## Complex multiplication: proof

In my previous post, I claimed that when multiplying two complex numbers, their lengths multiply and their angles add, like this: In particular, this means that there are always $n$ different complex numbers whose $n$th power is equal to $1$: they are equally spaced around the circumference of the unit circle. For example, here are the twelve $12$th roots of $1$: Did you figure out why? Here’s one way to understand it. If a complex number $z$ is at distance $r$ away from the origin, and makes an angle $\theta$ with the positive $x$-axis, then $z^n$ will be at distance $r^n$ from the origin (since lengths multiply), and will make an angle $n\theta$ with the positive $x$-axis (since angles add). First of all, for all the complex numbers on the unit circle, $r = r^n = 1$. So if we start with a complex number on the unit circle, all its powers will always stay on the unit circle. Now, if we have $n$ equally spaced points around the unit circle, the angles they make with the positive $x$-axis are $1/n$ of a full turn, $2/n$ of a full turn, $3/n$, and so on. So if we take their $n$th powers, the results have angles of $n \cdot (1/n) = 1$ full turn, $n \cdot (2/n) = 2$ full turns, and so on… but any whole number of turns will land exactly back on the positive $x$-axis.

So, let’s prove that this is how complex multiplication works! Let’s write $\langle r, \theta \rangle$ to represent the complex number at radius $r$ and angle $\theta$. From the picture, we can see that $\langle r, \theta \rangle$ is at one corner of a right triangle with hypotenuse of length $r$ and one angle $\theta$. Remembering some basic trigonometry, we can say that the horizontal leg of the triangle (the blue edge) has length $r \cos \theta$, and the vertical leg (the red edge) has length $r \sin \theta$. Therefore, $\langle r, \theta \rangle = r \cos \theta + i r \sin \theta = r (\cos \theta + i \sin \theta)$

So now let’s see what happens when we multiply. $\begin{array}{rcl} \langle r_1, \theta_1 \rangle \cdot \langle r_2, \theta_2 \rangle &=& r_1 (\cos \theta_1 + i \sin \theta_1) \cdot r_2 (\cos \theta_2 + i \sin \theta_2) \\ &=& r_1 r_2 (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2) \\ &=& r_1 r_2 (\cos (\theta_1 + \theta_2) + i \sin (\theta_1 + \theta_2)) \\ &=& \langle r_1 r_2 , \theta_1 + \theta_2 \rangle \end{array}$

Mostly this is just substituting and then distributing out the multiplication; but in the second-to-last step we have to remember some trig identities, specifically the angle sum identities. So I guess this proof seems a bit magical if you don’t already believe those; but here’s a nice proof of the angle sum identities. 