two base points to make a line, and the two arrowheads to make a line. Find where these two lines meet. Call this point \(S\).
2. At \(S\), draw the vectors \(\mathbf{a}\) and \(\mathbf{b}\) and join their endpoints. Call the point where this meets line the arrowhead line \(C\).
3. At \(S\), draw the vectors \(–\mathbf{a}\) and \(–\mathbf{b}\) and join their endpoints. Call the point where this meets the arrowhead line \(D\).
4. Then the two i-arrows from \(S\) to \(C\) and from \(D\) to \(S\) are both on the line joining \(P+\mathbf{a}i\) and \(Q+\mathbf{b}i\), and they are arranged head to tail.

From these two i-arrows arranged head-to-tail, I can find the real point on this complex line, and using that real point and any one of the i-arrows so far, I can create all the points on this complex line.

I wouldn’t be surprised if there was a more direct way to find the real point from the original i-arrows without going through the head-to-tail pair, but for now I am content. I do know a way, and it’s enough for me for now.

Conclusion

I have now found geometrical constructions that allow me to find any complex point on a line joining two complex points, whether the original points are real and real, real and unreal, or unreal and unreal. There are some lovely special cases with particularly simple constructions. I think my favourite bit is that spiderweb that neatly partitions the plane into eight slices where all the i-arrows are parallel.

I think I can finally let this go for a bit, and think about other things. It’s been an amazing ride. If you’re here at the end, I hope you’ve enjoyed it. These are all the other posts in this blog series, if you want to find them.

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows (YOU ARE HERE)

UPDATE: I came back to this two years later and you can read about it in the further blog post:

9. Complex lines with i-arrows again

Introduction

This is the last (for now) post in a series about about i-arrows, which are a way I have created of visualising the complex points on graphs in the real plane. As I’ve done for every other post since the first one, I will repeat the description here.

Every point with complex coordinates is represented as an arrow (which I call an “i-arrow”) from one place to another on top of the Cartesian plane.

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

In the previous blog post, I talked about the complex points on a real circle, with an equation like \((x-g)^2+(y-h)^2=R\] for a positive number R. In this blog post, I’ll talk about the complex points on an unreal circle.

Circles with negative area

The equation \(x^2+y^2 = R\) for positive number \(R\) represents a circle in the sense that the collection of points \((x,y)\) with real coordinates that satisfy the equation are indeed a circle. The area of that circle will be \(\pi R\) and its radius is \(\sqrt{R}\).

The equation \(x^2+y^2 = -R\) for positive number \(R\) represents nothing at all. There are no real numbers \(x\) and \(y\) that satisfy this equation, because when you square real numbers, you can’t get a negative result. And that’s that.

Except it’s not. That first equation equation \(x^2+y^2 = R\) has a large number of points \((x,y)\) with complex coordinates that satisfy it, and when you draw those points as i-arrows, they have a beautiful and surprisingly simple structure. There are definitely going to be points with complex coordinates that satisfy \(x^2+y^2 = -R\), and it would be completely unsurprising if they also had a beautiful and simple structure.

If the circle with equation \(x^2+y^2 = R\) has area \(\pi R\) and radius is \(\sqrt{R}\), then it seems fair to say that the area of the circle with equation \(x^2+y^2 = -R\) is \(-\pi R\) and its radius is \(\sqrt{-R}=\sqrt{R} i\). So I am going to say that this circle has negative area and imaginary radius.

Ever since I found a way of thinking about where complex points are, I’ve wanted to investigate what circles with negative area look like. Now I finally can do it properly, with i-arrows.

Complex points on a circle with negative area: the i-arrows at each point

Just like I did for real circles, I’ll start by substituting in a general complex point and see what happens.

Consider the equation of a circle with negative area, \(x^2+y^2 = -R\), for positive real number \(R\). Let \((p+si,q+ti)\) be a complex point, for real numbers, \(p\), \(q\), \(s\), \(t\), and suppose \((p+si,q+ti)\) satisfies the equation \(x^2+y^2 = -R\).

\[\begin{aligned}(p+si)^2+(q+ti)^2 &= -R\\
p^2 + 2psi-s^2 + q^2 + 2qti-t^2 &= -R\\
p^2 + q^2-s^2-t^2 + (2ps+2qt)i &= -R\end{aligned}\]

Equate real and imaginary parts.

Imaginary:
\[\begin{aligned}
2ps + 2qt &= 0\\
ps + qt &= 0\\
(p,q)\cdot(s,t) &=0\end{aligned}\]
This tells me that the vector \((s,t)\) is perpendicular to the vector \((p,q)\), which goes from the origin to the point \((p,q)\). This is the same as what happened for a real circle.

Real:
\[\begin{aligned}p^2 + q^2-s^2-t^2 &= -R\\
p^2 + q^2 +R &= s^2+ t^2\\
s^2 +t^2 &= p^2+ q^2+ R\\
\sqrt{s^2 +t^2} &=\sqrt{p^2 + q^2 + R}\\
||(s,t)|| &=\sqrt{p² + q² + R}\end{aligned}\]
Again this is almost the same as what happened for a real circle. I have the possible directions from the imaginary part, and a formula for the length from the real part, which is enough to locate the i-arrow. But the geometrical interpretation is just a little different.

It is still true that \(\sqrt{p^2 + q^2}\) is the length of the journey from the origin to \((p,q)\) and that this length, \{\sqrt{R}\) and \(||(s,t)||\) are related by a Pythagoras-theorem-like expression. But this time, the different parts are playing different roles. Now, \(||(s,t)||\) is the hypotenuse of the triangle, and \(\sqrt{R}\) and \(\sqrt{p^2 + q^2}\) are the short sides.

Last time, the right-angle told me there was a tangent to the circle, but this time that’s not going to happen. That right angle has to be at the origin, which is the centre of the circle, so it’s just not going to work. So my hope for a nice geometrical construction fades. After playing around, this is the best I can do.

To find the complex points on the circle of negative area \(x^2+y^2 = -R\) whose i-arrows are based at \(P=(p,q)\):

1. Draw the line from the origin to \(P\).
2. Draw a circle of radius \(\sqrt{R}\) centred at the origin.
3. Draw a diameter of this circle perpendicular to the line from the origin to \(P\) and mark where it meets the circle.
4. Draw a circle centred at \(P\) passing through the two points on that diameter just marked.
5. Draw a line through \(P\) perpendicular to the line from the origin to \(P\) and mark where it meets the circle just drawn.
6. The two points just marked are the arrowheads of the i-arrows based at \(P\).

Actually a fairly simple construction now I’ve seen the picture! I note that the construction will work no matter whether P is inside, outside or on the edge of that dotted-line circle of radius \(\sqrt{R}\), so for any point \(P\) there will always be complex points whose i-arrows are based there.

This bears out in the algebra, since \((p^2 + q^2 + R)\) is always going to be greater than zero — in fact it will always be greater than or equal to \(R\) — there will always be a result for \(\sqrt{p^2 + q^2 + R}\) and so always a solution for the length of \((s,t)\).

There is one special case I notice though: when \((p,q)\) is the origin. In that case the length of \((s,t)\) will still have an answer, which will be \(\sqrt{R}\). But also \((p,q)\cdot(s,t) =0\) for every possible vector \((s,t)\). So actually every vector based at the origin of length \(\sqrt{R}\) is an i-arrow on this circle! That is, given any point \(P\), there are two i-arrows on the unreal circle based there, unless \(P\) is the origin, in which case there are infinitely many i-arrows on the unreal circle based there, and their arrowheads form a circle of radius \(\sqrt{R}\). I think this is so cool! The imaginary circle really does contain an imaginary circle in its complex points.

I guess the last thing to do is to calculate the coordinates of the arrowheads for the i-arrows at a specific point \((p,q)\). This ought to be very similar to what happened for a real circle.

The vector \((s,t)\) is perpendicular to the vector \((p,q)\), so it’s some multiple of \((q,-p)\). Let’s say \((s,t)=m(q,-p)\).

The length of \((s,t)\) now has two ways to figure it out.

\[\begin{aligned}||(s,t)|| &= ||m(q,-p)||\\
&= |m| ||(q,-p)||\\
&= |m|\sqrt{p^2 + q^2}\end{aligned}\]
and
\[||(s,t)|| =\sqrt{p^2 + q^2+ R}\]

So
\[\begin{aligned}|m|\sqrt{p^2 + q^2} &=\sqrt{p^2+ q^2 + R}\\
|m| &=\sqrt{p^2 + q^2 + R}{\sqrt{p^2 + q^2}}\\
&=\sqrt{1 + \frac{R}{p^2 + q^2}}\end{aligned}\]

Therefore
\[(s,t) = \pm\sqrt{1 + \frac{R}{p^2 + q^2}} (q,-p)\]

And so the arrowheads of the i-arrows based at (p,q) are at
\[\left(p+q\sqrt{1 + \frac{R}{p^2 + q^2}}, q-p\sqrt{1 + \frac{R}{p^2 + q^2}}\right)\]
and
\[\left(p-q\sqrt{1 + \frac{R}{p^2 + q^2}}, q+p\sqrt{1 + \frac{R}{p^2 + q^2}}\right)\]

Yes, this is exactly the same as what happened for a real circle. Indeed, if I have the equation \(x^2+ y^2=R\) and allow the number \(R\) to be negative, I get this for the locations of the arrowheads of the i-arrows based at (p,q) either way:
\[\left(p+q\sqrt{1 – \frac{R}{p^2 + q^2}}, q-p\sqrt{1 – \frac{R}{p^2 + q^2}}\right)\]
and
\[\left(p-q\sqrt{1 – \frac{R}{p^2 + q^2}}, q+p\sqrt{1 – \frac{R}{p^2 + q^2}}\right)\]

How nice. What this means is it’s actually very easy to set up a GeoGebra activity that covers both real and unreal circles. This is precisely why I have been doing the equations as \(x^2 + y^2=R\) as opposed to \(x^2 + y^2=r^2\), by the way. So that I could smoothly transition from real to unreal without a major change in the algebra. Well I hoped it would turn out that way, and I’m glad it did.

Collections of complex points on an unreal circle

And now I really can investigate the complex points on an unreal circle properly and look at collections of them all at once.

All the i-arrows at once

First, all the i-arrows for the complex points on the circle \(x^2+y^2 = -4\) of negative area based at the points on a grid with spacing \(1/4\). (You can play around with this for different values of R, including both positive and negative, as well as different grid spacings, using this GeoGebra activity.

As you can see, there are a lot more lines and arrowheads than there were for the real circle of the matching real radius, due to there being i-arrows based at every point, rather than only ones outside a circle. I think I can at see a space in the middle there that is less red. When I look closer it’s not just that there are fewer arrows there, but specifically fewer arrowheads. But there are almost too many arrows for me to see clearly. Let me try again with a wider grid spacing. This picture is the same but with a grid spacing of \(1/2\).

I can more easily distinguish that circle of i-arrows at the origin from the rest of the i-arrows, and it’s making me even more sure that there are no arrowheads inside that circle. Let me see if I can confirm that…

The arrowheads of the i-arrows based at the non-origin point \((p,q)\) are given by
\[\left(p+q\sqrt{1 + \frac{R}{p^2 + q^2}}, q-p\sqrt{1 + \frac{R}{p^2 + q^2}}\right)\]
and
\[\left(p-q\sqrt{1 + \frac{R}{p^2 + q^2}}, q+p\sqrt{1 + \frac{R}{p^2 + q^2}}\right)\]

Phew that’s complicated. What if I let \(m =\sqrt{1 + \frac{R}{p^2 + q^2}}\) so that the arrowheads are at
\((p+qm,q-pm)\) and \((p-qm,q+pm)\).

These points will be outside the circle of radius R when their x and y coordinates satisfy \(x^2+y^2>R\). Let me see what I get for that calculation for the first point.

\[\begin{aligned}&x^2+y^2\\
&= (p+qm)^2+(q-pm)^2\\
&= p^2 + 2pqm + q^2m^2 + q^2 – 2qpm + p^2m^2\\
&= p^2 + q^2 + p^2m^2 + q^2m^2\\
&= p^2 + q^2 + (p^2 + q^2)m^2\\
&= (p^2+q^2)(1+m^2)\end{aligned}\]

Note the same result would have happened if I had used the other point, because the \(-2pqm\) and \(+2qpm\) would have appeared in the other order but still had a sum of zero.

Now \(m =\sqrt{1 + \frac{R}{p^2 + q^2}}\)
So \(m^2 = 1 + \frac{R}{p^2 + q^2}\).

Therefore
\[\begin{aligned} & x^2+y^2\\
&= (p^2+q^2)(1+m^2)\\
&= (p^2+q^2)(1+1 + \frac{R}{p^2 + q^2})\\
&=(p^2+q^2)(2 + \frac{R}{p^2 + q^2})\\
&= 2(p^2+q^2)+R\end{aligned}\]
which is definitely greater than or equal to \(R\).

I also notice that if this was a real circle with radius \(\sqrt{R}\), then the arrowhead’s coordinates substituted into \(x^2+y^2\) would have produced \(2(p^2+q^2)-R\), which is also greater than or equal to \(R\), since the only points with i-arrows are outside the circle where \(p^2+q^2>R\). Very nice.

So for an unreal circle with equation \(x^2+y^2 = -R\) (for positive \(R\)), the i-arrows for its complex points have arrowheads that are all outside the circle of radius \(\sqrt{R}\), plus the circle of i-arrows at the origin, whose arrowheads are the whole of the circle of radius \(\sqrt{R}\) itself.

The i-arrows on a vertical line

You can play around with what the i-arrows based at points along a vertical line look like in this GeoGebra activity. It will allow you to change the value of R to be both positive and negative to compare how both real and unreal circles work.

Here is GIF showing a selection of i-arrows on the unreal circle \(x^2+y^2 = -4\) based at points on a vertical line as the line moves back and forth. (The real circle with equation \(x^2+y^2 = 4\) has been shown for reference.)

It’s a bit jerky, I’m sorry, but GeoGebra was working hard, and also I moved the slider manually because I wanted it to show the circle of i-arrows at the origin for a moment as I went past. I just love that it’s there at that one moment. Actually, what happens at that moment is pretty cool.

As well as the circle of i-arrows at the origin, the arrowheads for all the i-arrows seem to be forming a hyperbola, like they did for a real circle, but the other way around. That definitely deserves investigating, and it ought to look very similar to how it looked for a real circle.

The vectors that are the i-arrows at the point \((p,q)\) are given by
\[\pm\sqrt{1 + \frac{R}{p^2 + q^2}}(q,-p)\]

If \(p=0\), these vectors are
\[\pm \sqrt{1+\frac{R}{q^2}}(q,0)=\pm\sqrt{q^2+R}(1,0)\]
This confirms that all the i-arrows based along the y-axis are horizontal.

The arrowheads of the arrowheads of the i-arrows based at the point \((p,q)\) are given by
\[(p,q)\pm\sqrt{1+\frac{R}{p^2+q^2}}(q,-p)\]

If \(p=0\), these points are
\[(0,q)\pm\sqrt{q^2+R}(1,0)=\left(\pm\sqrt{q^2+R},q\right)\]

When you take the x-coordinates and y-coordinates and square them, you get this:
\[\begin{aligned}
x^2&=q^2+R\\
y^2&=q^2\\
x^2-y^2&=R\end{aligned}\]
That is indeed the equation of a hyperbola with turning points at \((\sqrt{R},0)\) and \((-\sqrt{R},0)\).

Awesome. I am particularly loving how this unreal circle is highlighting the intimate relationship between a circle and a hyperbola.

The i-arrows on the lines through the origin

This hyperobola of i-arrow arrowheads should appear not just for i-arrows based along the y-axis, but along any line through the origin, which means I can cover all the i-arrows on the unreal circle systematically by looking at the ones attached to a rotating line through the origin. Here is the GeoGebra activity that lets you do this for circles of both positive and negative area. And here is the GIF of the i-arrows for the complex points on the unreal circle \(x^2+y^2 = -4\) based at equally-spaced points along a rotating line.

The i-arrows based at points around a circle

Jim Simons on twitter suggested for a real circle to organise the i-arrows by how far their base is from the centre, and it really is a good idea. Here is the GeoGebra activity that will let you investigate this selection of i-arrows for both real and unreal circles. And here is a GIF showing the i-arrows on the negative unit circle with equation \(x²+y² = -1\), organised by how far their base is from the centre.

I really love this GIF. I love how the i-arrows are longer and longer the further out from the centre they are based, but also I love how the circle of i-arrows at the origin seems perfectly natural now because that’s where the i-arrows based at points around a very small circle are converging towards.

Conclusion

I am very satisfied to have finally got a handle on what is happening with these circles of negative area. I love how the complex points on both the real and unreal circles have that hyperbola of arrowheads formed by the i-arrows based along a line through the circle’s centre. It’s just so simple and clean. Even more than this, I just love that circle of i-arrows based at the centre of an unreal circle. It’s just so very circley and simultaneously unreal, just like you would hope on an unreal circle. Just to draw all of this together, here is a GIF of the hyperbola of arrowheads for the i-arrows along a rotating line for circles \(x^2+y^2 = R\), with \(R\) ranging from -5 to 5.

I feel like this GIF really highlights the similarities and differences between the circles of positive and negative area. Also it’s just super cool. I could watch it all day long.

As an aside: one thing you can notice in the above GIF is the momentary transition between the positive and negative area circles. In that case, there is exactly one real point, which is the origin, and we’ve seen things like that before: in unreal lines. This actually does make sense, because \(x^2+y^2 = 0\) can be factorised into \((y+ix)(y-ix)=0\), which is the two lines \(y=ix\) and \(y=-ix\). It seems like one more way to make sense of a line of unreal slope is to think about it as half of a circle (or possibly an ellipse?) with zero area. But that’s another story and shall be told at another time.

This GIF shows the i-arrows on the positive and negative unit circles \(x^2+y^2 = 1\) and \(x^2+y^2 = -1\), with each frame showing the i-arrows at points around a circle. I feel this one really highlights the similarities and differences between positive and negative area circles. Again, I could watch it all day long.

I’ve certainly had a great time using i-arrows to investigate the complex points on a variety of shapes. If you’ve been following along, or just now come to just a part of it, I hope you’ve had a good time too. But for now, it’s time to sign off. They’ll still be there waiting when I come back one day.

One last time, these are all the other posts in this blog series, if you want to find them.

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows (YOU ARE HERE)
8. The line joining two complex points using i-arrows

Introduction

This is the second last (it was going to be the last, but it got too big so I made it a separate one) in a series of blog posts about i-arrows: a way I have created of visualising complex points on real graphs. The first blog post described how they work, and I repeat the description here.

Every point with complex coordinates is represented as an arrow (which I call an “i-arrow”) from one place to another on top of the Cartesian plane.

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

In this blog post, I’ll work through using i-arrows to represent the complex points on real circles. I did real circles about two years ago using my older idea of i-planes, but the i-arrows are so much better and will allow me to visualise things and talk about things much more easily.

The complex points on a real circle: i-arrows based at each point

I will investigate the complex points on the real circle with equation \(x^2+y^2 = R\) for positive number \(R\). In the previous post, I investigated the complex points on a parabola using the fact it was the graph of a function. I was able to get GeoGebra to do calculations using the function and show me the results. Unfortunately, circles are not the graphs of functions, so I won’t be able to do that here. Yes I know the two halves of a circle are the graphs of separate functions, but they have square roots, which can be a bit weird with complex numbers, so it’s safer not to do that. I will have to try to find the complex points on the real circle by seeing what happens when I substitute in a general point. I have done all this before, but I am happy to do it again. It is going to be ever so slightly different this time.

Let \((p,q)\) be a fixed real point, and suppose the complex point \((p+si,q+ti)\), for some real \(s\) and \(t\), satisfies the circle’s equation. So

\[\begin{aligned}(p+si)^2+(q+ti)^2 &= R\\
p^2 + 2psi – s^2 + q^2 + 2qti – t^2 &= R\\
p^2 + q^2 – s^2 – t^2 + (2ps+2qt)i &= R\end{aligned}\]

Equate real and imaginary parts.

Real:
\[\begin{aligned}p^2+ q^2 – s^2 – t^2 &= R\\
p^2 + q^2 – R &= s^2 + t^2\\
s^2+t^2 &= p^2+ q^2 – R\end{aligned}\]

Imaginary:
\[\begin{aligned}2ps + 2qt &= 0\\
ps + qt &= 0\end{aligned}\]

Looking at these two equations, they look like they’re telling me about the vector \((s,t)\). I have been trying to avoid talking about the i-arrows as vectors, but this is what the equations seem to be telling me to do.

The imaginary equation can be rewritten as a dot product:
\[\begin{aligned}ps + qt &= 0\\
(p,q)\cdot (s,t) &=0\end{aligned}\]
And this tells me that the point \((p,q)\) imagined as a vector is perpendicular to the vector \((s,t)\).

The real equation can have its square root done:
\[\begin{aligned}s^2+t^2 &= p^2+ q^2-R\\
\sqrt{s^2+ t^2} &= \sqrt{p^2+ q^2-R}\\
||(s,t)|| &=\sqrt{p^2+ q^2-R}\end{aligned}\]
And this tells me what the length of the vector \((s,t)\) is.

So I know the possible directions and the length of the vector (s,t), which is enough information to actually draw the i-arrows. Nice!

Except there are some details to sort out…

First, that length for the vector \((s,t)\) isn’t going to exist if \(p^2+q^2-R\) is a negative number. That is, if \(p^2+q^2<R\), then there will be no i-arrows with base at the point \((p,q)\). But \(p^2+q^2<R\) happens exactly when the point \((p,q)\) is inside the circle. So there are no i-arrows based at points inside the circle. If \(p^2+q^2=R\), then \((s,t)\) will have length 0 and there is no i-arrow at all. But that’s ok, because \((p,q)\) is then a point on the real circle anyway. Finally, if \(p^2+q^2>R\), then \(\sqrt{p^2+q^2-R}\) definitely exists, so there are definitely vectors of that length, and we’ll get two of them: one in each direction perpendicular to \((p,q)\).

So I now know there are two i-arrows based at every point outside the circle, and none based at any point inside the circle. I know they point perpendicularly to the journey from the origin to the point where they are based. And I have a formula for how long they are. Let me look closer at that formula.

The number \(\sqrt{p^2+q^2}\) is how far \((p,q)\) is from the origin, and the number \sqrt{R} is the radius of the circle. That means that \(\sqrt{p^2+q^2-R\}\) is the length of the other short side of a right-angled triangle with long side the length of \((p,q)\) and short side the radius of the circle.

The vertical edge in the picture is a radius of the circle, and the hypotenuse is the line from the origin to the point \((p,q)\). The horizontal edge is perpendicular to the radius, so it has to be a tangent to the circle. That means the correct distance is the length of the tangent from \((p,q)\) to the circle! Very cool.

I still haven’t actually found the i-arrow yet though. The top edge in that picture isn’t actually the i-arrow itself, it’s just the length of the i-arrow. The i-arrow has to be perpendicular to the line from the origin to \((p,q)\), and the top edge of that triangle is not. I need to draw arrows of that length in the right directions. To make sure they’re the right length, I can draw a circle of that radius. Then I can draw the arrows perpendicular to the line to \((p,q)\).

Interestingly, if I draw both tangents from \((p,q)\) to the circle, then the line joining the places where they meet the circle will also be perpendicular to the line from the origin to \((p,q)\). So instead of going perpendicular to the line from the origin to \((p,q)\), I could go parallel to the line joining the two points of tangency. This is kind of pleasant, because points inside the circle have no tangents of the circle through them, which would explain why the construction doesn’t work there.

So here’s a geometrical process for finding the i-arrows based at a point \((p,q)\) outside the circle:

• Draw the tangents from \((p,q)\) to the circle and mark the points of tangency.
• Draw a circle with centre \((p,q)\) through those points of tangency.
• Draw the line through those points of tangency.
• Draw a line parallel to this one through the point \((p,q)\).
• The intersections of this line just drawn with the circle just drawn are the two arrowheads of the i-arrows.

Of course, if I want GeoGebra to draw the points, it might be easier to actually figure out their coordinates…

The vector \((s,t)\) is perpendicular to the vector \((p,q)\), so it’s some multiple of \((q,-p)\).
Let’s say \((s,t)=m(q,-p)\).
The length of \((s,t)\) now has two ways to figure it out.

\[\begin{aligned}||(s,t)|| &= ||m(q,-p)||\\
&= |m| ||(q,-p)||\\
&= |m|\sqrt{p^2 + q^2}\end{aligned}\]
and
\[||(s,t)|| =\sqrt{p^2 + q^2 – R}\]

So
\[\begin{aligned}|m|\sqrt{p^2 + q^2} &=\sqrt{p^2 + q^2 – R}\\
|m| &= \frac{\sqrt{p^2 + q^2-R}}{\sqrt{p^2 + q^2}}\\
&=\sqrt{1-\frac{R}{p^2 + q^2}}\end{aligned}\]

Therefore
\[(s,t) = \pm\sqrt{1- \frac{R}{p^2 + q^2}} (q,-p)\]

And so the arrowheads of the i-arrows based at \((p,q)\) are at
\[\left(p+q\sqrt{1 – \frac{R}{p^2 + q^2}}, q-p\sqrt{1 – \frac{R}{p^2 + q^2}}\right)\]
and
\[\left(p-q\sqrt{1 – \frac{R}{p^2 + q^2}}, q+p\sqrt{1 – \frac{R}{p^2 + q^2}}\right)\]

Where a circle meets an external line

I was going to use the algebraic calculations above to investigate collections of i-arrows, but I have noticed that the geometrical construction I drew before could actually help me find the complex points where a line that doesn’t meet the circle in a real point, actually does meet the circle.The trick is to find the point where they are based, then do the above construction to find the arrowheads.

Here is the construction.

1. Draw the circle and the line.
2. Draw the tangent to the circle parallel to the line and closest to the line.
3. Draw the line from the centre of the circle to the point of tangency and find the point P where it meets the original line.
4. Draw the tangents from this new point P to the circle.
5. Draw a new circle with centre P and passing through the points of tangency.
6. The i-arrows are drawn from P along the line to where it meets this circle.

(That point P is also the nearest point on the line to the centre of the circle, which is a coordinate geometry thing you can do instead of finding that parallel tangent.)

I really love this. I love the idea of being able to find the complex points where a circle and a line meet. Plus I still love being reminded that the i-arrows on a real line line up with the line. It’s just so clean and neat.

Collections of complex points on a real circle

The calculations above are enough to get GeoGebra to draw i-arrows for me. Here is a picture of all the i-arrows on the circle \(x^2+y^2=4\) based at the points on a grid spaced with gaps of \(1/4\).

I adore this picture, and not just because it reminds me of a big school of fish swimming in a circle. I love how you can see so clearly the huge empty space inside the circle where there are no complex points, and how obvious it is that the further from the circle you get, the more and more unreal the points are. I knew these two things were true based on my previous model, but I had never seen them until now.

I also wanted to do what I did for lines and the parabola and look at all the i-arrows based along a moving vertical line, just to have clearer idea of some of the structure. The animated GIF below shows what happens. (In other posts, I have put in a GeoGebra activity to play with, but I just want to hold off on that for a moment, because… well, you’ll see in a later blog post.)

Well, that is VERY interesting. I can really see how the i-arrows avoid the space inside the circle, bending around it as the line moves past. Also the curves the arrowheads are following are very strange, ranging from almost straight lines far from the circle, to what looks like it might be a hyperbola on the y-axis. I don’t know if I feel like investigating those curves further, except that hyperbola.

Let me see… The vectors that are the i-arrows at the point (p,q) are given by

\[\pm \sqrt{1 – \frac{R}{p^2 + q^2}} (q,-p)\]

If \(p=0\), these vectors are
\[\begin{aligned}&\pm \sqrt{1 – \frac{R}{q^2}} (q,0)
&=\pm\sqrt{q² – R} (1,0)\end{aligned}\]

This confirms that all the i-arrows based along the y-axis are horizontal. I knew this to be true from the geometrical construction earlier, but it is nice for the algebra to back it up.

The arrowheads of the arrowheads of the i-arrows based at the point \((p,q)\) are given by
\[(p,q)\pm \sqrt{1 – \frac{R}{p^2 + q^2}}\]

If \(p=0\), these points are
\[\begin{aligned}&(0,q)\pm\sqrt{q² – R} (1,0)\\
&= (\pm\sqrt{q² – R},q)\end{aligned}\]

I see what’s happening here! Look what happens when you take the x-coordinate and y-coordinate and square them.
\[\begin{aligned}x^2&=q^2-R\\
y^2 &=q^2\\
y^2-x^2&=R\end{aligned}\]
That’s the equation of a hyperbola, which has its turning points at \((0,\sqrt{R})\) and \((0,-\sqrt{R})\). Exactly the hyperbola I thought it was based on the picture.

A similar thing would have happened if \(q=0\) and I had drawn all the i-arrows based along the x-axis. Only this time all the i-arrows would have been vertical and the equation would have been \(x^2-y^2=R\), so the hyperbola would be the other way around.

Considering that the i-arrows can be made via a geometrical construction using tangents to the circle, and the circle is the same shape in every direction, this same hyperbola would appear in the arrowheads of the i-arrows based along any line through the origin, rotated to match the axis we had chosen. I used GeoGebra to confirm this, letting the points \((p,q)\) be regular multiples of \((\cos(\theta),\sin(\theta))\). Here is the resulting animation of the i-arrows on a circle based along a rotating line.

It’s much less mysterious-looking than the previous GIF, but I suppose that’s a good thing. The whole purpose of the i-arrows was to make the complex points less mysterious, wasn’t it? I am loving how simple the structure of the complex points on a circle has actually turned out to be. Also I remember when I learned finite geometry that when I extend a field, all the circles become hyperbolas, and this seems to be confirming that quite clearly!

UPDATE: Jim Simons on Twitter has suggested another way to organise the i-arrows — draw all the i-arrows based at the points on a circle themselves. Here is a GIF showing the i-arrows on the unit circle organised this way. Each frame shows the i-arrows based at 40 points all the same distance from the origin.

You can really see how the i-arrows get longer and longer the further out the base is, so basically the complex points on the circle get more and more unreal the further out they are from the centre.

The complex points on a circle not centred at the origin

All of the calculations so far have been based on a circle with equation \(x^2+y^2=R\), which means the centre of the circle is at the origin. The geometrical construction I described above ought to work for any circle anywhere, but I just want to be absolutely sure.

Consider the circle with equation \((x-g)^2+(y-h)^2=R\), for positive number R and real numbers g and h. This is a circle with radius \(\sqrt{R}\), centred at the point \((g,h)\). Let \((p,q)\) be a fixed real point, and suppose the complex point \((p+si,q+ti)\), for some real \(s\) and \(t\), satisfies the circle’s equation. So

\[\begin{aligned}(p+si-g)^2+(q+ti-h)^2 &= R\\
(p-g+si)^2+(q-h+ti)^2 &= R\\
(p-g)^2+ 2(p-g)si-s^2 + (q-h)^2 + 2(q-h)ti-t² &= R\\
(p-g)^2+ (q-h)^2-s^2-t^2 + (2(p-g)s+2(q-h)t)i &= R\end{aligned}\]

Equate real and imaginary parts.

Real:
\[\begin{aligned}
(p-g)^2 + (q-h)^2-s^2-t^2 &= R\\
(p-g)^2+ (q-h)^2-R &=s^2 + t^2\\
s^2+t^2 &= (p-g)^2 + (q-h)^2-R\\
\sqrt{s^2 +t^2} &= \sqrt{(p-g)^2 + (q-h)^2-R}\\
||(s,t)|| &=\sqrt{(p-g)^2 + (q-h)^2-R}\\
&=\sqrt{||(p-g,q-h)||^2 -R}\\
&=\sqrt{||(p,q)-(g,h)||^2-R}\end{aligned}\]
So the length of \((s,t)\) is the short side of a right-angled triangle that has as one side \(\sqrt{R}\) and the other short side as the journey from \((g,h)\) to \((p,q)\).

Imaginary:
\[\begin{aligned}2(p-g)s + 2(q-h)t &= 0\\
(p-g)s + (q-h)t &= 0\\
(p-g,q-h)\cdot(s,t) &= 0\\
[(p,q) – (g,h)]\cdot(s,t) &= 0\end{aligned}\]
So the vector \((s,t)\) is perpendicular to the vector journey from \((g,h)\) to \((p,q)\).

So basically the picture of how to find the i-arrow \((s,t)\) based at \((p,q)\) is exactly the same as it was before! This means everything I’ve done will still work for a circle not centred at the origin. I wasn’t worried it wouldn’t be, but I just wanted to do the algebra all the same.

Conclusion

I feel like I have a really good understanding of how the complex points on a circle work. That rotating hyperbola of i-arrow arrowheads is my favourite part I think, with the geometrical construction of where a circle meets an external line coming in a close second. I am also happy that I have finally confirmed that things work even for circles not centred at the origin, which I had never gotten around to doing until now. Ultimately, I am just loving how the i-arrows make it so much easier to talk about the locations of these complex points and to draw enough of them at once to feel like I can see what’s happening.

There is one blog post left. Every time I’ve thought about this stuff, I have really wanted to properly investigate the points on an unreal circle, with an equation like \(x^2+y^2 = -R\), for positive number \(R\). I hope to get it done before the weekend is over and the rest of life kicks in.

These are all the other posts in this blog series, if you want to find them easily.

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows (YOU ARE HERE)
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows

Introduction

In an earlier post in this same series, I presented the following introduction, and I think it’s good enough to just repeat here…

The Cartesian plane is pretty cool. You think up an equation like \(y=x^2+1\) and find all the points \((x,y)\) whose coordinates satisfy it, and you get a shape (in this case a parabola). Different kinds of shapes have different kinds of equations, and finding the places where shapes meet becomes solving equations simultaneously. Geometry becomes deeply connected to algebra and everything is lovely.

Then we create the complex numbers so that equations that weren’t previously solvable become solvable. Most notably the equation \(x^2+1=0\) gets the solutions \(x=i\) and \(x=-i\). The simultaneous solutions to \(y=x^2+1\), \(y=0\) are supposed to be the places where that parabola met the x-axis, only it doesn’t meet the x-axis. If the complex numbers are to be believed, that parabola meets the x-axis at the points \((i,0)\) and \((-i,0)\). But I can’t see those points in the original Cartesian plane! Maybe there is a way to add the new complex points to my existing Cartesian plane so that the graphs do look like they meet.

And there is a way. In 2016, I created such a way, and I have just made it better, producing a thing called an i-arrow. This is the full description, which I am shamelessly cutting-and-pasting into every post in this series.

Every point with complex coordinates is represented as an arrow from one place to another on top of the Cartesian plane.

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

At this point in a previous post in this series, I reneged on my promise to illustrate where the parabola with equation \(y=x^2+1\) met the x-axis and instead talked about the complex points on a line. Now I will actually fulfill the promise. (Note I am not sorry that I didn’t do the parabola first. It was actually important to investigate the complex points on a line, because the x-axis itself is a line, so if I want to find out the complex points it shares with another object, I am going to need to know what complex points it has all on its own.)

Noticing things in the i-arrows on the graph of a quadratic function

When I first investigated complex points on parabolas, I wasn’t thinking of them as functions but as the sets of points that satisfy a condition. Only later did I consider them as functions. This time, I’ll go the other way around, and think of it as a function first. I’ll go one further, actually, and let my friend GeoGebra do some work for me to start with. GeoGebra is able to do complex number arithmetic, so I’ll sub in a selection of complex inputs, and draw the resulting (input,output) points as i-arrows and see what it looks like.

This is a view of the i-arrows that go with the complex points on the parabola that is the graph of the function \(f(x)=x^2+1\), made from substituting in \(x=p+si\) for both \(p\) and \(q\) ranging from to \(-5\) to \(5\) in steps of \(\frac14\). It doesn’t show the full length of every i-arrow, but a zoomed-in view so I can see the details I need to.

Have a close look at this picture and see what you can notice. Here are some things I notice:

• There are no i-arrows with base above the parabola, or with arrowhead above the parabola. In fact, the line segment that is each i-arrow is fully contained in the space below the parabola.
• Each base-point with an i-arrow has two i-arrows based there. I notice this most when looking at the base-points nearby to \((0,0.75)\).
• Many of the points have two arrowheads there.
• There seem to be several upwards-opening parabolas made of base points, and several made of arrowheads too.
• I can also see several straight lines of base-points. Following those lines, they seem to be tangent to the parabola.
• There are collections of parallel i-arrows, and they get longer the further down they go. Also, their base-points are lined up vertically. I notice the set of horizontal i-arrows most, followed by a set nearby on either side.
• I can just see two horizontal i-arrows lying on the x-axis with their bases at the origin and arrowheads at \((-1,0)\) and \((1,0)\). That means they represent the complex points \((0,-i)\) and \((0,i)\), and since they line up with the x-axis, they are complex points on the x-axis. So these are where the parabola meets the x-axis, just as I expect.

There’s a lot to unpack there! I am extremely happy that this i-arrow representation afforded so many things to notice in one diagram. Such is the power of a good model, hey? Before I try to do the algebra that will explain the things I see, let me do another GeoGebra thing to investigate just one of these things more.

I noticed that those sets of parallel i-arrows had their base-points lined up vertically. All the i-arrows here are for complex points of the form \((p+si,f(p+si))\), with base point at \((p,Re[f(p+si)])\). A collection of base-points lining up vertically means that they belong to points with all the same value of \(p\). That is, all the inputs have the same real part. And so what I did is I made GeoGebra draw the i-arrows for a selection of points made by inputting numbers with the same real part. Then I could look at just these ones. This is the animated GIF that came out of that GeoGebra activity, but you can play with the original yourself here.

Have a look at that GIF (or play around with the GeoGebra tool) and see what you notice. Here are some things I notice:

• Every base point has two i-arrows, one in each direction.
• The i-arrows are shorter closer to the parabola and longer further away.
• The i-arrows for a specific real part are all parallel. In fact, it looks like they might be parallel with the tangent to the parabola at the matching point on the parabola.
• The arrowheads for the i-arrows for a specific real part look like they form a parabola of their own. This parabola is inverted relative to the original one, and the two appear to be tangent to each other.
• You can clearly see x-intercepts in the frame where the arrows are horizontal. The parabola of arrowheads looks like it is exactly the same as the real parabola, inverted and sharing the same vertex.

That’s a lot to notice, and my list of things to investigate is getting longer. I think it is this that is causing the collections of parallel lines in the original image, though, so that’s nice. Before I start on the algebra, I do want to do one more thing: I want to investigate what happens if I keep the imaginary part the same instead of the real part. Here’s the GeoGebra activity, and here’s the animated GIF.

One last time, look to see what you notice. Here’s what I notice:

• The arrows momentarily disappear in the middle (which makes sense because that’s when the imaginary parts of the inputs are all zero).
• The base points in each frame follow a parabola, as do the arrowheads in each frame. These two parabolas look like they might be the same as the real parabola but shifted downwards.
• The three parabolas get further apart the further the imaginary part is from zero.

So those parabolas I saw in the first image seem to be caused by the outputs with all the same imaginary part. That’s good to have that mystery solved.

And now I have a long list of things that I want to confirm with some algebra.

Algebra (and some geometry)

Here goes! I’ll start by investigating the parabola as the graph of a quadratic function. I think some of the things I noticed above will need to use a different perspective, but for now I’ll stick with the function.

Let \(f(x)\) be a real quadratic function. That is, let \(f(x) = ax^2+bx+c\) for real numbers \(a\), \(b\), \(c\).
Consider the input \(x=p+si\) for real numbers \(p\) and \(s\).

\[\begin{aligned}&f(p+si)\\
&= a(p+si)^2+b(p+si)+c\\
&= a(p^2+2psi-s^2)+b(p+si)+c\\
&= ap^2+2apsi-as^2+bp+bsi+c\\
&= ap^2+bp+c-as^2+2apsi+bsi\\
&= ap^2+bp+c-as^2+(2ap+b)si\end{aligned}\]

Looking at that equation, I can see \(ap^2+bp+c\), which is the value of \(f\) at \(p\), and I can see \(2ap+b\), which is actually the value of the derivative of \(f\) at \(p\)! (This happened last time I investigated complex points on real parabolas and it’s just fascinating!)

So it looks like if \(f(x) = ax^2+bx+c\),
then \(f(p+si) = f(p) – as^2 + f'(p)si\).

That looks like something I can work with!

The i-arrows based along a vertical line

Fix a specific value of \(p\), and consider all inputs \(x=p+si\) with all possible values of \(s\). It’s worth noting that i-arrows whose bases lie along a vertical line all must have come from such an input, since the real part of the x-coordinate of the matching complex point must all be the same.

I would like to show the following:

• The i-arrows produced this way are all parallel.
• The slope of the i-arrows produced this way are parallel to the tangent to the parabola at \(x=p\).
• The arrowheads of the i-arrows produced this way form a parabola themselves. This parabola is inverted relative to the original and they are tangent to each other at the point where \(x=p\).
• The i-arrows are horizontal when \(x=p\) passes through the axis of the original parabola, and when this happens, the parabola the arrowheads make is the same as the original, but inverted and sharing the same vertex.

Let’s see what I can do.

The complex point corresponding to the input \(p+si\) is \((p+si,f(p)-as^2+f'(p)si)\). Its i-arrow has base \((p,f(p)-as^2)\) and extends along the vector \((s,f'(p)s)\), to finish at the arrowhead \((p+s,f(p)-as^2+f'(p)s)\).

The vector \((s, f'(p)s) = s(1,f'(p))\), which is parallel to \((1,f'(p))\). So the i-arrows are definitely all parallel since they’re parallel to this one vector. The slope of this vector is \(f'(p)\), which is indeed the slope of the tangent to the parabola at the point where \(x=p\). Also note that the two i-arrows for opposite values of \(s\) are the same length and in opposite directions. Well, that was remarkably straightforward!

The arrowheads have coordinates \((p+s,f(p)-as^2+f'(p)s)\) for all values of \(s\). For this to be a parabola, the y-coordinates must be a quadratic function of the x-coordinates.

Let \(x=p+s\) and \(y=f(p)-as^2+f'(p)s\).
Then \(x-p=s\), so
\(y=f(p)-a(x-p)^2+f'(p)(x-p)\)
\(y= -a(x-p)^2+f'(p)(x-p)+f(p)\).

This is a perfectly good quadratic function, and when I sub in \(x=p\), the value comes out to \(f(p)\). Its leading coefficient is \(-a\) compared to the original leading coefficient of \(a\), so it’s inverted relative to the original parabola. Finally, the derivative of this function is \(\frac{dy}{dx} = -2a(x-p)+f'(p)\), which does become \(f'(p)\) when \(x=p\), so the two parabolas are indeed tangent at the point where \(x=p\).

That just leaves what happens when the i-arrows are horizontal, which is supposed to happen when \(x=p\) passes through the vertex of the original parabola.

The i-arrows are parallel to \((1,f'(p))\), which is horizontal exactly when \(f'(p)=0\). And \(f'(p)=0\) exactly at the turning point of the parabola, where the tangent is horizontal. So yes the i-arrows are horizontal exactly when \(x=p\) passes through the turning point of the parabola.

When \(x=p\) does pass through the vertex, \(f'(p)=0\) and so the equation of the parabola of arrowheads becomes
\(y= -a(x-p)^2+0(x-p)+f(p)= -a(x-p)^2+f(p)\).
This is the equation of a parabola with vertex at \((p,f(p))\) and leading coefficient the exact opposite of the original. So yes it is the same parabola, inverted through the vertex

I wonder if all those other arrowhead parabolas are the same shape as the original too? Actually yes they have to be. All quadratic functions with leading coefficient a can be rewritten as \(y=a(x-c)^2+h\) for some \(c\) and \(h\), and are so the same shape as \(y=ax^2\). All quadratic functions with leading coefficient -a can be rewritten as \(y=-a(x-c)^2+h\) for some \(c\) and \(h\), and are so the same shape as \(y=-ax^2\), which is the same shape as \(y=ax^2\) but inverted.

So everything I thought was true really is.

Aside: relationship to other representations

I will pause at this point to notice that you can abstract a graphical procedure for finding the complex roots of a quadratic function: invert the parabola at its vertex, then draw two equal-length i-arrows along the x-axis inside this new parabola from the axis of symmetry to the edge. The picture here shows the two complex roots of \(f(x)=x^2+6x+13\), which are \(3-2i\) and \(3+2i\).

I have to say I do find such a diagram a bit unedifying. I much prefer to have all the i-arrows drawn in along that vertical line, to show that the curve itself is actually made of all the arrowheads of all those i-arrows.

The reason I drew the first picture is that I have seen it before in other people’s descriptions of how to visualise the complex solutions to real equations (though without the arrows on the picture). I’ve seen a pure description like the one I just gave (find where the inverted parabola meets the x-axis, x-coord of vertex is real part, distance from there to inverted parabola is imaginary part) just presented raw with only an algebraic explanation. But I’ve also seen a notable physical representation in Philip Lloyd’s phantom graphs. In those graphs, you imagine 3D graph with the complex plane as the x-y-axis and the real part of the output as the z-axis. Only the whole of this 3D graph is never drawn, just the curve of a parabola perpendicular to the original.

The thing I like most about my i-arrows is that the construction literally draws an actual representation of the actual complex points themselves, and also shows all the other complex points nearby, and indeed, you can draw all the complex points everywhere if you wanted to. As cool as Lloyd’s phantom graphs are, they are very phantom in the sense that they only exist in a 3D drawing outside the real plane. My i-arrows are drawn right there on the real plane itself. Also the Lloyd’s phantom graphs only represent a part of the actual complex point itself (the real part of the output) even if we did draw the whole 3D surface. My i-arrows show the complex part of the output too.

I’m sure I’m biased towards my own representation because it’s mine, but I don’t think I’m biased when I say it has a LOT of really cool features that make it useful beyond what phantom graphs are usually used for. I mean, the fact that you can find the complex points on a straight line by simply drawing an arrow on the line itself is just super amazing.

By the way, based on everything I know is true, I actually have a procedure for finding the complex points where the graph of a quadratic function meets any line it doesn’t meet in the real plane.

1. Draw both the parabola and the line.
2. Find the tangent to the parabola parallel to the line.
3. Rotate the parabola 180° around the point where the tangent meets the parabola.
4. Draw a vertical line from the parabola to the original line and find where it meets this line.
5. Draw two i-arrows from this intersection point to the rotated parabola. These i-arrows are the complex points of the quadratic function on this line.

This collection of pictures illustrates this process for the function \(f(x)=x^2+6x+13\) and the line with equation \(y=5-2x\). The intersection points are \((2-2i,1+4i)\) and \((2+2i,1+4i)\).

The i-arrows for inputs with the same imaginary part

Let me return to the quadratic function \(f(x) = ax^2+bx+c\) for real numbers \(a\), \(b\), \(c\), and the inputs \(x=p+si\) for real numbers \(p\) and \(s\). As before, the complex point corresponding to the input \(p+si\) is \((p+si,f(p)-as^2+f'(p)si)\). Its i-arrow has base \((p,f(p)-as^2)\) and extends along the vector \((s,f'(p)s)\), to finish at the arrowhead \((p+s,f(p)-as^2+f'(p)s)\). This time, let me fix \(s\) and vary \(p\).

The set of base points are of the form \((p,f(p)-as^2)\). The function \(g(p)=f(p)-as^2\) is still a quadratic function of \(p\). Indeed, its graph is the same as that of \(y=f(x)\), just shifted downwards by \(as^2\).

The set of arrowheads are of the form \((p+s,f(p)-as^2+f'(p)s)\). For this to be a parabola, just like last time, we need the y-coordinates to be a quadratic function of the x-coordinates. But this time, s is a constant.
Let \(x=p+s\).
Then \(x-s=p\).
And so \(y=f(x-s)-as^2+f'(x-s)s\).

The function \(f'(x-s)\) is linear, so subtracting it from the quadratic \(f(x-s)\) till makes a quadratic function like I noticed before. Let me do some more working…

\[\begin{aligned}y&=f(x-s)-as^2+f'(x-s)s\\
&=a(x-s)^2+b(x-s)+c-as^2+(2a(x-s)+b)s\\
&=a(x^2-2sx+s^2)+b(x-s)+c-as^2+2as(x-s)+bs\\
&=ax^2-2asx+as^2+bx-bs+c-as^2+2asx-2as^2+bs\\
&=ax^2+bx+c-2as^2\\
&=f(x)-2as^2\end{aligned}\]

Wow. It’s actually the exact same function as \(f(x)\), shifted down by \(2as^2\) — exactly twice as far as the parabola of base points. So cool. That’s all the observations that went with that GIF.

The i-arrows with a certain base or arrowhead

There were several observations right at the beginning about how many i-arrows were based at a certain point, and how many had their arrowhead at a certain point, and whether any part of an i-arrow passed through them. These are the last ones I want to think about.

Consider the function \(f(x)=ax^2+bx+c\) for real \(a\), \(b\), \(c\) and suppose \((p,q)\) is a point anywhere in the real plane. If there is an i-arrow in the graph of this function with base \((p,q)\), then that means there are \(s\) and \(t\) such that the complex point \((p+si,q+ti)\) satisfies the equation. So \(q+ti=f(p+si)=f(p)-as^2+f'(p)si\). Therefore

\[\begin{aligned}
q&=f(p)-as^2\\
t&=f'(p)s\end{aligned}\]

In this scenario, \(p\) and \(q\) are fixed and we must find \(s\) and \(t\).

From the first equation,

\[\begin{aligned}
q &=f(p)-as^2\\
as^2 &=f(p)-q\\
s^2 &=\frac{f(p)-q}{a}\end{aligned}\]

Depending on whether \((f(p)-q)/a\) is positive, negative or zero, there will be two, one or zero solutions for \(s\), and then once \(s\) is known the second equation from earlier will give one value of \(t\). Therefore every point has either two, one or zero complex points of the parabola based there.

If \(a\) is positive, then \((f(p)-q)/a\) is positive when \(f(p)-q>0\), which is when \(q<f(p)\). The number q is the y-coordinate of the base point in question, and \(f(p)\) is the y-coordinate of the point on the parabola with x-coordinate \(p\). Since \(q<f(p)\), that means the point \((p,q)\) is below the parabola! So every point below the parabola has exactly two i-arrows on the parabola’s graph based there.

If \(a\) is positive, then \((f(p)-q)/a\) is negative when \(f(p)-q<0\), which happens when the point \((p,q)\) is above the parabola. In that case, there are no solutions for \(s\), and so no i-arrows based there.

If \(a\) is negative, then the situation is reversed, and there are no i-arrows based at the points below the parabola, and two i-arrows based at the points above the parabola.

Regardless of whether \(a\) is positive or negative, \((f(p)-q)/a\) can only be zero if \(q=f(p)\), which means \((p,q)\) is a point actually on the real parabola itself. But also, the solution for \(s\) will be \(s=0\), so the solution for \(t\) will be \(t=0\), so there will actually be no i-arrow there but only the real point which we know is on the parabola already.

Putting all this together, whether \(a\) is positive or negative, this means that none of the points inside the curve of the parabola have i-arrows based there, and all of the points outside the curve of the parabola have exactly two i-arrows based there.

What about the arrowheads of the i-arrows?

If the point \((g,h)\) is an arrowhead for some point on the parabola, then it is \((p+s,f(p)-as^2+f'(p)s)\) for some complex number \(p+si\). This sets up two equations.

\[\begin{aligned}
p+s &=g\\
f(p)-as^2+f'(p)s &=h\end{aligned}\]

I’ve seen something like this before, most recently when I was investigating i-arrows for inputs with the same imaginary part.

Write \(p=g-s\) and substitute into the second equation to get \(h=f(g-s)-as^2+f'(g-s)s\), which rearranges to \(h=f(g)-2as^2\), based on my earlier experience. So,
\[\begin{aligned} h &=f(g)-2as^2\\
2as^2 &=f(g)-h\\
s^2 =\frac{f(g)-h}{2a}\end{aligned}\]
And there are two, one or zero solutions for \(s\), depending on whether \((f(g)-h)/(2a)\) is positive, zero or negative. Just like last time, these happen when the point \((g,h)\) is outside the curve of the parabola, inside the curve, or on the parabola. Amazing. So glad I was able to reuse that algebra from before. Just to be clear, every point outside the curve of the parabola has two i-arrow arrowheads meeting there, and every point inside the curve has none.

Finally, every i-arrow is based at a point \((p,q)\) outside the curve of the parabola (for example, below the curve if it opens upwards), and follows an arrow parallel to the tangent to the parabola at \(x=p\). Since the whole real parabola is on the other side of the tangent from \((p,q)\), the whole i-arrow must be on this side of the tangent and so miss the real parabola entirely.

Conclusion

So, I have confirmed pretty much everything I noticed before.

• Given a quadratic function, there are two of its i-arrows based at every point outside its curve, and two of its i-arrows with arrowhead at every point outside its curve.
• All of the i-arrows based at points along a vertical line are parallel to the tangent to the original parabola at the point where that line meets the parabola.
• Taking all the arrowheads for i-arrows based along a vertical line produces a parabola the same shape as the original, but rotated 180° around the point where the line meets the original parabola.
• You can find the unreal points where a line meets the parabola by first finding the tangent to the original parabola that is parallel to that line and using both of the two previous observations.
• In particular, the i-arrows based along the parabola’s axis of symmetry are all horizontal, and their arrowheads form a parabola the same shape as the original, but rotated around its vertex, and this allows you to find the complex x-intercepts of the original function.

All in all, I am very very happy. I really feel I understand a lot better where the complex points are on a parabola. But also I’m proud of both the algebra and geometry moves I made today, as well as how I structured this blog post around observation first and proof second.

There is only one promised blog post left in the series, which is about the unreal points on a circle, but it may take a while to finish, since I come out of quarantine tomorrow.

These are all the other posts in this blog series, if you want to find them easily:

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows (YOU ARE HERE)
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows

On 26th July 2022, I wrote a blog post about using i-arrows to make sense of the complex points on both real and unreal lines. And at the end I mentioned how there were still some things mysterious to me. But of course I kept thinking about them and now I know more things.

I’ll keep posting updates here in this blog post as long as I have them, or until I have to stop and do other stuff.

UPDATE  1 (about 9:30pm 26/7/2022)

I have at least been able to prove the that the further the base of your i-arrow is from the real point on an line of unreal slope, the longer the arrow is. Here goes…

Suppose \(Ax + By = C\) is the equation of a line of unreal slope, where \(A\), \(B\) and \(C\) are complex numbers. (This version of the equation of the line was easier than the functional form this time.) And suppose the real point \((c,d)\) and the unreal point \((p+si,q+ti)\) are both on this line. Note that the i-arrow for this unreal point has base \((p,q)\) and arrow extending along the vector \((s,t)\).

Now since both points are on the line, they satisfy the equation, so

\[\begin{aligned}
Ac + Bd &= C \\
A(p+si)+B(q+ti) &= C\end{aligned}\]

Subtracting the first equation from the second produces

\[A(p-c+si) + B(q-d+ti)=0\]

Multiplying this by the real number k produces

\[A(k(p-c)+ksi) + B(k(q-d)+kti)=0\]

Adding this to the very first equation produces

\[A(c+k(p-c)+ksi) + B(d+k(q-d)+kti)=C\]

And so the point \((c + k(p-c)+ksi,d+k(q-d)+kti)\) is also on this line.
The i-arrow for this complex point has base \((c + k(p-c),d+k(q-d))\),
which is \((c,d)+k[(p,q)-(c,d)]\). This is a point somewhere on the line joining \((c,d)\) and \((p,q)\), \(k\) times as far from \((c,d)\) as \((p,q)\) is. The arrow itself is the vector \((ks,kt) = k(s,t)\), which is \(k\) times the length of \((s,t)\).

Therefore, if the base point of an i-arrow on the line is \(k\) times as far from \((c,d)\) as \((p,q)\), then the arrow is \(k\) times as long as \((s,t)\).

You can see this very clearly if you freeze one of the earlier animations at the moment you can see the real point. The arrows based at points twice, three times, etc, far from the real point are twice, three times, etc as long, and pointing in opposite directions on opposite sides.

So that’s very satisfying. I’m sure more will come later.

UPDATE 2 (about 5:00am 27/7/2022)

I thought of  this one very late last night and had to wait until this morning to add it here. The previous realisation doesn’t just say that the i-arrows are longer if the base point is further from the real point on the line of unreal slope. It also says that the i-arrows with base points along a line through the real point are all parallel to each other.

Even more so, if two i-arrows on a line of unreal slope are parallel, then the real point on that line must be on the line joining their bases, and also on the line joining their arrowheads. That’s what I realised looking at the picture above.

Here’s the calculation that proves it works.

Suppose \(Ax + By = C\) is the equation of a line of unreal slope, where \(A\), \(B\) and \(C\) are complex numbers, and suppose \((p+si,q+ti)\) and \((a+ksi,b+kti)\) are both on this line, for real numbers \(p\), \(q\), \(s\), \(t\), \(a\), \(b\), \(k\). Then:

\[\begin{aligned}A(p+si)+B(q+ti)&=C\\
A(a+kti)+B(b+ksi)&=C\end{aligned}\]

Subtracting the first from the second:

\[A(a-p+(k-1)si)+B(b-q+(k-1)ti)=0\]

Multiplying this by \(\frac{1}{k-1}\)…

(Aside: I know \(k-1\) can’t be 0, because the two i-arrows for \((p+si,q+ti)\) and say \((a+si,b+ti)\) are the same length and direction, which means their base points and arrowhead points are on two parallel lines, and that would mean the line is actually a line of real slope.)

… gives:

\[A\left(\frac{a-p}{k-1}+si\right)+B\left(\frac{b-q}{k-1}+ti\right)=0\]

Subtracting this from the very first equation gives:

\[A\left(p-\frac{a-p}{k-1}\right)+B\left(q-\frac{b-q}{k-1}\right)=C\]

The point \(\left(p-\frac{a-p}{k-1},q-\frac{b-q}{k-1)}\right)\) is a real point that satisfies the equation of the line! Even more,

\[\left(p-\frac{a-p}{k-1},q-\frac{b-q}{k-1)}\right)=(p,q)-\frac{1}{k-1}[(a,b)-(p,q)]\]

so it’s a point on the line joining \((p,q)\) and \((a,b)\).

I didn’t actually do the calculation last night when I thought of this idea. I realised it was true, and then drew a diagram of the geometrical construction that would find the real point from two parallel i-arrows. If you join the two bases, and also join the two arrowheads, the intersection of those two lines will be the real point on the line of unreal slope! Any arrow drawn parallel to these from the line of bases to the line of arrowheads will necessarily be an i-arrow on the graph of this line. So cool!

Here are some clean accurate diagrams drawn in GeoGebra:

UPDATE 3 (about 6:00am 27/7/2022)

I really want to figure out a construction from any two i-arrows that will find the real point on the line of unreal slope that joins them. I mean, what I really want to do is figure out not just the real point, but all the other i-arrows too, but one step at a time.

I’ve been staring at earlier diagrams of large collections of i-arrows on a line to see if I can notice anything, and I have noticed something: if you draw a line through the bases of some i-arrows, and step along this line in equal steps, then the arrowheads will step along another line in equal steps too.

This bears out with similar calculations to those above:

Suppose \(Ax + By = C\) is the equation of a line of unreal slope, where \(A\), \(B\) and \(C\) are complex numbers, and suppose \((p+si,q+ti)\) and \((a+ui,b+vi)\) are both on this line, for real numbers \(p\), \(q\), \(s\), \(t\), \(a\), \(b\), \(u\), \(v\). Note the i-arrows for these points have bases \((p,q)\) and \((a,b)\), and arrowheads \((p+s,q+t)\) and \((a+u,b+v)\). Now:

\[\begin{aligned}A(p+si)+B(q+ti)&=C\\
A(a+ui)+B(b+vi)&=C\end{aligned}\]

Subtracting these two gives:

\[A((a-p)+(u-s)i)+B((b-q)+(v-t)i)=0\]

Multiplying this by any real number \(k\) and adding it to the first equation gives:

\[A(p+k(a-p)+[s+k(u-s)]i)+B(q+k(b-q)+[t+k(v-t)]i)=C\]

Therefore the point \((p+k(a-p)+[s+k(u-s)]i,q+k(b-q)+[t+k(v-t)]i)\) is also a point on this complex line.

The base of its i-arrow is
\((p+k(a-p),q+k(b-q))\)
\(=(p,q)+k[(a,b)-(p,q)]\)
which is a point on the line joining \((p,q)\) and \((a,b)\).

The arrowhead of its i-arrow is
\((p+k(a-p)+s+k(u-s),q+k(b-q)+t+k(v-t))\)
\(= (p+s+k[a+u-(p+s)], q+t+k[b+v-(q+t)])\)
\(= (p+s,q+t)+k[(a+u,b+v)-(p+s,q+t)]\)
which is a point on the line joining \((p+s,q+t)\) and \((a+u,b+v)\).

So yes it really is true that i-arrows whose bases lie on a line also have their arrowheads lying on a line too. Stepping along the line of bases a multiple \(k\) of the space between \((p,q)\) and \((a,b)\) will step along the line of arrowheads a multiple \(k\) of the space between the matching arrowheads. Here is a picture illustrating how such a process produces a collection of i-arrows.

That’s what’s causing those lines of arrowheads in the animated GIFs I shared before. I’m glad I have algebra to back them up now.

UPDATE 4 (about 9:00am 27/7/2022)

I have been thinking hard about how to use two i-arrows, or a real point and one i-arrow to reconstruct all of the i-arrows on a line of unreal slope. Given the points \((p+si,q+ti)\) and \((a+ui,b+vi)\) on the line, you can definitely create the equation in “point-slope form”, same as you do for real lines: \(y=(q+ti)+\frac{(b+vi-q-ti)}{(a+ui-p-si)}(x-p-si)\). But I want a geometrical way to do it, especially because I don’t like the look of dividing by that complex number there.

I was comparing the pictures I drew with two parallel i-arrows to the ones I was sketching with two general i-arrows. They were very similar. In both, I draw a line joining the bases and a line joining the arrowheads, and in both, there are i-arrows going from one line to the other. Surely the non-parallel ones should tell me something about where the real point on the unreal line is? At the same time I was looking at the swirl of i-arrows around the real point in those pictures of a heap of i-arrows on an unreal line, and suddenly something occurred to me: none of the drawings of the i-arrows pass through the real point! Is that actually true, or is it an artifact of my choices of which i-arrows to draw?

I started doing calculations like this: suppose \((c,d)\) and \((p+si,q+ti)\) both satisfy the unreal line equation \(Ax+By=C\). Then

\[\begin{aligned}Ac+Bd&=C\\
A(p+si)+B(q+ti)&=C\end{aligned}\]

Consider the line segment that is the drawing of the i-arrow for \((p+si,q+ti)\). This contains all points of the form \((p,q)+k(s,t)\), for \(k\) between 0 and 1. Can \((c,d)\) be any of these points?

But then I realised I didn’t need to do any calculations!

For \((c,d)\) to be on the drawing of the i-arrow, it has to at least be on the line joining \((p,q)\) and \((p+s,q+t)\). But I know what that line is: it’s the unique real line containing the unreal point \((p+si,q+ti)\). And that would mean that the line joining \((c,d)\) and \((p+si,q+ti)\) is a real line, which I know it’s not!

So I wasn’t imagining it, the i-arrows don’t pass through the real point.

Which means when I have two non-parallel i-arrows on the same unreal line, then the space between the line joining their bases and the line joining their arrowheads can’t contain the real point on that unreal line, since that whole space is covered by i-arrows. Cool!

But I can do even more than that. The argument above actually shows the real point can’t be on any of the whole lines that are the extensions of the i-arrows. To put it more simply, none of the i-arrows point at or point directly away from the real point on the line. That’s so cool! Like, really cool!

And so even more of the plane can’t be places where the real point is. Let me see what GeoGebra has to say…

Well would you look at that. It looks like we can wipe out one whole side of a parabola. We haven’t pinned down exactly where that elusive real point is, but we’ve done a pretty fair job. I do wonder what that parabola is, but I don’t really mind too much. I’m just happy I have a bit more hope of finding that real point from a random two i-arrows.

Other work is pressing on me now, and I want to get down all the other stuff I had planned. So here is where I will leave it for now. I’m glad I got as far as I did. I’ve noticed some cool things, and properly confirmed them with the algebra, and made some nice constructions. Overall I think I have a much better feeling for how unreal lines work than I did before.

FURTHER UPDATE: A week or so later, and then again two years later, I found how to find all the complex points on a line joining two i-arrows.

These are all the other posts in the series, so you can find them easily.

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows (YOU ARE HERE)
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows

Introduction

The Cartesian plane is pretty cool. You think up an equation like \(y=x^2+1\) and find all the points \((x,y)\) whose coordinates satisfy it, and you get a shape (in this case a parabola). Different kinds of shapes have different kinds of equations, and finding the places where shapes meet becomes solving equations simultaneously. Geometry becomes deeply connected to algebra and everything is lovely.

Then we create the complex numbers so that equations that weren’t previously solvable become solvable. Most notably the equation \(x^2+1=0\) gets the solutions \(x=i\) and \(x=-i\). The simultaneous solutions to \(y=x^2+1\), \(y=0\) are supposed to be the places where that parabola met the x-axis, only it doesn’t meet the x-axis. If the complex numbers are to be believed, that parabola meets the x-axis at the points \((i,0)\) and \((-i,0)\). But I can’t see those points in the original Cartesian plane! Maybe there is a way to add the new complex points to my existing Cartesian plane so that the graphs do look like they meet.

And there is a way. In 2016, I created such a way, and I have just made it better, producing a thing called an i-arrow. This is the full description, which I am shamelessly cutting-and-pasting into every post in this series.

Every point with complex coordinates is represented as an arrow from one place to another on top of the Cartesian plane.

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

In this blog post, I’ll work through using i-arrows to represent the complex points on various line equations. (I’ve done this before using my old iplane concept, but I am now dowing it with i-arrows, and it’s better.) Let me look at real lines, unreal lines with real slope, and unreal lines with unreal slope.

Complex points on a real line

Consider the line with equation \(ax + by = c\) for some real numbers \(a\), \(b\), \(c\) with \(a\) and \(b\) not both zero. And suppose the complex point \((p+si, q+ti)\) satisfies this equation. Then…

\[\begin{aligned}
a(p+si) + b(q+ti) &= c\\
ap + asi + bq + bti &= c\end{aligned}\]

Equate real and imaginary parts to get…

Real parts: \[ap + bq = c\]
Imaginary parts: \[as + bt = 0\]

The real parts equation says that if \((p,q)\) isn’t on the original line, then there will be no solutions at all. And if it is, then anything that satisfies the imaginary parts equation will be a solution for \(s\) and \(t\).

So far this is all the same as what I did originally. But now I am imagining the complex points as i-arrows, based at \((p,q)\) and with arrowhead at \((p+s,q+t)\). I wonder if I can just locate the points \((p+s,q+t)\)? I can, if I add those two equations I had before:

Real parts: \[ap + bq = c\]
Imaginary parts: \[as + bt = 0\]
Sum: \[a(p+s) + b(q+t) = c\]

Well would you look at that, the coordinates of the arrowhead ALSO satisfy the original equation! So all the i-arrows for all the complex points on the real line have their base on the line AND their arrowhead on the line too! That is, the collection of all the complex points on a real line is all the i-arrows drawn from one point on the line to another. So much easier to describe than the previous version of this!

But while writing this I also realised it means something else too. The ONLY way a complex point can lie on a real line is if its i-arrow goes from one point on that line to another. Since an i-arrow is essentially a line segment, that means an i-arrow lies on EXACTLY ONE REAL LINE. So an unreal complex point lies on exactly one real line, and it’s the one that its i-arrow is part of. I think this is so cool!

For example, the complex point \((3+2i,5-i)\) has its i-arrow base at \((3,5)\) and arrowhead at \((3+2,5-1)=(5,4)\). And there is exactly one real line this complex point is on, which is the line joining \((3,5)\) and \((5,4)\). So clean. So simple. So cool.

Complex points on an unreal line with real slope

Now let me consider a line with real slope, but with no real points. That is, a line with equation \(ax+by=c+\gamma i\) for \(a\), \(b\), \(c\), \(\gamma\) real (\(a\) and \(b\) not both zero) and \(\gamma\) not zero. (This also covers vertical lines, which I still consider to have real slope.) I know this has no real points, because I can’t sub in real numbers for x and y and end up with a result that isn’t real! Let me look at the points \((p+si,q+ti)\) satisfying this equation…

\[\begin{aligned}a(p+si)+b(q+ti)&=c+\gamma i\\
ap+asi+bq+bti &=c+\gamma i\end{aligned}\]

Real parts: \[ap + bq = c\]
Imaginary parts: \[as+bt=\gamma\]
Sum: \[a(p+s)+b(q+t)=c+\gamma\]

The real parts equation tells me that the i-arrow’s base \((p,q)\) satisfies the equation \(ax+by=c\). That is, the base \((p,q)\) lies on the real line made by taking the real coefficients. If the base is not on that line, then the complex point is not on my original line.

The sum equation tells me that the i-arrow’s head \((p+s,q+t)\) satisfies the equation \(ax+by=c+γ\). This is the equation of a real line too. It’s parallel to the line where the base is, but just offset by a bit. Given a specific base \((p,q)\), every point on this second line will work as an arrowhead.

What this means is that the complex points on the complex line with real slope \(ax+by=c+\gamma i\) are the i-arrows with base any point on the line with equation \(ax+by=c\) and arrowhead any point on the line with equation \(ax+by=c+\gamma\). For example, this is a picture of a heap of i-arrows for complex points on the line with equation \(x+2y=13+4i\). All the i-arrow bases are on the line \(x+2y=13\), and all the i-arrow arrowheads are on the line \(x+2y=17\).

Would you just look at that picture? It’s so evocative. Even without putting the dotted line in for the bases, you can imagine all the arrows from one line to the other. It’s beautiful. You can even see the x-intercept and y-intercept on this graph too. The x-axis is a real line like any other, and a complex point is on that line when its i-arrow lies along the line. So the i-arrow along the x-axis from the line with equation \(x+2y=13\) to the line with equation \(x+2y=17\), that’s the x-intercept. It’s the point \((13+4i,0)\). Similarly the y-intercept is \((0,\frac{13}{2}+2i)\).

The x-intercepts and y-intercepts have made me realise something too. If you have two complex points, their i-arrow bases will define a line, and their i-arrow arrowheads will define another line. If those two lines are parallel, then these two i-arrows are part of the graph of a COMPLEX line with real slope like I drew above. Two points can only be in one line, so they must be in that exact line. For example, the points \((1+4i,2+i)\) and \((7+i,2i)\). Their i-arrow bases \((1,2)\) and \((7,0)\) are on the line \(x+3y=7\), while their i-arrow arrowheads \((5,3)\) and \((8,2)\) are on the line \(x+3y=14\). Those two lines are parallel, so the complex line joining them has equation \(x+3y=7+7i\), and its i-arrows are all arrows with base on one line and arrowhead on the other. Nice.

Ooh! A special case is when the two complex points have i-arrows with the same base. In that case, you are always able to set up the right pair of parallel lines to make a line of real slope containing both i-arrows. For example, these two complex points have i-arrows with bases \((1,2)\). Their arrowheads are at \((4,7)\) and \((5,3)\) which are joined by the line with equation \(4x+y=23\). The line \(4x+y=6\) passes through both bases. That means both i-arrows are on the graph of the line with equation \(4x+y=6+17i\), so that must be the unique complex line they are both on. Very nice.

Wait. This would also happen if the two i-arrows had the same arrowhead. Cool. I actually never expected such a clean geometric way to find the llne joining two complex points. Technically all of this was possible before with the i-plane model. It’s just that the i-arrow model allows me to look at more than one complex point at a time and see their whole story, both the imaginary and real parts, and this has let me realise things I didn’t before.

Complex points (and one real point) on an unreal line with unreal slope

This leaves us with one more type of line. We’ve covered all the real lines, which have heaps of complex points on them, which are are all the i-arrows drawn literally on top of the line. And we’ve covered unreal lines with real slope, which are made by taking a pair of parallel real lines and drawing all the i-arrows with base on one line and arrowhead on the other. That leaves unreal lines with unreal slope.

I already know that two complex points whose i-arrows share a base lie together on a line of real slope. So no two i-arrows for points on a line of unreal slope can share a base. And for the same reason, no two i-arrows for points on a line of unreal slope can share an arrowhead either. I never noticed this reasoning before and I am very happy about it.

Anyway, let me do the calculations I’ve done before.

Consider a line with equation \(y=(c+\gamma i) + (m+\mu i)x\) for real numbers \(c\),\(\gamma\), \(m\), \(\mu\), with \(mu\) not 0. I know I can write it in this form because it hasn’t got real slope. I could have kept it in the standard form, but this one will make it easier for me to do some GeoGebra things later.

Suppose the point \((p+si,q+ti)\) satisfies this equation. Then…

\[\begin{aligned}q+ti &=(c+\gamma i) + (m+\mu i)(p+si)\\
q+ti&=c+\gamma i + mp + msi +\mu pi -\mu s\end{aligned}\]

Real part:

\[\begin{aligned}
q &= c + mp – \mu s\\
\mu s &= c+mp – q\\
s &= (c+mp – q)/\mu\end{aligned}\]

Imaginary part:
\[t =\gamma + ms +\mu p\]

Given a specific point \((p,q)\), the real part equation tells us a unique answer for \(s\), and substituting this into the imaginary part equation tells us a unique answer for \(t\). So every point \((p,q)\) is the base for exactly one i-arrow. This confirms what I noticed earlier, that there couldn’t be two i-arrows on this graph with the same base.

Though I suppose it might not be an actual i-arrow if \(s\) and \(t\) are both zero. When does that happen? For \(s\) to be zero, I’d need \(q=c+mp\), and for \(t\) to be zero at the same time as \(s\), I’d need \(0=\gamma + \mu p\). That second one gives me a unique answer for \(p\), and substituting into the first gives me a unique answer for \(q\). So there is exactly one combination of \(p\) and \(q\) so that the real point \((p,q)\) is on the line.

That makes sense from a different perspective. The equation \(y=(c+\gamma i) + (m+\mu i)x\) is a function equation with \(x\) as input and \(y\) as output. If I put the real number \(p\) in for \(x\), it can only produce as its output a real number when \(\gamma i+\mu i p = i(γ+μp)=0\), and the only \(p\) that does that is \(\frac{-\gamma}{\mu}\). (This is one of the reasons I wrote the equation this way, since the input-output frame can be useful like this.)

I do also notice that \(q=c+mp\) is the equation of the line \(y=c+mx\) with \((p,q)\) substituted in, and \(y=c+mx\) has the real parts of the coefficients from the original equation. I do actually know this is a thing, even if the original equation was in standard form. But I am yet to produce any deep insights on how to use this fact to geometrically draw all the i-arrows. I’m sure I’ll get there one day.

For now, though, I do have equations I can use to draw the i-arrows using GeoGebra. Even without them, GeoGebra can do complex arithmetic, so I can actually just put in a list of inputs and get it to draw the i-arrows that go with the (input,output) complex points.

Here are the i-arrows for the complex points on the line with equation \(y=ix\) and bases at all the 1/4 grid points from -4 to 4 on both axes. It is pleasantly swirly. You can just see the single real point \((0,0)\) on the line in the middle, and how the complex points get longer and longer the further away from the centre their bases are.You can also believe that every point has exactly one arrowhead pointing at it.

A different view of this is to take all the (input, output) pairs that go with a collection of inputs and draw their i-arrows, then move the list dynamically. That’s what this animation does. An interesting feature is how the arrowheads that go with the bases along a vertical line all fall on a diagonal line. I’d like to prove that this is a thing, but it will probably be some other time.

These are the same pictures, but for a more general unreal line with equation \(y=(2+i)+(\frac12-i)x\). Again you can see the real point clearly in both the static and animated pictures. In the static picture you can see again how all the i-arrows seem to swirl around the real point the line is on, and are longer the further their base is from that real point. The shape of the swirl is elongated in one direction, rather than being perfectly symmetrical, which is interesting. Actually I can see the arrows get longer faster in one direction than another. I’m sure I can figure out how to predict that too. One day.

In the dynamic picture, again the arrowheads for the i-arrows with bases along a vertical line themselves lie on a diagonal line. The second dynamic picture has the bases attached to a diagonal line, corresponding to inputs all with the same imaginary part. Again, in both, the arrowheads for the i-arrows with bases along a line also fall along a line themselves. This feels right, but I haven’t found a way to wade through the calculations to be sure of it.

The structure of these unreal lines with unreal slope is fascinating to me, and mysterious. I want to understand them better. And one day I will. For now, I am happy to play around with these representations that the i-arrows have afforded me. (You can play too, with these GeoGebra pages: static i-arrows on complex line and dynamic i-arrows on complex line ).

Conclusion

The i-arrow model has really helped me to see complex lines better. I would never have had the insights I did without being able to draw so many of the complex points all at once. The unreal line with real slope in particular took on a whole new meaning for me, much more meaningful than the old picture I used to have of them from the i-planes. Even though the lines of unreal slope are still mysterious, I feel I know them better, and actually their mysteriousness is somehow nearer than it was before. In particular, I’ve had more questions than I would have had without being able to see so many complex points all at once, and without having a terminology like “arrowheads” to ask them.

I am loving these i-arrows.

Postscript

I have actually made some breakthroughs in the few hours since I wrote this blog post, and rather than break the story in this one, I’ve made a new blog post with all the updates. I’ll keep adding to it while I still have inspiration.

UPDATE: Even later, I made even more breakthroughs for unreal lines, and you can read about them here.

Oh, and these are all the other posts in this blog series, if you want to find them easily.

1. Where the complex points are: i-arrows
2. The complex points on a line using i-arrows (YOU ARE HERE)
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows

Problems with i-planes

Once upon a time in 2016, I created the idea of iplanes, which I consider to be one of my biggest maths ideas of all time. It was a way of me visualising where the complex points are on the graph of a real function while still being able to see the original graph.

This is the idea: at every real point \((p,q)\) of the real plane, there is a planes-worth of complex points attached, all of which have coordinates \((p+si, q+ti)\). The collection of all the points with real part \((p,q)\) I called the iplane at \((p,q)\), and I imagined the iplane as a transparent sheet attached at the point \((p,q)\) which I could flatten out to see its points when I needed to.

This was groundbreaking for me and I was so excited when it helped me find the complex points. There was a lot of cool stuff, mainly to do with taking an equation like \(y=x^2+1\), finding all the complex points in each iplane, then flattening out all the iplanes attached along a line and looking at where the points in them land. But you can already see a problem with it: the thing I want, which is to see where the complex points are (or at least look like they are) is several steps away from locating them. And when I do this process, I tend to lose the connection between the complex point and the point where its iplane is attached.

However, in my original series of blog posts, I actually already created a solution to this problem!

I can draw a complex number as an arrow on the real line, which starts at the real part and extends in the length and direction of the imaginary part.

This representation is not particularly useful for visualising operations on complex numbers. Addition is not too bad; multiplication is not at all obvious. What it does make obvious is conjugation. Conjugation is drawing the arrow from the same starting point in the opposite direction. This model is in fact the first time that conjugation has seemed to me like a natural thing to do.

Anyway, combining this arrow model of a complex number from an x-coordinate and a y-coordinate produces an arrow in the plane. The point \((p+si,q+ti)\) is an arrow based at the point \((p,q)\) and extending along the journey \((s,t)\) from there.

This is the representation I need. I have decided to call them i-arrows. I don’t want to call them i-vectors because that sounds too much like I want them to be the general powerful things vectors are. I want it to be clearly tied to the picture itself. (Plus “i-vector” is a terminology in computer science where an AI is trying to interpret spoken language, and I don’t want the wrong thing showing up when you google it.) I can show various complex points themselves more easily all on the same diagram and not lose the real point their iplane is attached to. I can still think about iplanes if I want, but I can talk about them as a collection of all possible i-arrows based at a certain point. And I can also talk about the locations of the arrowheads as opposed to the “real image of the complex point when the iplane is flattened”. I will still tend to draw the arrows in red or pink to maintain the old connection to the cellophane. I am very happy.

Introducing i-arrows

To sum up, here’s the technical details for the i-arrow representation of complex points:

• Real points are dots on the Cartesian plane, the same as they have always been.
• The complex point \((p+si,q+ti)\) is represented as an i-arrow, which is based at the point \((p,q)\) and extends along the vector \((s,t)\) to have its arrowhead at the point \((p+s,q+t)\).
• If we need it, the collection of all i-arrows based at \((p,q)\) is the i-plane based at \((p,q)\).

In the picture below, there are three examples of i-arrows.

• The complex point \((1+4i,2+i)\) has been drawn as an i-arrow. Its base is at the point \((1,2)\) and its arrowhead is at the point \((1+4,2+1)=(5,2)\).
• The complex point \((7,2i)\) has been drawn as an i-arrow. Its base is at the point \((7,0)\) and its arrowhead is at the point \((7,0+2)=(7,2)\).
• The complex point \((12-2i,3)\) has been drawn as an i-arrow. Its base is at the point \((12,3)\) and its arrowhead is at the point \((12-2,3)=(10,3)\)

Working and playing with i-arrows

When I started this blog post, I was going to show you everything I’ve been working on and playing with using i-arrows, but when I came to organise it, I thought more about it and made more stuff, and it got bigger and bigger. So I am putting all that other stuff into later blog posts.

The titles of the eight posts in the series are:

1. Where the complex points are: i-arrows (YOU ARE HERE)
2. The complex points on a line using i-arrows
3. Updates on the complex points on an unreal line using i-arrows
4. The complex points on a line in finite geometry using i-arrows
5. The complex points on a parabola using i-arrows
6. The complex points on real circles using i-arrows
7. The complex points on unreal circles using i-arrows
8. The line joining two complex points using i-arrows

I have to admit to a little sadness at my original idea of iplanes fading into the background as this new idea asserts its usefulness. But it may become useful again one day and it will still hold a special place in my heart. And really, looking at what the i-arrows can do, I am feeling very happy too.

Acknowledgements

Thanks to Dr David Roberts for listening to me talk about all of this, and suggesting the terminology “base”.

UPDATE: In 2024, I came back to the i-arrows concepts and further investigated the line joining two complex points:

9. Complex lines with i-arrows again