# Complex rectangles 2

Suppose we have four complex numbers $$A, B, C, D$$ on the unit circle such that $$A + B + C + D = 0$$. We want to show that these points must form a rectangle. How can we do this?

An intuition I had last time was that we might want to draw the vector-loop picture of these points. Let's do that now.

Figure 1 shows the situation. Here, $$P = 0, Q = A, R = A + B$$, and $$S = A+B+C$$. We draw each complex number as a vector joining this four points, and we also draw the diameter $$A + B$$ connecting $$P$$ with $$R$$.

The key idea here is to relate these as two isoceles triangles, which forces the angles to be equal. Because $$A, B, C$$, and $$D$$ lie on the unit circle, they are all equal in magnitude, so that the outer sides are all equal. But the diameter $$A + B$$ is common, so the triangles $$PQR$$ and $$PSR$$ are side-side-side congruent. This means the corresponding angles must be equal. The angle of $$A$$ with $$B$$ and the angle of $$C$$ with $$D$$ are equal, to start with. But because these are isoceles triangles, the base angles are equal to each other, so the angles of $$B$$ with $$C$$ and the angle of $$A$$ with $$D$$ are equal. This is true of the end-to-end angles, but also the base-to-base angles because the base-to-base angles are the corresponding complementary angles. So, the angles work out equal.

In fact, we can go further: The angle between $$A$$ and $$B$$ is complementary with that between $$B$$ and $$C$$. We know this because the two base angles of each isocoles triangle taken together must be complementary to the third. Here, the third angles are $$\angle PQR$$ and $$\angle PSR$$, also known as the angles between $$A$$ and $$B$$ and between $$C$$ and $$D$$. But the two base angles together are equal to the angle between$$B$$ and $$C$$, or equivalently between $$A$$ and $$D$$. So, the angle between $$A$$ and $$B$$ is complementary with that between $$C$$ and $$D$$. For the same reason the angle between $$C$$ and $$D$$ is complementary with the angle between $$D$$ and $$A$$. So, the angles work out as complementary.

This shows almost exactly what we want. Clearly, the angles being complementary means we should have $$A = -C$$ and $$B = -D$$. That in turn implies that $$A + B = -C - D$$ hence $$|A + C| = |C + D|$$ so that the four points $$A, B, C, D$$ form a rectangle.

But this connection between the geometry and the complex numbers is a little tenuous for my taste. I would like to spell it out in more detail. This seems tricky. Can we relate the geometry back to statements about complex numbers?

Actually, I think we can. I think we can restate the complementary angles claim as saying that the products of certain complex numbers must be -1, the complex number with angle $$\pi$$. That's because complex numbers adds their angles. The question is what complex number expresses the right angles. I think the answer is $$(B/A)(C/B)$$. Inverting a complex number negates the angle, and we're interested in the angles between complex numbers.

Let's make this a bit more formal. Say that we write $$A, B, C, D$$ as $$e^{i \alpha}, e^{i \beta}, e^{i \gamma}, e^{i \delta}$$ where $$0 \leq \alpha < \beta < \gamma < \delta < 2\pi$$. I think we can state the complementary angles  Saying that the angle between $$A$$ and $$B$$ are complementary with the angle between $$B$$ and $$C$$ is equivalent to saying that $$(\beta - \alpha) + (\gamma - \beta) = \gamma - \alpha = \pi$$. But $$B/A = e^{i (\beta - \alpha)}$$ and similarly for the other configurations. So as statements about complex numbers this is equivalent to saying $$(B/A)(C/B) = C/A = -1$$, or $$A = -C$$.

But we've already proven this. The geometric approach we've just taken shows exactly that $$(\beta - \alpha) + (\gamma - \beta) = \pi$$ as desired.

This feels a little disorienting. It feels like I'm still missing a step. I think I'm missing the step where we explicitly relate the angles in the diagram to the complex numbers $$A, B, C$$, and $$D$$.

I think we need another diagram.

This pretty clearly shows that the relationship just described should hold.

Can we make this same kind of argument algrebraically? Tune in next time.