# The line joining two complex points using i-arrows

## 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 an arrow 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). ## 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 a, so that the complex point (p,q)+(s,t)i becomes P+ai. 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 care 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+ai and Q+bi. Then all the complex points on the line joining them are given by P+ai + (m+μi)[(Q+bi) – (P+ai)] for some complex number m+μi. Doing some algebra on this point…

P+ai + (m+μi)[(Q+bi) – (P+ai)]
= P+ai + (m+μi)[Q+bi – P-ai]
= P+ai + (m+μi)[Q- P + bi –ai]
=  P+ai + (m+μi)[(Q- P) + (b –a)i]
= P+ai + m(Q- P) + m(b –a)i +μ(Q- P)i – μ(b –a)
= P + m(Q-P) – μ(b –a) + [a + μ(Q- P) + m(b – a)]i

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

Notice how the new base and arrow are the old base and arrow for P+ai adjusted by some multiples of the vectors Q-P and ba. 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+bi. (That is, the point P+ai in the formula above is actually the real point R, so that P=R and a=0.) Then the other points on the line become
R + m(Q-P) – μb + [μ(Q- R) + mb]i which has i-arrow with
base R + m(Q-R) – μb,
arrow μ(Q- R) + mb,
and arrowhead P + m(Q-R) – μb + μ(Q- R) + mb.

If μ=0, then this becomes
R + m(Q-R) + mbi, which has i-arrow with
base P + m(Q-R),
arrow mb,
and arrowhead R + m(Q-R) + mb.

The point R+m(Q-R) is a point somewhere on the line joining R and Q.
The vector mb is parallel to b.
The point R+m(Q-R)+mb = R+ m(Q+b – R) is a point somewhere on the line joining R and Q+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 b until you reach the line joining R and the arrowhead Q+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+bi 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+bi, all the complex points on the line joining them were given by
R + m(Q-R) – μb + [μ(Q- R) + mb]i which has i-arrow with
base R + m(Q-R) – μb,
arrow μ(Q- R) + mb,
and arrowhead R + m(Q-R) – μb + μ(Q- R) + mb.

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

How ridiculously simple is THAT?!

If I take a real point R and a complex point Q+bi, 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+b and is the vector b.
The next arrow goes from Q+b to R+2and is the vector (R+2b) – (Q+b) = (R – Q) + b.
The next arrow goes from R+2b to R+2((R – Q) + b) = R + 2(R – Q) + 2b, and is the vector (R+2(R – Q) + 2b) – (R+2b) = 2(R – Q).
The next arrow goes from R + 2(R – Q) + 2b to  R + 2(2(R – Q)) =  R + 4(R-Q), and is the vector (R + 4(R-Q)) – (R + 2(R-Q) + 2b) = 2(R – Q) – 2b.
The next arrow goes from R + 4(R-Q) to R + 2(2(R – Q) – 2b) = R + 4(R-Q) – 4b, and is the vector (R + 4(R-Q) – 4b) – (R + 4(R-Q)) = 4b.

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+bi, I can construct the complex point R+2b+2(R-Q)i. Even better, I can follow parallel i-arrows inwards to get the complex point R+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+bi:

R + m(Q-R) – μb + [μ(Q- R) + mb]i which has i-arrow with
base R + m(Q-R) – μb,
arrow μ(Q- R) + mb,
and arrowhead R + m(Q-R) – μb + μ(Q- R) + mb.

If m=0 and μ=-1 (corresponding to the number m+μi being -i), then the point is
R +b – (Q- R)i = R+b+(R-Q)i which has i-arrow with base
base R+b,
arrow R-Q,

This is a remarkably simple construction to produce that spiderweb from real point R and complex point Q+bi.

1. Copy vector b and place two of them one after the other starting at R. Draw the i-arrow from Q+b to this new point R+2b. This is a second i-arrow on the line.
2. Copy vector R-Q and place it at R+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+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+bi be an unreal point. As calculated before, all the complex points on the line joining them are given by
R + m(Q-R) – μb + [μ(Q- R) + mb]i which has i-arrow with
base R + m(Q-R) – μb,
arrow μ(Q- R) + mb,
and arrowhead R + m(Q-R) – μb + μ(Q- R) + mb.

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+ρb for some number ρ between 0 and 1.
That is, R + m(Q-R) – μ= Q+ρb.
I can make that happen by setting m=1 and μ=-ρ.
This makes the arrow -ρ(Q- R) + 1bb + ρ(R-Q),
and the arrowhead Q+ρ+  b + ρ(R-Q) = Q+b + ρ(R-Q+b).

The vector R-Q+b is the vector for the i-arrow with base Q+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 ρ along the first i-arrow, then the arrowhead is proportion ρ along the second. The arrowhead itself being b + ρ(R-Q) gives me a geometrical construction for finding the arrowhead. The vector 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 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+bi and the real point R. Go another copy of the vector b from Q+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+bi.

1. Start at Q+b and draw the i-arrow that is the vector b+(R-Q).
2. From the arrowhead of this i-arrow, go –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) – 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 a complex point Q+bi, 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+bi.
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+bi and P+ai are arranged so that the arrowhead of Q+bi’s i-arrow is the base of P+ai’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 -2b away from the arrowhead of P+ai’s i-arrow. Just to be sure, let me do the calculations…

The points on the line joining P+ai and Q+bi are of the form:
P + m(Q-P) – μ(b –a) + [a + μ(Q- P) + m(b – a)]i.
This point’s i-arrow has base at P +m(Q-P) – μ(b –a),
arrow a + μ(Q- P) + m(b – a),
and arrowhead at  P + m(Q-P) – μ(b –a) + a + μ(Q- P) + m(b – a).

If the base of P+ai is the arrowhead of Q+bi, then that means P = Q+and b=P-Q. So the point becomes:
P + m(-b) – μ(b –a) + [a + μ(-b) + m(b – a)]i = P – mb – μ(b –a) + [a – μb + m(b – a)]i
whose i-arrow has base at P – mb– μ(b –a),
arrow a – μb + m(b – a),
and arrowhead at  P – mb – μ(b –a) + a – μb + m(b – a).

If this point is real, then
a – μb + m(b – a) = 0
a – μb + mb – m0
(1-m)a +(1-μ)0

If and 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 μ= 1.
So the base (which is now the real point’s location) is P-b – (– a) = P+a+2b, 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+ai and Q+bi are of the form:
P + m(Q-P) – μ(b –a) + [a + μ(Q- P) + m(b a)]i.
This point’s i-arrow has base at P +m(Q-P) – μ(b – a),
arrow a + μ(Q- P) + m(b – a),
and arrowhead at  P + m(Q-P) – μ(b – a) + a + μ(Q- P) + m(b a).

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

This is a point on the line joining P+a and Q+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+a to Q+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  a + m(b a), which is the vector  a, plus a multiple of (b a). If I draw the vectors and b at the base point, then the vector (b a) is the vector from the end of a to the end of 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 a + m(b a) backwards, which means following the vector –a – m(b – a) = –a + m(-b -(-a)) . But this vector can be found by placing the vectors –a and –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+ai and Q+bi, 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 a and b and join their endpoints. Call the point where this meets line the arrowhead line C.
3. At S, draw the vectors –a and –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+ai and Q+bi, 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 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.

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