Making Your Own Sense

Reflections on maths, learning and maths learning support, by David K Butler

Tag: complex numbers

  • Where the complex points are: on a complex line (again)

    It’s been four years since I came up with the idea of iplanes as a way to organise the complex points on a graph, and in the intervening time I have thought about them on and off. For some reason right now I am thinking about them a lot, and I thought I would write down some of what I am thinking.

    A recap of the iplane idea

    At every point in the real plane, there are a real-plane’s-worth of complex points attached. The complex points attached to the real point \((p,q)\) are all of the form \((p+si,q+ti)\). That is, the real parts of the two coordinates are \((p, q)\). I imagine them as a plane, with x-axis showing what imaginary part we’ve added to the x-coordinate, and y-axis showing what imaginary part we’ve added to the y-coordinate. The real point itself is in the centre of this plane. I call this plane of complex points the iplane at \((p,q)\). Every complex point is in one of these iplanes. The complex point \((3+i,5-2i)\) is in the iplane attached to the real point \((3,5)\); the complex point \((2i, -6+7i)\) is in the iplane attached to the real point \((0,-6)\); and the complex points \((i,0)\) and \((-i,0)\) are both in the iplane attached to the real point \((0,0)\).

    A pair of coordinate axes with background in pink. The axes are labelled i sub x and i sub y. Various points are marked on the plane in blue. At the origin, the point is labelled (3,2). At the point marked 3 on the horizontal axis, the point is labelled (3+31,2). At the point marked 4 on the vertical axis, the point is labelled (3, 2+4i). Ohter points on neither axis are labelled similarly, all with real part 3 and 2.

    I imagine the iplanes as transparent sheets attached at each real point, that are usually rolled up like an umbrella, but can be flattened out to sit on top of the ordinary real plane when you need to. And yes I imagine them as transparent red or pink, because that’s the colour cellophane I had in my original model.

    The same cork from the previous picture with a pair of coordinate axes attached. The paper is almost completely covered with a large number of pieces of cellophane, all scrunched like flowers.

    The points on a complex line

    Last time, I did a thorough investigation into the points on a complex line, stepping through the points on a real line, a line with real slope but non-real intercept, and a line with non-real slope. But I’ve always been a little unsatisfied with my treatment of it. I really wanted to have a unified approach to all three types of lines, since they are all just complex lines. I was sure there was a way of dealing with them that could allow me to just take a complex line with a standard equation and know what it should look like.

    Well there is a way and I’ve done it. I am very proud of it, and not least of the things making me proud is a GeoGebra page that allows you to choose any complex line and then see what points are on the line in the iplane attached to a moving point, but with all kinds of complex lines together in one GeoGebra tool, rather than separate like last time. But we’ll get to that soon…

    First, the calculations.

    Consider a line \(L\) with equation \((a+\alpha i)x+(b+\beta i)y=(c+\gamma i)\), for real numbers \(a\),\(\alpha\), \(b\), \(\beta\), \(c\), \(\gamma\) with with \(a\),\(\alpha\), \(b\), \(\beta\) not all zero. This is the most general form of a complex line. Given a real point \((p,q)\), I am going to find all the complex points in the iplane attached at \((p,q)\) that are on the line \(L\). I will suppose that the point \((p+si,q+ti)\) is on \(L\) and try to find the imaginary parts of its coordinates (\(s\) and \(t\)).

    Since \((p+si, q+ti)\) is on the line \(L\), it satisfies the equation, and so:

    \[\begin{aligned} (a+\alpha i)(p+si)+(b+\beta i)(q+ti) &=(c+\gamma i)\\
    ap + asi +\alpha pi +\alpha si^2+ bq + bti +\beta qi +\beta ti^2 &= c+\gamma i\\
    ap-\alpha s + bq-\beta t + (as+\alpha p+bt+ \beta q)i &= c+\gamma i\end{aligned}\]

    Equating real and imaginary parts, I get two equations:

    \[\begin{aligned}ap-\alpha s + bq-\beta t &= c\\
    as+\alpha p+bt+\beta q &= \gamma\end{aligned}\]

    Which I can rearrange into:

    \[\begin{aligned}\alpha s + \beta t &= ap + bq – c\\
    as+bt &= \gamma-\alpha p-\beta q\end{aligned}\]

    Whatever points \((p+si,q+ti)\) inside the iplane at \((p,q)\) are on the line \(L\), they must have their \(s\) and \(t\)’s satisfy these two equations.

    But look closer at them! These two equations are the equations of lines! So whatever points in the iplane lie on both of these lines are the points of the original complex line \(L\) in this iplane. Two lines meet in exactly one point, which means there is exactly one point in the iplane on the complex line \(L\)!

    They are not just any lines either: the coefficients are chosen from the real and imaginary parts of the original line equation. Equation 1 has \(s\) and \(t\) coefficients which are the imaginary parts of the original line’s coefficients, while Equation 2 has \(s\) and \(t\) coefficients which are the real parts of the original line’s coefficients. That means these two lines inside the iplane are parallel to the “real part” and “imaginary part” of the original line equation, with equations \(ax+by=c\) and \(\alpha x+\beta y=\gamma\) respectively.

    And those two equations appear again in the equations inside the iplane at \((p,q)\). The constant term in Equation 1 is \(ap+bq-c\), which you would get if you substituted the point \((p,q)\) into the real part of the original equation. The constant term in Equation 2 is \(\gamma – \alpha p – \beta q\), which you would get if you substituted \((p,q)\) into the imaginary part of the original equation. If \((p,q)\) satisfies both of these equations, then Equation 1 and 2 both have constant 0, and so have \(s=0\), \(t=0 \)as a solution. But if \(s=0\) and \(t=0\), that means the point in the iplane is \((p,q)\), which is a real point! That is, there is one real point on the complex line \(L\) with equation \((a+\alpha i)x+(b+\beta i)y=(c+\gamma i)\), and this one real point is the intersection of the two real lines with equations \(ax+by=c\) and \(\alpha x + \beta y = \gamma\). I think that is so cool!

    Technicalities

    Of course, I have glossed over a few technical details. What if the two lines with equations \(\alpha s + \beta t = ap+bq-c\) and \(as+bt=\gamma-\alpha p-\beta q\) are parallel? Then there would be no intersection and so no points of the line \(L\) at all in this iplane. When would this happen? This would happen when the coefficients \(\alpha\) and \(\beta\) were the same multiple of \(a\) and \(b\), so \(\alpha = ka \) and \(\beta = kb\) for some \(k\). And then the original equation for \(L\) would have been

    \[\begin{aligned}(a+kai)x + (b+kbi)y &= c+\gamma i\\
    a(1+ki)x+b(1+ki)y &=c+\gamma i\\
    ax+by=(c+\gamma i)/(1+ki)\end{aligned}\]

    So the line \(L\) can be rewritten to have real coefficients for x and y, which means it has real slope! So a line with real slope will have many iplanes with no points of the line at all. Which ones will have actual points? Well, two parallel lines will share points (indeed all their points) if they are actually the same line after all. This would happen when as well as \(\alpha=ka\) and \(\beta=kb\), we also have the constant terms being related in the same way:

    \[\begin{aligned} ap + bq – c &= k(γ – \alpha p -\beta q)\\
    ap + bq – c &= k\gamma – k\alpha p – k\beta q\\
    ap + bq – c &= k\gamma – kka p – kkb q\\
    ap+k^2 ap + bq + k^2 bq &= k\gamma+c\\
    ap(1+k^2) + bq(1+k^2) &= k\gamma+c\\
    ap + bq = \frac{k\gamma+c}{1+k^2}\end{aligned}\]

    This is the equation of a straight line parallel to the one with equation \(ax+by=c\), which is the real part of the original line \(L\)’s equation! So either an iplane will have no points of \(L\) at all, or it will have a whole line of points of \(L\), whenever the point \((p,q)\) is on a specific line parallel to the real part of the original line.

    If the line \(L\) itself was a real line all along, then this could only happen if \((c+\gamma i)/(1+ki)\) was real, so \(\gamma\) would have to be \(kc\) as well, and the original line equation becomes \(ax+by=c\). There will be points in the iplane at \((p,q)\) when

    \[\begin{aligned}
    ap+bq &= \frac{k\gamma+c}{1+k^2}\\
    &= \frac{kkc+c}{1+k^2}\\
    &= \frac{(k^2+1)c}{1+k^2}\\
    &= c\end{aligned}\]

    This is the actual real line again! So a real line has points only in those iplanes attached to the real line itself, and in those iplanes, the points form a line parallel to the real line.

    There is one final technical detail. What if both \(a\) and \(b\) are zero, or both \(alpha\) and \(\beta\) are zero? If both \(\alpha\) and \(\beta\) are zero, then the two equations inside the iplane become

    \[\begin{aligned}0 &= ap + bq – c\\
    as+bt&=\gamma\end{aligned}\]

    If \(ap+bq-c\) is not zero, then Equation 1 has no solutions and there will be no points of the original line \(L\) in this iplane; if \(ap+bq-c\) is zero, then Equation 1 is 0=0, which is always true, and so any point satisfying Equation 2 is a point in this iplane. But wait, the points satisfying \(ap+bq-c=0\) are exactly the points on the real line with equation \(ax+by=c\). And since \(\alpha\) and \(\beta\) are zero, the original equation was \(ax+by=c+\gamma i\) anyway. So this matches with what we found before: a line of real slope has points only in the iplanes attached to a real line parallel to the real part of the original line. In this case, not just parallel but equal to the real part of the original line!

    The same thing happens when both \(a\) and \(b\) are zero. Then the two equations inside the iplane become

    \[\begin{aligned}\alpha s + \beta t &= -c\\
    0 &= \gamma – \alpha p – \beta c\end{aligned}\]

    And now there are no points of \(L\) in the iplane unless \(\alpha p + \beta q = \gamma\), which is still the equation of a real line. Indeed, the original equation is \((\alpha i)x+(\beta i)y=c+γi\), which is the same as \(\alpha x + \beta y = -ci+\gamma\), so the line with equation \(\alpha p + \beta q = \gamma\), is actually the real part of one version of the original line’s equation.

    Conclusion

    So after all this, we find that the complex line \(L\) with equation \((a+\alpha i)x+(b+\beta i)y=(c+\gamma i)\) behaves like this in the iplanes:

    1. The points \((p+si,q+ti)\) of the complex line \(L\) in the iplane attached at \((p,q)\) satisfy the two equations \(\alpha s+\beta t=ap+bq-c\) and \(as+bt=\gamma – \alpha p – \beta q\).
    2. If the complex line \(L\)’s equation cannot be turned into one with real coefficients for both x and y, then every iplane has one point of \(L\), at the intersection of two lines inside the iplane. There is exactly one real point on the line \(L\), which is the intersection of the real lines with equations \(ax+by=c\) and \(\alpha x+\beta y=\gamma\).
    3. If the equation can be turned into one with real coefficients for both x and y, then there are no points of the line \(L\) in most iplanes. The only iplanes with points on \(L\) are those attached to a real line parallel to the real (or imaginary) part of the original line equation.
    4. If the equation can be turned into one with real coefficients for x and y and real constant term, then \(L\) is a real line and the only iplanes with points on \(L\) are attached to the points on \(L\) itself. The points of \(L\) in each iplane form a line appearing to lie on top of \(L\) itself.

    You can investigate all of this in this GeoGebra applet, which allows you to choose coefficients for the equation of \(L\) and move the point \((p,q)\) to see the point(s) of \(L\) in the iplane attached there.

    A GeoGebra graph. In the top left is a place to enter the real and imaginary parts of the coefficients for a complex line. The equation is displayed top centre. A blue dotted line shows the real line whose coefficients come from the real parts of the complex line's coefficients. A pink dotted line shows the real line whose coefficients come from the imaginary parts of the complex line's coefficients. A gold point labelled A is the centre of an iplane represented by a pink square. Inside this square are two red dotted lines parallel to the pink and blue dotted lines outside the iplane. The point where they meet is coloured bright red.

    Thanks for being here for my mathematical scribblings. I worry that it might not make any more sense than the old version, but I feel a deeper sense of satisfaction having tried it both ways (and having one GeoGebra applet with all the types of lines at once).

  • Stop hating on cis(θ)

    I met with some lovely Electrical and Electronic Engineering lecturers yesterday about their various courses and how I can help their students with the maths involved. And of course complex numbers came up, because they do come up in electronics. (I have not the slightest clue how they come up, but I am aware that they do.)

    I asked them what notation they used for complex numbers, and they told me about using j instead of i, and then about polar form using either reᶿʲ and r∠θ. I told them that the students will have seen the notation r cis(θ) at school for those and I was met with righteous indignation. What a horrible notation! How could anyone think that was a good idea!

    It was rather a surprise considering their wonderful openness and friendliness up to this point. But this is not the first time I have been met with this sort of response to cis from more-or-less nice people. Many a mathematician has expressed to me their disdain for this little notation, and I have to say I am bamboozled. Why do people hate it so very much?

    I actually rather like cis. I like how cuddly-looking it is with all those curvy letters. I like how wonderfully easy it is to write simply with no special symbols and with no powers that have to be typeset at a higher level than the text. I like how it sounds like you’re singling out one of your many sisters when you say it aloud.

    I think cis(θ) is friendlier than eⁱᶿ because it doesn’t require you to suddenly believe that imaginary powers of real numbers are a thing and believe they ought to be this unusual combination of cos and sin. I think it’s also friendlier than ∠θ because we’re not co-opting a symbol we saw in geometry for a new and strange usage. (Not to mention the fierce pointiness of ∠ compared to the cuddly roundness of cis.)

    Finally, the thing I like the most is how cis(θ) is a function that takes real numbers and produces complex numbers, and those numbers are on the unit circle. For me It’s such a nice thing right from the start to recognise that there are such things as functions that produce complex-number outputs, and that they can have rules to manipulate them just like ordinary functions. Indeed, writing it with a multiple-letter name like all their other familiar functions forges this lovely connection. I also  love that multiplying a complex number by cis(θ) rotates it around the origin θ anticlockwise, because of course it does since cis(θ) is a position on a circle. (If anything, I think it wouldn’t hurt it if we renamed it “cir” to highlight its connection to a circle rather than highlighting its connection to cos and i and sin, even if that means losing its familial pronunciation and a little roundness.)

    Here’s a little geogebra applet to play with cis as a function from the real numbers to the complex numbers: https://ggbm.at/MWnm5TJA 

    Two graphs side-by-side. The left-hand one is labelled INPUT and shows just an x-axis with a point marked A. The right-hand one is labelled OUTPUT and shows an real and imaginary axis and a point marked cis(A) somewhere in the first quadrant.

    I know it hides the thing about complex powers of real numbers, but I think it’s important to hide that for a bit, lest the eⁱᶿ feels too much like a fancy trick. Plus I think it’s good to be familiar with this friendly little function for a while before discovering that it also solves a problem of complex powers. Indeed, I would love it if students were familiar with this friendly little function even before introducing the concept of polar form.

    So there are the reasons why I actually rather like cis. But there is one more reason I think mathematicians and professional maths users shouldn’t hate on cis: because it means hating on the students themselves. Our students put a lot of effort into understanding cis, and to loudly announce how much you hate it is telling them that they wasted all that effort and that their way of doing things is less sophisticated and babyish. Not the best way to start your relationship with them as their university teacher. I’d prefer it if people started off with acknowledging the coolness cis and then pointed out that the other notations are more convenient for doing what they want to do, or allow them to write formulas in more magical-looking ways, or allow them to do some cool stuff with powers. Then maybe we might not get them offside early in the piece.

    So please, stop hating on cis(θ)!

  • Where the complex points are: where the idea came from

    This is the last (for now) in a series of posts about Where the Complex Points Are. To catch you up, I discovered a way of visualising where the complex points are in relation to the points of the real plane. All the complex points \((p+ri,q+si)\) are in a plane attached to the real plane at the point \((p,q)\). I call these planes the “iplanes”, and I imagine it as a transparent sheet that can be unfolded to lie flat on the real plane in order to see its complex points. Using this model, I’ve investigated where the complex points are on a line, on a parabola, and on the graph of a complex function.

    Here are links to all the other posts in the series:

    1. Where the complex points are: introduction to the iplane concept
    2. Where the complex points are: on a line
    3. Where the complex points are: on a parabola
    4. Where the complex points are: the graph of a function
    5. Where the complex points are: where idea came from (YOU ARE HERE)

    There is one more thing I want to talk about and that’s where the idea came from, because I think it’s an interesting story and I want it recorded somewhere.

    Between not beyond

    Ever since I first learned about complex solutions to quadratic equations, I have wondered where the complex points are. I never really thought about it deeply, but the general picture in my head was that the complex points were somehow “beyond” the real plane. I imagined them as over there beyond the edge of the real plane, surrounding it so that the real plane was a small plane inside some much bigger plane.

    But something doesn’t quite work about this picture. When I studied conics in depth much much later I learned that a “degenerate circle” with equation \(x^2+y^2 = 0\) wasn’t just the origin \((0,0)\) as I had originally thought. In fact it was a pair of complex lines that each happen to have one real point. (These lines have equations \(x+iy=0\) and \(x-iy=0\).)

    Only recently did this start to bother me. Suddenly I asked myself, if the complex points are beyond the real plane, how can a line made almost entirely of complex points meet the real plane in the origin, right there in the centre?

    It occurred to me that the line must somehow travel over or between the real points to arrive at the origin in the middle.

    And the seed of an idea began to grow. If the lines have to travel between the other points to reach the origin, then the complex points must somehow be bewteen the real points, not beyond them.

    Finite planes — the smallest plane

    Once I realised that the complex points had to be between the other points, I started thinking about the problem in terms of finite planes. In a finite geometry, the points are discrete points, and there is plenty of space between them to put extra points, if indeed that’s where they are.

    I started out with the smallest possible affine plane (our ordinary geometry is a kind of affine plane). It has four points with coordinates taken from the set \(\{0,1\}\) (also known as \(\mathbb{Z}_2\)) and six lines. The lines have equations \(x=0\), \(x=1\), \(y=0\), \(y=1\), \(x+y=0\), \(x+y=1\). One name for this plane is \(AG(2,2)\) – the Affine Geometry of dimension 2 and order 2.

    A drawing in crayon on brownish paper. The drawing shows a square with its two diagonals drawn in. At the corners of the square are four big blue dots. The dots are labelled from bottom left anticlockwise as (0,0), (1,0), (1,1) and (0,1). The lines are each labelled with an equation, which are listed in the paragraph nearby.

    What I needed was to find where the “complex points” were in relation to this plane. Unfortunately, if you’re using \(\mathbb{Z}_2\), \(1+1=0\) so -1 is the same as 1. This means the equation \(x^2 = -1\) already has a solution, so I can’t add the number \(i\). You actually have to add the solution to \(x^2 + x = 1\). I called this number \(\alpha\) so that I ended up with a set of four numbers to use for coordinates: \(\{0,1,\alpha, 1+\alpha\}\).

    From this I can build \(AG(2,4)\), which has 16 total points, and 20 total lines. I wrote down the equations of all the new lines and tried to figure out how to add the points to my diagram so that the new lines looked like lines.

    I knew the lines in my new plane would have to have four points — one for each “number” in the coordinates — and this meant each line already existing needed two extra points. It seemed reasonable to exend them outwards. It also seemed reasonable to try and make the new lines \(x=\alpha\) and \(x=1+\alpha\) vertical and the new lines \(y=\alpha\) and \(y=1+\alpha\) horizontal.

    Fiddling around with this gave me a nice square arrangement for the points. Then I investigated what all the new lines looked like and I got something like this. (Note I was drawing in crayon on the paper tablecloths at my sister’s restaurant at the time. I didn’t take a photo on the day, but this is a reasonable facsimile.) You can see the lines a little more clearly in this GeoGebra applet.

    A drawing in crayon on brownish paper. Four blue dots are arranged in a square, and around them are twelve blue stars filling out a four by four grid. The points are all given coordinates using 0, 1, alpha and 1+alpha. Many straight and curved lines are drawn through the dots and stars. In green are the four vertical grid lines. In red are the four horizontal grid lines. In orange are four arches, two up and two down. In pink are these same arches rotated 90 degrees. In blue are the main diagonals of the big square and two big ovals oriented at 45 degrees either way.

    Unfortunately the extra points were outside the original ones, and not between them as I had hoped. I did notice that each new line that wasn’t part of the original plane met the original plane in exactly one point. I knew this would have to happen because it’s a theorem from finite geometry, but I was glad that everything was working the way it should.

    Finite planes — noughts and crosses saves the day!

    After the half-success of the smallest affine plane, it was natural to move on to the second-smallest affine plane \(AG(2,3)\). This is the plane you get when you add in all the wraparound diagonals to a Noughts and Crosses board. When drawing it on a noughts-and-crosses board, the twelve lines look like this:

    Twelve 3 by 3 grids of squares arranged around the edges of a rectangle. The three grids in the middle of the top edge each have a different row of squares coloured dark blue. The three grids in the middle of the bottom edge each have a different column of squares coloured dark blue. The grid in the middle of the left edge has the downward diagonal squares coloured in dark blue, while the grid in the middle of the right edge has the upward diagonal in dark blue. The grids in the four corners each have three squares coloured dark blue, one in the corner and the two on either side of the opposite corner.

    If you draw the points as dots and the lines as (curved) lines, you can make it look like this. You can see the individual lines more clearly in this GeoGebra applet.

    As you can see, it’s already quite invovled, and I didn’t really like to imagine what would happen when I went up to the “complex” plane — that would mean going from 9 points to 81 points and from 12 lines to 90 lines. The 20 lines of the affine plane of order 4 were enough of a mess!

    But I did start to investigate. \(AG(2,3)\) has its coordinates taken from \(\mathbb{Z}_3\), which is \(\{-1,0,1\}\) with the rule that \(1+1=-1\). This field doesn’t have any solutions to \(x^2=-1\), so it was safe to attach the number \(i\) this time. I drew my extra points clustered around the existing points and started to investigate the lines I would need. In order to get the lines with equations like \(x=i\) to be vertical, I needed to make sure that the points with coordinates \((i,something)\) were lined up above each other. This required me to make sure that within each cluster of points the same imaginary parts were added to the coordinates.

    Here’s a picture of my ideas book around this time:

    A spiral-bound notebook with unlined paper open to a double-page spread. Both pages are filled with diagrams and calculations. A date in the top left corner of the left-hand page says 31 / 7 / 2016.

    After drawing it on paper, it was time to investigate it more fully, so I worked on creating a GeoGebra applet that would show the points on a specific complex line. It took some fancy footwork with the spreadsheet to make it do what I wanted.

    81 points are arranged in a 3 by 3 grid of 3 by 3 grids. The centre point of each small grid is a blue diamond, while all other points are black dots. The main upwards diagonal of nine points is highlighted using a pink ring around each point. At the top is a note saying "line equation:" and then boxes to enter values for the equation. The equation currently entered is (1+0 i) x + (minus 1 + 0 i) y = 0 + 0 i.

    I noticed now that the extra points on every real line lined up neatly with the real line. This was totally awesome, and I was inspired to move on to the real plane to see what was happening there.

    Finally iplanes are found

    I was all satisfied with how my complex points worked in finite planes. The final step was to think about how this applied to the infinite real plane.

    There was a big problem, of course. In the finite planes, there was plenty of space between the points to fit the extra points. In the contiuous real plane we’re already full. Where do I fit those extra points?

    My first idea was that they were all squished into each point of the plane — “fat” points, as it were. But this was somehow unsatisfying because I couldn’t actually see where the points were. Then it occurred to me that if I couldn’t squish them into each point, they’d have to be above the plane but attached at the real point. And the iplane was born.

    At the start, I was sketching and calculating on scrap paper, drawing a little square around the real point to visualise the iplane. This was giving me enough insight to get a feel for what was going on with lines. The GeoGebra applets in the previous posts are based on this visualisation. It wasn’t until I got out the cellophane and the tap-tap board that I really felt like I had a proper feel for what was going on. This pinkness has transferred to all of my diagrams and all of my thinking about iplanes, but you’ve seen the results of all that thinking already…

    Four pictures in progression. The first picture shows an x-y plane with a parabola. The second picture has a cellophane sheet all scrunched up at the origin. In the third picture, the cellophane sheet has been opened out partly. In the third picture, the sheet has been laid flat on top of the x-y-plane and on it there are the points (-i,0) and (i,0) marked.

    Conclusion

    So what is the point of this story? I don’t know. I just thought you (and future me) might like to see my train of thought in creating this idea.

    I suppose for me the main moral is that finite geometry helps give me insights into real geometry, and that my habit of making physical models of things is still a worthwhile thing to foster.

  • Where the complex points are: the graph of a function

    This is the fourth in what has turned out to be a series on Where the Complex Points Are.

    1. Where the complex points are: introduction to the iplane concept
    2. Where the complex points are: on a line
    3. Where the complex points are: on a parabola
    4. Where the complex points are: the graph of a function (YOU ARE HERE)
    5. Where the complex points are: where idea came from

    This one wasn’t part of the original plan, but some interesting thoughts have come up while discussing the previous posts on Twitter, and I needed to process them. So here we are doing an extra blog post before my final epilogue.

    A problem of viewpoints

    In the introduction, I talked about how I’ve always wondered where the complex points are when you talk about a parabola meeting the x-axis in complex points. I introduced the concept of an iplane: at each point \((a,b)\) in the real plane, there is an iplane attached, which contains all the points \((a+ci,b+di)\). These points are arranged with the x-axis showing the imaginary part added to the x-coordinate and the y-axis showing the imaginary part added to the y-coordinate. I imagine them as being a transparent sheet attached to the point itself and being able to be flattened out to sit on top of the real plane.

    A pair of coordinate axes with background in pink. The axes are labelled i sub x and i sub y. Various points are marked on the plane in blue. At the origin, the point is labelled (3,2). At the point marked 3 on the horizontal axis, the point is labelled (3+31,2). At the point marked 4 on the vertical axis, the point is labelled (3, 2+4i). Ohter points on neither axis are labelled similarly, all with real part 3 and 2.
    Four pictures in progression. The first picture shows an x-y plane with a parabola. The second picture has a cellophane sheet all scrunched up at the origin. In the third picture, the cellophane sheet has been opened out partly. In the third picture, the sheet has been laid flat on top of the x-y-plane and on it there are the points (-i,0) and (i,0) marked.

    This representation of the complex points allowed me to investigate the complex points that lie on lines (both complex lines and real lines) and to investigate the complex points that lie on a vertical parabola, and also how lines meet such a parabola. I shared these thoughts on Twitter, and waxed lyrical about them in person to basically every person I met. People were impressed, but there was something in the way they spoke that niggled at me and I couldn’t figure out what it was.

    Yesterday I figured out the problem: I was thinking about the lines an parabolas in totally different ways to other people! In my head, the lines and parabolas were sets of points whose coordinates satisfy a condition. To decide if a point is part of the set, you sub its coordinates into the equation and see if it’s true: if it’s true, then the point is on the object, if it’s not true then the point isn’t on the object. I was looking to see which points were on the parabola by simply subbing the coordinates in to see which ones worked. I did this one iplane at a time because it seemed the most reasonable way of organising my search.

    But what was reasonable to me was not reasonable to others! Taking each iplane one at a time and looking to see which points in the iplane satisfy the equation was totally alien to quite a few people I talked to. These people totally understood the iplane concept and they thought it was a great idea, but they had no idea what I was getting at by looking in each iplane separately. Yesterday I realised why: they were thinking of the lines and parabolas as an organisation of the input/output relationship for a function. They were looking for an x-axis and a y-axis to “line up” the input and the output of the function formula.

    This is of course a perfectly valid way to think of many graphs, especially for the parabolas I was looking at, whose equations all look like “y = something in terms of x”. It’s just that I wasn’t thinking about them that way so my iplanes weren’t designed to match with this viewpoint. So I started thinking about how to make it match.

    The complex line

    There is one key issue with making the graph relate more closely to the idea of lining up the input on the x-axis with the output on the y-axis: the input is a complex number and it doesn’t really live on the x-axis. But, I dealt with that problem already in the plane by adding an iplane to each point of the plane. Surely I could do the same thing to the line, right? Why can’t I add an iline to each point on the real line? The iline attached at the number a would contain all the numbers of the form \(a+bi\).

    With the iplanes, I flattened them out to sit on top of the real plane so I could see the complex points, but I only flattened out enough to see the point I was looking for. I wanted to do the same thing with the ilines, but it’s not really the same feel. Somehow I can’t tell so easily what’s going on with the complex number. Then I realised what I could do is just draw an arrow from the point representing the real part to the complex number in its iline. That is, I could draw the number a+bi as the point a on the number line and an arrow of length and direction b starting from that point. So the following representation was born:

    A number line. A blue point at -1 has a red arrow pointing along the number line to 2. This arrow is labelled u = minus 1+3i. Another blue point at 5 has a red arrow pointing along the number line to 3. This arrow is labelled v = 5 minus 2i.

    You can investigate what complex-number addition looks like in this representation with this GeoGebra applet, and what complex-number multiplication looks like with this GeoGebra applet. (This representation is not so good for complex multiplication, but that’s ok — different representations are useful for different things.)

    A complex x-axis and y-axis

    So I was now able to locate a point on the x-axis as an input for a function, and locate its output on the y-axis. Then something rather wonderful happened: I realised that I could combine this point-and-arrow x-coordinate and this point-and-arrow y-coordinate into one point-and-arrow in the plane. The point (p+ri,q+si) became the point (p,q) and the arrow (r,s) starting at that point.

    But the point \((p,q)\) is the centre of the iplane in which the complex point \((p+ri,q+si)\) lives! And the arrow \((r,s)\) is the vector-version of the coordinates \((r,s)\) within that iplane! So it turns out that my point-and-arrow x-coordinates and y-coordinates matched up perfectly with my iplanes! I couldn’t be happier!

    A coordinate plane. On the x-axis at 2 is a blue point with a red arrow pointing from there 3 units to the right. The arrow is labelled x = 2 + 3i. On the y-axis at 5 is a blue point with a red arrow pointing from there 6 units up. The arrow is labelled y = 3 + 6i. An iplane represented by a pink rectangle, anchored at the gold point (2,3). An arrow from that gold point goes diagonally 3 to the right and 6 up. This arrow is labelled (x,y) = (2 + 3i, 3 + 6i).

    The complex graph of a real function

    So now I was able to investigate graphs of complex functions from the perspective of input-output relationships. It’s a whole different way of thinking about things, and I haven’t investigated it to any great depth yet, but so far it is most interesting. One particular thing I’ve noticed is that quadratic functions are rather special when it comes to how the real images of the complex points are arranged — the tangents seem to play a much bigger role than they do for higher polynomials.

    I’ve made a GeoGebra applet that will allow you to enter a real function and move the input around on the x-axis to see what happens to the output and the matching complex point on the graph. You can turn on the trace to see what shape is made when you do move the input around. I now have a host of new questions rising in my mind which I hadn’t even considered before, and it will probably take years to sort them out.

    A screenshot of a GeoGebra graph. In the top left corner is a box to enter a polynomial of degree up to 6. Currently a cubic is entered in the box. Below that is a tick box to turn the trace on. A black curve shows the graph of the cubic. On the x-axis is a blue point with an arrow from there to a red point also on the xaxis. This is labelled with x = a complex number. You can't see it, but in the GeoGebra graph, you can move the blue and red points to change this arrow. On the y-axis is another blue point with a red arrow along the y-axis. You can't see it, but this moves when you move the x-axis points. An iplane is represented by a pink rectangle centred on a gold dot. A black arrow inside the iplane extends from this gold dot and is labelled (x,y). The gold dot and its arrow line up with the blue dots and red arrows on the axes. A collection of gold dots make a vertical line from the black curve downwards. A collection of green dots marks where the ends of the black arrows have been and form a parabola.

    But for now I’ll just be content that I have a picture that works for both views of a graph: as a collection of points that satisfy a condition and as an encoding of the input/output relationship of a function.

  • Where the complex points are: on a parabola

    This is the third in what has turned out to be a series on Where the Complex Points Are.

    1. Where the complex points are: introduction to the iplane concept
    2. Where the complex points are: on a line
    3. Where the complex points are: on a parabola (YOU ARE HERE)
    4. Where the complex points are: the graph of a function
    5. Where the complex points are: where idea came from

    In the introduction, I talked about how I’ve always wondered where the complex points are on a parabola, especially wondering where the x-axis intercepts are if the parabola doesn’t meet the parabola.

    A set of coordinate axes with a parabola drawn in blue. The parabola has its lowest point on the y-axis at y=1. The equation y=x squared plus 1 is written next to the parabola. Handwritten in red writing at the bottom of the picture is the question "Where are the complex points?"

    Then I described the place where I imagine the complex points are: at each point \((a,b)\) in the real plane, there is an iplane attached, which contains all the points \((a+ci,b+di)\). These points are arranged with the x-axis showing the imaginary part added to the x-coordinate and the y-axis showing the imaginary part added to the y-coordinate. I imagine them as being a transparent sheet attached to the point itself and being able to be flattened out to sit on top of the real plane.

    A pair of coordinate axes with background in pink. The axes are labelled i sub x and i sub y. Various points are marked on the plane in blue. At the origin, the point is labelled (3,2). At the point marked 3 on the horizontal axis, the point is labelled (3+31,2). At the point marked 4 on the vertical axis, the point is labelled (3, 2+4i). Ohter points on neither axis are labelled similarly, all with real part 3 and 2.
    Four pictures in progression. The first picture shows an x-y plane with a parabola. The second picture has a cellophane sheet all scrunched up at the origin. In the third picture, the cellophane sheet has been opened out partly. In the third picture, the sheet has been laid flat on top of the x-y-plane and on it there are the points (-i,0) and (i,0) marked.

    In the second post, I talked about where the complex points are on a line. Now I’m finally ready to talk about where the complex points are on a parabola.

    The points in each iplane

    When I looked at lines with complex slope, I looked at each iplane individually and found out which points in that plane were part of the line. (There was one in each.) Let’s do the same for our parabola.

    Consider a parabola with equation \(y = ax^2+ bx + c\) and look at the complex points in the iplane attached to the point \((p,q)\). That is, the points \((p + ri, q + si)\)…

    \[\begin{aligned}
    q+si &= a(p+ri)^2+b(p+ri)+c\\
    &= a(p^2+2pri-r^2)+b(p+ri)+c\\
    &= ap^2+2apri-ar^2+bp+bri+c\\
    &= (ap^2+bp+c-ar^2)+(2ap+b)ri\end{aligned}\]

    Equate real and imaginary parts:

    \(q=ap^2+bp+c-ar^2\) and \(s=(2ap+b)r\).

    From the first equation:

    \[\begin{aligned}
    ar^2 &= ap^2+bp+c-q\\
    r^2 &= \frac{ap^2+bp+c-q}{a}\end{aligned}\]

    So we have an expression for \(r^2\). Since \(r\) is a real number, we’ll get a different number of solutions depending on whether this value is negative, zero or positive. But let’s look closer, when is this number negative, zero or positive?

    Well, \(ap^2 + bp + c\) is the y-coordinate of the point on the parabola with x-coordinate \(p\).
    So \(ap^2 + bp + c – q\) is the vertical distance between the point \((p,q)\) and the parabola. If the point is below the parabola, this value is positive; if it’s above the parabola, this value is negative; and if it’s on the parabola, this value is zero. The value of the coefficient a being positive or negative will change whether the whole expression is positive or negative. But it will also change whether the parabola itself is upward or downward curving. So overall we get the following possibilities.

    Point \((p,q)\) below parabolaPoint \((p,q)\) on parabolaPoint \((p,q)\) above parabola
    Parabola upward-curvingTwo solutions for \(r\)One solution for \(r\)No solutions for \(r\)
    Parabola downward-curvingNo solutions for \(r\)One solution for \(r\)Two solutions for \(r\)

    Since the second equation only has one \(r\), each solution for \(r\) will give one solution for \(s\). This means that we now know many points of the parabola there are in each iplane, depending on where the iplane is attached.

    • If the iplane is attached to a point inside the curve of the parabola, then there are no points of the parabola in the iplane.
    • If the iplane is attached to a point on the parabola, then this is the only point of the parabola in the iplane.
    • If the iplane is attached to a point outside the curve of the parabola, then there are two points of the parabola in the iplane.

    This GeoGebra applet allows you to investigate the complex points on a parabola in the iplanes attached nearby.

    A screenshot of a GeoGebra applet. There is a set of coordinate axes with a a parabola drawn in blue. In the top left corner is a place where you can enter the coefficients of the parabola's equation. There is an iplane represented by a pink square centred on a moveable gold dot. Currently the gold dot is below the parabola and the pink square contains two red points on opposite sides of the gold dot.

    Collecting together iplane points

    If you’ve played with that GeoGebra applet for a bit, you might have noticed that when you move the point where the iplane is attached, the complex points on the parabola move in interesting ways across the screen. It’s actually very interesting.

    In a completely technical sense, the points we see in the iplanes in my GeoGebra applets aren’t really where they look like they are. They are up in iplane which only touches our real plane at the centre. It’s just that when we flatten a part of the iplane out and lay it on the real plane it’s easier to see the points. I’m going to make a new terminology to describe where the complex point looks like it is when you do this flattening out..,

    Given a complex point, when the iplane it is in is flattened out to lay on top of the real plane, the real point where it appears to be is called the real image of the complex point. It’s quite straightforward to find the real image of a complex point directly from its coordinates: the complex point \((p+ri,q+si)\) has real image \((p+r,q+s)\).

    Anyway, what happens when you take all iplanes attached along a vertical line and collect together the real images of the parabola’s complex points in those iplanes? (There’s got to be a way to define terminology to make it easier to say what I just said! But right now I can’t think of anything good.) You can investigate this question visually in this GeoGebra applet.

    A screenshot of a GeoGebra applet. There is a set of coordinate axes with a a parabola drawn in blue. In the top left corner is a place where you can enter the coefficients of the parabola's equation. There is a vertical green line showing the axis of symmetry of the parabola. There is an iplane represented by a pink square centred on a moveable gold dot. The gold dot is currently on the green vertical line. A tick box has a tick in it and a label saying "Trace on". The trace of the red points as the gold dot moves up and down the green line is a bold red parabola with vertex touching the blue parabola but facing the other way.

    Look at that! The real images of the points in the iplanes attached to that vertical line form a parabola of their own! The line I’ve shown there is the axis of symmetry of my parabola, and it appears that this parabola-of-real-images is just the original parabola flipped upside-down.

    Let’s do some calculations. Think about all the points \((x,y)\) that are the real images of the parabola’s complex points in the iplanes attached to the vertical real line \(x=p\). These complex points are \((p+ri,q+si)\) for some \(q\), \(r\), \(s\) and have real images \((x,y)=(p+r,q+s)\), where \(q=ap^2+bp+c-ar^2\) and \(s=(2ap+b)r\).

    Now \((x,y)= (p+r,q+s)\),
    so \(y=q+s = ap^2+bp+c-ar^2+(2ap+b)r\).

    But \(x=p=r\), so \(r=x-p\),
    which gives \(y=ap^2+bp+c-a(x-p)^2+(2ap+b)(x-p)\).

    This is the equation of a parabola.

    So given any vertical line, if you look in all the iplanes attached to the line and collect together the images of the complex points in those iplanes, you make a parabola! What if the line is the axis of symmetry of the parabola? That is, what if \(p = -b/(2a)\)? Then \(2ap+b=0\) and the equation becomes \(y=ap^2+bp+c-a(x-p)^2\).

    This is the equation of a parabola with vertex \((p, ap^2 + bp + c)\) — the same vertex as our original parabola. Its leading coefficient is \(-a\) — as opposed to \(+a\) for the original parabola — so it’s the same shape as the original parabola, but upside-down.

    What this means is, if you want to find the complex points where a parabola meets, say, the x-axis, just flip it upside-down at the vertex and see where this parabola meets the x-axis. These are the real images of the complex points of the parabola on the x-axis, and they are in the iplane attached to the point where the axis of symmetry meets the x-axis.

    iplanes and tangents

    Did you notice the second equation we got above? It said that when \((p+ri,q+si)\) is on the parabola then \(s = (2ap+b)r\). The number \(2ap+b\) is the slope of the tangent to the parabola at the point where \(x=p\). The equation \(s = (2ap+b)r\) is an equation of a line of this slope inside the iplane through the centre of the iplane. When the iplane is laid flat on the real plane, you get a line of slope \((2ap+b)\) through the point \((p,q)\). This means that the complex points of the parabola in the iplane at \((p,q)\) are both on a line parallel to the tangent to the parabola at the point where \(x=p\).

    A screenshot of a GeoGebra graph. There is a set of coordinate axes with a a parabola drawn in blue. In the top left corner is a place where you can enter the coefficients of the parabola's equation. There is an iplane represented by a pink square centred on a gold dot. There is a vertical line drawn through the gold dot, with a black point where it meets the parabola. The tangent to the parabola at that point is drawn. A line parallel to that tangent is drawn through the gold dot. In the iplane there are two red points, which both lie on the line through the gold dot that is parallel to the tangent.

    But wait, there’s more! We just went vertically to the parabola from our iplane centre point. What happens when we go vertically to the parabola from those two complex points in our iplane? (Well, from their real images anyway.) The real image of these complex points are \((p+r,q+s)\). The matching points on the parabola have x-coordinate \(x=p+r\). Let’s find the y-coordinate:

    \[\begin{aligned}
    y &= a(p+r)^2 + b(p+r)+c\\
    &= a(p^2+2pr+r^2)+b(p+r)+c\\
    &= ap^2 + 2apr+ar^2+bp+br+c\\
    &= ap^2 + bp +c+2apr+br+ar^2\\
    &= ap^2 + bp + c + (2ap+b)r + ar^2\\
    &= ap^2 + bp+c – ar^2 + (2ap+b)r+2ar^2\\
    &= q+s+2ar^2\end{aligned}\]

    Now \(q+s\) is the y-coordinate of the real image of our complex point, and we already figured out above that \(ar^2\) is the vertical distance from the point \((p,q)\) to the parabola. So the y-coordinate is the y-coordinate of the real image of our complex point, plus twice the distance from the iplane centre to the parabola. That’s pretty cool. Even cooler is the fact that both complex points are the same vertical distance from their matching real parabola point, so the line joining these points is parallel to the line joining the complex points.

    What this means is we now have a geometric way of finding the complex points on the parabola in the iplane at P=\((p,q)\):

    1. Draw a vertical line from P to the parabola and mark the point on the parabola. Let’s call it A.
    2. Draw the tangent to the parabola at the point A.
    3. Copy the distance from A to P inside the parabola from A. Let’s call this new point B.
    4. Draw a line through B parallel to the tangent at A and mark where it meets the parabola. Let’s call these points C and D.
    5. Draw vertical lines through C and D.
    6. Draw a line through P parallel to the tangent at A. Mark where it meets the vertical lines through C and D.

    These final two points are the real images of the complex points on the parabola in the iplane at P. You can see this construction in this GeoGebra applet. Isn’t that cool?

    A screenshot of a GeoGebra applet. There is a set of coordinate axes with a a parabola drawn in blue. In the top left corner is a place where you can enter the coefficients of the parabola's equation. There is an iplane represented by a pink square centred on a gold dot, which is labelled P. There is a green vertical line drawn through the gold dot, with a black point labelled A where it meets the parabola, and a black point labelled B the same distance above A as A is above P. The tangent to the parabola at the point A is drawn. A line parallel to this tangent is drawn through B, meeting the parabola at C and D. Gren vertical lines are drawn through D and C. A line parallel to the tangent at A is drawn through P, meeting the vertical lines through C and D at two red points in the iplane.

    It also means we can do a similar construction to find where a particular line meets the parabola. Obviously if it meets the parabola in real points then there they are in the picutre. If it has no real points on the parabola, then we know that all the complex points on that line have real images on the line too. We also know that the complex points of the parabola in any iplane are on a line parallel to a tangent of the parabola. This means we have this geometric way of finding the complex points where a line meets a parabola:

    1. Find the tangent to the parabola parallel to your line. Mark the point where it meets the parabola A.
    2. Draw a vertical line through A. Mark the point where it meets the original line P — this is the centre of the iplane we need.
    3. Follow the construction above to find the complex points in the iplane at P.

    If your line is a horizontal line, such as the x-axis, then this process is particularly simple because the only point with a horizontal tangent is the vertex.

    Conclusion

    So, in every iplane centred at a point outside the parabola there are two complex points on the parabola, and the real images of those points can be found via a geometric construction using tangents. Isn’t that totally awesome?

    In the final post, I’ll talk about where this idea came from, because I think it’s totally worth mentioning.

  • Where the complex points are: on a line

    This is the second in what has turned out to be a series on Where the Complex Points Are.

    1. Where the complex points are: introduction to the iplane concept
    2. Where the complex points are: on a line (YOU ARE HERE)
    3. Where the complex points are: on a parabola
    4. Where the complex points are: the graph of a function
    5. Where the complex points are: where idea came from

    In the introduction, I described the place where the complex points are: at each point \( (a,b)\) in the real plane, there is an iplane attached, which contains all the points \((a+ci,b+di)\). These points are arranged with the x-axis showing the imaginary part added to the x-coordinate and the y-axis showing the imaginary part added to the y-coordinate. I imagine them as being a transparent sheet attached to the point itself and being able to be flattened out to sit on top of the real plane.

    A pair of coordinate axes with background in pink. The axes are labelled i sub x and i sub y. Various points are marked on the plane in blue. At the origin, the point is labelled (3,2). At the point marked 3 on the horizontal axis, the point is labelled (3+31,2). At the point marked 4 on the vertical axis, the point is labelled (3, 2+4i). Ohter points on neither axis are labelled similarly, all with real part 3 and 2.
    Four pictures in progression. The first picture shows an x-y plane with a parabola. The second picture has a cellophane sheet all scrunched up at the origin. In the third picture, the cellophane sheet has been opened out partly. In the third picture, the sheet has been laid flat on top of the x-y-plane and on it there are the points (-i,0) and (i,0) marked.

    In this post, I want to talk about where the complex points are on a line. This is the first reason why this representation of complex points is so very cool.

    A real line

    Every real line has an equation like \(ax + by = c\) for some real \(a\), \(b\), and \(c\). For example, \(2x + 9y = 7\), \(-3x + 5y=8\), \(x – 2y = 3\), \(x=4\) and \(y=5\). (I’ve used whole numbers here but there’s nothing stopping us using fractions or square roots or transcendental numbers like e.)

    But we know that at every point there is an iplane attached which holds all the complex points. Which of these complex points are part of the line? First we need to remind ourselves that a point is on a line when it satisfies the equation of the line. That’s fundamentally what an equation is for: deciding if a point is part of a set or not.

    Let’s consider the line \(x – 2y = 3\)…

    • The point \((3,0)\) is on this line because \(3 – 2\times 0 = 3\).
    • The point \((1,-1)\) is on this line because \(1 – 2\times (-1) = 1+2 = 3\).
    • The point \((9,3)\) is on this line because \(9 – 2\times 3 = 9-6 = 3\).
    • The point \((3+2i,i)\) is on this line because \(3+2i – 2\times i = 3+2i-2i = 3\).
    • The point \((1+4i,-1+2i)\) is on this line because \(1+4i – 2\times(-1+2i) = 1 + 4i +2 – 4i = 3\).
    • The point \((9-10i,3-5i)\) is on this line because \(9 -10i -2\times(3-5i) = 9 – 10i – 6 + 10i = 3\).

    There is a pattern forming here: in order to get my complex point to be on the line, I need to make sure the imaginary parts are the right sizes so that they subtract away when I put them in the formula. The real parts have to create the number 3 I need as well, which means the real parts have to satisfy the original equation.

    Let’s do this with algebra:

    Suppose the point \((p+ri, q+si)\) is on the real line with equation \(ax+by=c\).

    Then

    \[\begin{aligned}
    a(p+ri)+b(q+si) &= c\\
    ap+ari+bq+bsi &= c\\
    (ap+bq)+(ar+bs)i &= c+0i\end{aligned}\]

    Equating real and imaginary parts gives:

    \(ap + bq = c\) and \(ar+bs=0\).

    So \((p,q)\) satisfies \(ax+by=c\)
    and \((r,s)\) satisfies \(ax+by=0\).

    What does this mean? If the point \((p+ri,q+si)\) is on the line with equation \(ax + by = c\), then the point \((p,q)\) has to be on the line. This means that all the complex points on the real line are in the iplanes at the points on the real line. That’s very pleasant — the complex points on this line are in some way attached to the line itself, because they’re in the iplanes attached to this line.

    But it gets better! The complex point \((p+ri,q+si)\) is the point with coordinates \((r,s)\) in the iplane at \((p,q)\). And the point \((r,s)\) has to satisfy the equation \(ax + by = 0\). That means the point \((r,s)\) lies on a line parallel to our original line passing through the centre of the iplane. When you take the iplane at the point \((p,q)\) and lay it flat on the real plane, the complex points in that iplane that are part of the line will lie on top of the existing real line! This is because they are on a line parallel to this line passing through a point on this line.

    So in our representation, the complex points on a real line aren’t just in the iplanes attached to the real line, they look like they are actually on the line! Isn’t that cool?!!

    A screenshot of a GeoGebra activity. There is a coordinate plane with x and y axes. At the top left is a place to enter the coefficients of the equation of a line. This line is drawn in blue on the plane. There are two iplanes represented by a pink square with its attachment point maked in gold. One attachment point is on the line and it shows red points along the line itself. Another attachment point is off the line and no red points in the iplane are shown.

    You can play around with the iplanes attached to real lines in the following GeoGebra applet .

    A line with real slope

    There are a whole lot of lines in the complex plane that aren’t real lines. The first kind I want to deal with is those with real slope. That is, their equations are of the form \(ax + by = c + fi\) for some non-real number \(c+fi\) and real numbers \(a\) and \(b\). This kind of line has no real points at all – it’s not possible to put real numbers into \(ax + by\) and get a non-real result. But where actually are all the non-real points on this line?

    Suppose the point \((p+ri,q+si)\) is on the line with equation \(ax + by = c + fi\).

    Then

    \[\begin{aligned}
    a(p+ri)+b(q+si) &= c+fi\\
    ap+ari+bq+bsi &= c+fi\\
    (ap+bq)+(ar+bs)i &= c+fi\end{aligned}\]

    Equating real and imaginary parts gives:

    \(ap + bq = c\) and \(ar+bs=f\).

    So \((p,q)\) satisfies \(ax+by=c\)
    and \((r,s)\) satisfies \(ax+by=f\).

    So the points on the line with equation \(ax+by=c+fi\) are in the iplanes attached to the real line with equation \(ax+by=c\), and in each in each iplane they form a line parallel to this real line. In fact, if you look at where they are across all the iplanes, they all form the same line parallel to the real line.

    A screenshot of a GeoGebra activity. There is a coordinate plane with x and y axes. At the top left is a place to enter the coefficients of the equation of a line, whose constant term is complex. A dotted line in blue is drawn on the plane. There are two iplanes represented by a pink square with its attachment point maked in gold. One attachment point is on the dotted line and it shows red points along a parallel line offset from the dotted line. Another attachment point is off the line and no red points in the iplane are shown.

    You can investigate these kinds of lines using this GeoGebra applet .

    Of particular interest for our purposes later on are the “vertical” complex lines with equations of the form x=fi. The points on this line are in the iplanes attached to the real line x=0 — that is, they’re in the iplanes attached to the y-axis. Inside each iplane they form a vertical line, which looks like it’s part of the real line x=f.

    A screenshot of a GeoGebra activity. There is a coordinate plane with x and y axes. At the top left is a place to enter the coefficients of the equation of a line, which this time is 1x+0y=0+4i. There are two iplanes represented by a pink square with its attachment point maked in gold. Both attachment points are on the y-axis. In both planes, red points form a vertical line offset from the y-axis passing through x=4.

    A line with unreal slope

    The last kind of line is a line with unreal slope. It would have an equation like \((a+di)x + (b+ei)y = c+fi\). You’d need to make sure that \(a+di\) and \(b+ei\) weren’t related by a real multiple or you could divide by some complex number to produce real coefficients for the x and y — we’ve already covered lines like that. In fact, it must therefore be possible to arrange it so that the y-coefficient is 1. In that case, we might as well write it in “slope-intercept” form, which might be a bit easier to process. So let’s take a look at the line with equation \(y = (a+di)x + (c+fi)\).

    Suppose the point \((p+ri,q+si)\) is on the line with equation \(y = (a+di)x + (c+fi)\) for \(d\neq 0\).

    Then

    \[\begin{aligned}
    q+si &= (a+di)(p+ri)+ c+fi\\
    &= ap + ari + dpi – dr +c+fi\\
    &= (ap-dr+c)+(ar+dp+f)i \end{aligned}\]

    Equating real and imaginary parts gives:

    \(q=ap-dr+c\) and \(s=ar+dp+f\).

    Hmm. That’s not particularly edifying. What does that mean? Why don’t we pick an iplane and see what the points on our line look like in this iplane:

    Consider the points in the iplane at the specific point \((p,q)\):

    \[\begin{aligned}
    q &= ap -dr+c\\
    dr &= ap – q + c\\
    r &= \frac{ap-q+c}{d}\end{aligned}\]

    This means that given \((p,q)\), we can find a unique value of \(r\). Substituting this into the other equation gives us a unique value of \(s\). That is, every iplane has exactly one point of this complex line! You can investigate what these points look like with this GeoGebra applet.

    A screenshot of a GeoGebra activity. There is a coordinate plane with x and y axes. At the top left is a place to enter the coefficients of the equation of a line in slope-intercept form, whose coefficients are complex. There is an iplanes represented by a pink square with its attachment point maked in gold. One red point is marked in the iplane.

    Note that there is in fact exactly one real point every line with complex slope. Using the notation above, it’s the point \(\left(-\frac{f}{d}, -\frac{af}{d}+c\right)\). It’s quite hard to find the point by trial and error in the GeoGebra applet!

    Conclusion

    So there you have it. Using iplanes, the lines with real slope actually look like lines, and the lines with unreal slope are a cloud of points with one in each iplane. In the next post, I’ll talk about parabolas…

  • Where the complex points are

    When you first learn complex numbers, you find out that they give you ways to solve equations that were previously unsolvable. The classic example is the equation equation \(x^2 + 1 = 0\), which if you’re only using real numbers has no solutions, but with complex numbers has the solutions \(x=i\) and \(x=-i\).

    As someone who likes to imagine the physical reality of everything, this has always caused me considerable difficulties. The equation \(x^2 + 1 = 0\) can be thought of as the equation that tells you where the parabola with equation \(y = x^2 + 1\) meets the x-axis.

    Only the parabola with equation \(y = x^2 +1\)doesn’t meet the x-axis. If our complex number solutions are to be believed, then it meets the x-axis in the points \((i,0)\) and \((-i,0)\), but I certainly can’t see those points on my graph. Where are they?

    A set of coordinate axes with a parabola drawn in blue. The parabola has its lowest point on the y-axis at y=1. The equation y=x squared plus 1 is written next to the parabola. Handwritten in red writing at the bottom of the picture is the question "Where are the complex points?"

    Presumably there are a whole host of points with complex coordinates, which are points where various things meet that don’t look like they meet. These points must be somewhere, and they must be some place that is somehow related to the graphs I see in the real plane. But where is this place?

    Well, about a week ago, I finally found the place where the complex points are!

    (At this point I need to make a few things clear: I have made this up entirely on my own. I have not learned it from anyone else, or from reading anyone else’s work. I am not so vain as to believe that no-one has ever thought of this before, but as yet I can’t find search terms in Google that have shown me anything close. If you have heard of it before I’d love to know!)

    Here’s the idea: at every point in the real plane, there are a real-plane’s-worth of complex points attached. The complex points attached to the real point \((a,b)\) are all of the form \((a+ci,b+di)\). That is, the real parts of the two coordinates are \((a,b)\). I imagine them as a plane, with x-axis showing what imaginary part we’ve added to the x-coordinate, and y-axis showing what imaginary part we’ve added to the y-coordinate.The real point itself is in the centre of this plane.

    I’m going to call this plane attached to the point \((a,b)\) the “iplane at \((a,b)\)”. In the picture below, I’ve shown a few points in the iplane at \((3,2)\). I’ve coloured it pink so you can tell it’s not the ordinary real plane, which as you saw above, I usually colour in white.

    A pair of coordinate axes with background in pink. The axes are labelled i sub x and i sub y. Various points are marked on the plane in blue. At the origin, the point is labelled (3,2). At the point marked 3 on the horizontal axis, the point is labelled (3+31,2). At the point marked 4 on the vertical axis, the point is labelled (3, 2+4i). Ohter points on neither axis are labelled similarly, all with real part 3 and 2.

    When I say “attached” I mean it literally — I actually do imagine the iplane being attached to the point in a physical sense, like in the photos here, which are at different levels of spread-out-ness:

    Two images. In the left image, a flat piece of cork has a white piece of paper on top of it, on which is drawn a pair of coordinate axes with the point (3,2) marked. At the point (3,2), there is a piece of red cellophane attached, most of which is scrunched upwards like a flower. In the right hand picture, that piece of cellophane has been opened out to be almost flat, and you can just see a pin in the centre holding it to the point marked (3,2). The label (3,2) can just be seen through the transparent cellophane.

    Of course, every point in the real plane has its own iplane, so the complex plane as a whole looks more like this:

    The same cork from the previous picture with a pair of coordinate axes attached. The paper is almost completely covered with a large number of pieces of cellophane, all scrunched like flowers.

    And that’s where the complex points are. The complex point \((3+i,5-2i)\) is in the iplane attached to the real point \((3,5)\); the complex point \((2i, -6+7i)\) is in the iplane attached to the real point \((0,-6)\); and the complex points \((i,0)\) and \((-i,0)\) are both in the iplane attached to the real point \((0,0)\)

    It’s easier to see them if you spread the appropriate iplane out and flatten it so it sits on top of the real plane. You don’t want to do the whole plane of course; you just want a big enough slice so that you can see where the point is that you want. Also, I imagine the iplane being transparent so you can see the real plane through it — hence the cellophane in all the photos of the physical models.

    In the below picture, a slice of the iplane at \((0,0)\) is being flattened out to show the points \((i,0)\) and \((-i,0)\).

    Four pictures in progression. The first picture shows an x-y plane with a parabola. The second picture has a cellophane sheet all scrunched up at the origin. In the third picture, the cellophane sheet has been opened out partly. In the third picture, the sheet has been laid flat on top of the x-y-plane and on it there are the points (-i,0) and (i,0) marked.

    This GeoGebra applet allows you to enter the coordinates of a complex point and it will show you the iplane it’s in and the real point where the iplane is attached. Give it a go to get a feel for what’s happening here.

    A screenshot of a GeoGebra graph. It shows a coordinate plane in white. At the top left is a space to enter the real and imaginary parts of the two coordinates of a complex point. There is a gold point and a pink square showing the location of the iplane and a red point showing the location of the point itself.

    So why am I so very very excited by this idea? Well, it all has to do with what happens when you figure out the complex points that are part of some familiar graphs. So far, I’ve thought a lot about lines and a bit about parabolas. It will take me a while to write about it all, so I’ll save those topics for the subsequent blog posts.

    The titles of the five posts in the series are:

    1. Where the complex points are (5 August 2016)
    2. Where the complex points are: on a line (5 August 2016)
    3. Where the complex points are: on a parabola (8 August 2016)
    4. Where the complex points are: the graph of a function (10 August 2016)
    5. Where the complex points are: where the idea came from (23 August 2016)

    And then two more posts were done several years later:

    1. Where the complex points are: on a complex line (again) (3 May 2020)
    2. Where the complex points are: on a real circle (31 August 2020)

    UPDATE: There was a later series of blog posts in 2022 where I modified the idea from iplanes to i-arrows, and an even later blog post in 2024 further investigating the line joining two complex points using i-arrows.

    These comments were left on the original blog post:

    Mark Pettyjohn 25 January 2017

    When I saw “the biggest maths idea I’ve ever had” I didn’t think I had a prayer of understanding it. That I do is a testament to your understanding of the problem and solution and your ability to articulate it so clearly.

    This post inspired a lot of thoughts that I tried to condense into something coherent here. I look at your “biggest idea ever” through the lens of the Standards for Mathematical Practice we use in the states k-12. The broader ideas are looking at and contrasting what mathematicians (like you!) do and what school mathematics is often like.

    Thanks for the inspiration!

    My blog post

    Robert Low 16 April 2017

    You’ve invented a way of visualizing R^4 by attaching an R^2 to each point of R^2, so in some sense the tangent bundle of R^2. It’s pretty neat, though, how you’ve made it concrete!

    David Butler 18 April 2017

    I’ve never known what a tangent bundle is before now, so I’m glad if I ever look into it more closely I will have something to work from!

    Martin Flashman 20 July 2018

    An alternative to understanding functions (of either real or complex variable) is to use mapping diagrams.
    For quadratic functions (with either real or complex coefficients) the roots of the quadratic equations are visualized by arrow(s) from point(s) in the source complex plane to the point 0+0i in the target plane.
    See for example my work with GeoGebra https://www.geogebra.org/m/JZYgq4Mx#material/aYeGJpS4  and MD to visualize solving linear and quadratic equations in https://www.geogebra.org/m/CX32kuXG#chapter/274869 .

    David Butler 24 July 2018

    Yes I am aware of this sort of thing and it’s an extremely useful way to think about things. I find it especially useful for derivatives. Thanks for sharing!

  • Complex is not the same as complicated

    The Complex Numbers are unfortunately named. Most people take the word complex to mean “difficult to understand”, so the very name we give this family of numbers sets students up to think it’s going to be a lot of hard work to understand them. This is sad, because they really are very very cool and not quite as difficult as people make them out to be.

    It turns out, though, that the word complex has only recently attained the connotations of confusion.  The word complex according to my dictionary means simply “composed of multiple parts”, which is plainly true of complex numbers: they have precisely two parts – a real part and an imaginary part.

    My dictionary has another meaning for the word complex. The meaning above is the one used when we’re using it as an adjective, but you can also use it as a noun. In that case a complex is an object which is composted of multiple parts. For example, a cinema complex is a building composed of multiple cinemas. In this sense, the phrase “complex number” is much more akin to “house boat” – a number which is a complex, like a boat which is a house. (I recently read this idea in “The Joy of x” by Steven Strogatz.)

    I quite like both of these ideas. When people think of them as complicated numbers, it feels like they are making a value judgement, but with this older meaning of complex, it is value-free. It’s a simple statement of fact about the structure of the numbers themselves. Sometimes you need to reclaim language to have a better perspective on the mathematical meaning.