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.

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About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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