The complex points on a line using i-arrows


The Cartesian plane is pretty cool. You think up an equation like y=x²+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 x²+1=0 gets the solutions x=i and x=-i. The simultaneous solutions to y=x²+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 arrow (which I call 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).

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…

a(p+si) + b(q+ti) = c
ap + asi + bq + bti = c

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+γi for a, b, c, γ real (a and b not both zero) and  γ 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…


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

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 coefficient. 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+γi 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+γ. 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 (0,13/2+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+γi) + (m+μi)x for real numbers c, γ, m, μ, with μ 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…

q+ti=(c+γi) + (m+μi)(p+si)
q+ti=c+γi + mp + msi +μpi – μs

Real part:
q = c + mp – μs
μs = c+mp – q
s = (c+mp – q)/μ

Imaginary part:
t = γ + ms + μp

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=γ + μp. 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+γi) + (m+μi)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 γi+μi p = i(γ+μp)=0, and the only p that does that is -γ/μ. (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 a tool 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. (I’ve just put a link to the twitter post in because that’s easier.) 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)+(1/2-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).


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.


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.

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.

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