# Complex rectangles 1

There's an exercise that's bugged me for a while from Tristan Needham's Visual Complex Analysis. It's a proof exercise: Prove that for complex numbers $$A, B, C, D$$ lying on the unit circle (so magnitude 1), if $$A + B + C + D = 0$$ then the points form a rectangle.

How can we approach this?

I think the first thing to do is order them. Without loss of generality, assume the points are in counterclockwise order $$A, B, C, D$$ starting from the positive real axis. Next, we can write down the "target." I think to prove they form a rectangle it is enough to show that the edges are at right angles and the diagonals are equal in length. We want to rewrite this in terms of statements about complex numbers.

If we view the sides as vectors for example $$B - A$$, we can express "right angle" by saying the dot product of consecutive edge vectors is zero. How to translate that cleanly into statements of complex arithmetic, though? I think we want to multiply each complex difference with the conjugate of the next one, and say that the result is... well, nonzero complex numbers won't multiply to zero because they're too well-behaved. But we can say the result is pure imaginary and I think that will mean effectively the same thing.

Translating our "goals" or "want to shows" into statements about the complex numbers:

\begin{align} (B - A)(\overline{C - B}) + (\overline{B - A})(C - B) &= 0 \\ (C - B)(\overline{D - C}) + (\overline{C - B})(D - C) &= 0 \\ (D - C)(\overline{A - D}) + (\overline{D - C})(A - D) &= 0 \\ (A - D)(\overline{B - A}) + (\overline{A - D})(B - A) &= 0 \\ |A - C| &= |B - D| \end{align}

Can we show all these statements given only that our points are on the unit circle with $$A + B + C + D = 0$$?

Well, the last one can be rewritten. The left side is $$A\bar{A} - A\bar{C} - \bar{A}C + C\bar{C} = 2 - A\bar{C} - \bar{A}C$$. Does this help us? Probably not yet, no. We can rewrite the right side for the same reasons and it's not terribly useful yet for the same reasons.

What about the four right-angle equations? $$(B - A)(\overline{C - B}) + (\overline{B - A})(C - B)$$ expands to $$B\bar{C} - A\bar{C} - B\bar{B} + A\bar{B} + \bar{B}C - \bar{A}C - \bar{B}B + \bar{A}B$$. The $$B\bar{B}$$ parts are $$1$$, as before. But... the other six products don't look especially friendly.

I have a hunch: What if we do (A + B + C + D)(\overline{A + B + C + D}) = 0\)? And what if we add up the four equations? Or all five? Maybe this gives us something useful.

Let's work through what we might get. The four initial equations should give us the four positive products $$A\bar{B}, B\bar{C}, C\bar{D}, D \bar{A}$$ and their conjugates, for a total of 8 distinct positive products. The total product should have 4 * 4 = 16. Where do we get the remaining 8? Well, there are the 4 self-with-conjugate products which we know are 1, so those are easy. Then there are the 2 products $$A\bar{C}, B\bar{D}$$ and their conjugates.

So yes, it looks like we should get something known out of this. Whether it's useful remains to be seen.

The complete sum is:

\begin{align} % the four differences && B\bar{C} - A\bar{C} - B\bar{B} + A\bar{B} + \bar{B}C - \bar{A}C - \bar{B}B + \bar{A}B \\ &+& C\bar{D} - B\bar{D} - C\bar{C} + B\bar{C} + \bar{C}D - \bar{B}D - \bar{C}C + \bar{B}C \\ &+& D\bar{A} - C\bar{A} - D\bar{D} + C\bar{D} + \bar{D}A - \bar{C}A - \bar{D}D + \bar{C}D \\ &+& A\bar{B} - D\bar{B} - A\bar{A} + D\bar{A} + \bar{A}B - \bar{D}B - \bar{A}A + \bar{D}A \\ % the two square? magnitudes % TODO may need to add the conjugates to these two: &+& A\bar{A} - A\bar{C} - \bar{A}C + C\bar{C} \\ &+& B\bar{B} - B\bar{D} - D\bar{B} + D\bar{D} \\ && \text{(Gather the positive terms...)} \\ % now gather the positive terms at the start &=& A\bar{B} + B\bar{C} + \bar{A}B + \bar{B}C - A\bar{C} - 2B\bar{B} - \bar{A}C \\ &+& B\bar{C} + C\bar{D} + \bar{B}C + \bar{C}D - B\bar{D} - 2C\bar{C} - \bar{B}D \\ &+& C\bar{D} + D\bar{A} + \bar{C}D + \bar{D}A - C\bar{A} - 2D\bar{D} - \bar{C}A \\ &+& A\bar{B} + D\bar{A} + \bar{A}B + \bar{D}A - \bar{D}B - 2A\bar{A} - D\bar{B} \\ &+& A\bar{A} + C\bar{C} - A\bar{C} - \bar{A}C \\ &+& B\bar{B} + D\bar{D} - B\bar{D} - D\bar{B} \\ && \text{(Looks a little redundant...)} \\ && \text{(Simplify...)} \\ &=&-A\bar{A} - B\bar{B} - C\bar{C} - D\bar{D} \\ &+& 2 (A\bar{B} + \bar{A}B) - 3 (A\bar{C} + \bar{A}C) + 2 (A\bar{D} + \bar{A}D) \\ &+& 2 (B\bar{C} + \bar{B}C) - 3 (B\bar{D} + \bar{B}D) \\ &+& 2 (C\bar{D} + \bar{C}D) \\ &+& \\ \end{align}

Hmm, so it looks like adding the two magnitude equations might have screwed things up. What about the self-conjugate expression of $$A + B + C + D = 0$$? Write it out:

\begin{align} && (A + B + C + D)(\overline{A + B + C + D}) = 0 \\ &=& A\bar{A} + B\bar{B} + C\bar{C} + D\bar{D} \\ &+& (A\bar{B} + \bar{A}B) + (A\bar{C} + \bar{A}C) + (A\bar{D} + \bar{D}A) \\ &+& (B\bar{C} + \bar{B}C) + (B\bar{D} + \bar{B}D) \\ &+& (C\bar{D} + \bar{C}D) \\ &=& 0 \end{align}

And, hmm, adding twice this into the sum-of-four-equations would give:

\begin{align} &&-2A\bar{A} - 2B\bar{B} - 2C\bar{C} - 2D\bar{D} \\ &+& 2 (A\bar{B} + \bar{A}B) - 2 (A\bar{C} + \bar{A}C) + 2 (A\bar{D} + \bar{A}D) \\ &+& 2 (B\bar{C} + \bar{B}C) - 2 (B\bar{D} + \bar{B}D) \\ &+& 2 (C\bar{D} + \bar{C}D) \\ &=& 4 (A\bar{B} + \bar{A}B) + 4 (A\bar{D} + \bar{A}D) \\ &+& 4 (B\bar{C} + \bar{B}C) \\ &+& 4 (C\bar{D} + \bar{C}D) \\ &+& \\ \end{align}

I don't know if this actually helps me, unfortunately... oh, wait, this implies that:

\begin{align} &&-2A\bar{A} - 2B\bar{B} - 2C\bar{C} - 2D\bar{D} \\ &-& 2 (A\bar{B} + \bar{A}B) - 2 (A\bar{C} + \bar{A}C) - 2 (A\bar{D} + \bar{A}D) \\ &-& 2 (B\bar{C} + \bar{B}C) - 2 (B\bar{D} + \bar{B}D) \\ &-& 2 (C\bar{D} + \bar{C}D) \\ &=& 0 \end{align}

Which, wait a second, is identical to the $$A + B + C + D = 0$$ self-conjugate expression. So this is no progress at all. Argh!

Maybe it is better to frame this in terms of angles. We can write these complex numbers as $$e^{i\alpha}, e^{i\beta}, e^{i\gamma}, e^{i\delta}$$ where $$\alpha < \beta < \gamma < \delta$$ are real numbers. We know that:

$e^{i\alpha} + e^{i\beta} + e^{i\gamma} + e^{i\delta} = 0$

Now can we do something? No, this doesn't look like a useful angle either.

What's important here? Maybe it's useful to think of this geometrically using the vector addition image. Think of each number/point as a vector and arrange them end to end. The sum being zero makes this sequence of arrows a closed loop. It's definitely important here that the points have identical magnitude/lie on the unit circle. If they weren't, then the sum being zero shouldn't force them to form a rectangle.

It seems like if the four points form a rectangle, we should have $$A = -C$$ and $$B = -D$$. But how would we prove it?

Can we use this vector-loop picture to do anything? We know from this logic that the four "vectors" should form a closed trapezoid. In fact I think it should be a diamond, because we know every side is equal in length. We can split it down the middle to form two isoceles triangles.

This feels like it might go somewhere. More to come?