Geometric Interpretation of Complex Multiplication

You probably know that multiplication of two complex numbers results in a third complex number whose angle with the horizontal can be found by adding the two input angles, and whose magnitude can be found by multiplying the two input magnitudes.

It works out really nicely but... Why? How does this geometric interpretation come about by using the simple defintion that $i=\sqrt{-1}$?

Let $z$ be a complex number of the form $a + bi$ where $a, b \in \mathbb{R}$, and $i$ is the imaginary unit.

The magnitude of $z$ is $\sqrt{a^2 + b^2}$. We will call this $r$ for radius.

$r = ||z|| = \sqrt{a^2 + b^2}$

We can now divide $z$ by $r$ resulting in a unit vector (or unit complex number I guess).

$z = r\frac{z}{r} = r(unit\_vector)$

Now let's think of a way to rewrite "unit_vector" because it's not doing us much good right now... Well, $\cos{\theta} + \sin{\theta}$ is guaranteed to lie on the unit circle, so this is promising! Of course we're dealing with the complex plane, so we'll multiply $\sin{\theta}$ by $i$ so that $\sin{\theta}$ will represent the imaginary component which is perpendicular to the real component.

Rewriting our new and improved equation, we get:

$z = r\frac{z}{r} = r(\cos{\theta} + i\sin{\theta})$

Ok great! We've written it in a form where the magnitude and direction are separated so it's easy to keep track of each one. Maybe this will let us discover what happens to magnitude and direction during multiplication!

Let's go ahead and multiply two complex numbers written in this form and see what happens:

$r_1(\cos{\theta} + i\sin{\theta}) \cdot r_2(\cos{\beta}+ i\sin{\beta})$
$= r_1 r_2(\cos{\theta} + i\sin{\theta})(\cos{\beta} + i\sin{\beta})$
$= r_1 r_2(\cos{\theta}\cos{\beta} + i\cos{\theta}\sin{\beta} + i\sin{\theta}\cos{\beta} - \sin{\theta}\sin{\beta})$
$= r_1 r_2(\cos{\theta}\cos{\beta} - \sin{\theta}\sin{\beta} + i(\sin{\theta}\cos{\beta} + \cos{\theta}\sin{\beta}))$

It's pretty obvious that the magnitudes are being multiplied which explains half of the puzzle! But it looks like the angles have become a little... messy.

Let's take an intermission from this problem, and give linear algebra a quick visit. In fact let's just derive the angle addition formula, just a totally random unrelated activity to get our minds off complex numbers.

Recall the 2D rotation matrix:

$\left[\begin{matrix}\cos{\theta} & -sin{\theta}\\ sin{\theta} & cos{\theta}\end{matrix}\right]$

So then if we first rotate by angle $\beta$ followed by a rotation of angle $\theta$, we can write this either as a rotation by angle $\theta + \beta$ or as the multiplication of two matrices representing the rotation of angle $\beta$ and $\theta$ respectively:

$\left[\begin{matrix}\cos(\theta + \beta) & -sin(\theta + \beta)\\ sin(\theta + \beta) & cos(\theta + \beta)\end{matrix}\right]$
$= \left[\begin{matrix}\cos{\theta} & -sin{\theta}\\ sin{\theta} & cos{\theta}\end{matrix}\right]
\left[\begin{matrix}\cos{\beta} & -sin{\beta}\\ sin{\beta} & cos{\beta}\end{matrix}\right]$
$= \left[\begin{matrix}\cos{\theta}\cos{\beta} - \sin{\theta}\sin{\beta} &
-cos{\theta}\sin{\beta} - \sin{\theta}\cos{\beta}\\
sin{\theta}\cos{\beta} + \cos{\theta}\sin{\beta} &
-\sin{\theta}\sin{\beta} + \cos{\theta}\cos{\beta}\end{matrix}\right]$

Therefore,

$\cos(\theta + \beta) = \cos{\theta}\cos{\beta} - \sin{\theta}\sin{\beta}$
$sin(\theta + \beta) = sin{\theta}\cos{\beta} + \cos{\theta}\sin{\beta}$

We just derived the angle addition formulas!! And it looks like we can use it to make some substitutions where we left off on our complex number multiplication!

...
$= r_1 r_2(\cos{\theta}\cos{\beta} - \sin{\theta}\sin{\beta} + i(\sin{\theta}\cos{\beta} + \cos{\theta}\sin{\beta}))$
$= r_1 r_2(\cos(\theta + \beta) + i\sin(\theta + \beta))$

And that's it! We can now see clearly that multiplying two complex numbers will result in a new complex number whose magnitude is multiplying the input magnitudes, and whose angle is the addition of the input angles.

This can also be shown using Euler's Formula:

$e^{ix} = \cos{x}+i\sin{x}$

Just multiply by some radius so you're no longer restricted to unit vectors, multiply two of them together and watch the magic.

$r_1 e^{i\theta} \cdot r_2 e^{i\beta} = r_1 r_2 e^{i(\theta + \beta)}
= r_1 r_2(\cos(\theta + \beta) + i\sin(\theta + \beta))$

Euler's Formula also gives rise to the ubiquitous Euler's Identity: $e^{i\pi}=-1$.
There is much more to be said about Euler's Formula and its derivation, but this post is long enough.