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 nth power is equal to 1: they are equally spaced around the circumference of the unit circle. For example, here are the twelve 12th 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 nth 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.


About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in geometry, pictures and tagged , , , , , . Bookmark the permalink.