Argument of Product equals Sum of Arguments

Theorem
Let $z_1, z_2 \in \C$ be complex numbers and $\arg$ the argument operator.

Then:
 * $\arg \left({z_1z_2}\right) = \arg \left({z_1}\right) + \arg \left({z_2}\right)$

This means that when one multiplies two complex numbers, one must add their respective "angles", measured from the real axis.

Proof
Let $\theta_1 = \arg \left({z_1}\right),\theta_2 = \arg \left({z_2}\right)$.

Then the polar forms of $z_1,z_2$ are:
 * $z_1 = \left|z_1\right| \left(\cos {\theta_1} + i\sin {\theta_1}\right)$
 * $z_2 = \left|z_2\right| \left(\cos {\theta_2} + i\sin {\theta_2}\right)$

By the multiplication definition, factoring $\left|z_1\right|\left|z_2\right|$ from all terms, we have:
 * $z_1z_2 =

\left|z_1\right|\left|z_2\right| \left[ \left(\cos {\theta_1}\cos {\theta_2}-\sin {\theta_1}\sin {\theta_2}\right) + i\left(\cos {\theta_1}\sin {\theta_2}+\sin {\theta_1}\cos {\theta_2}\right) \right]$

using the formulas for addition of sines and cosines, we have:
 * $z_1z_2 =

\left|z_1\right|\left|z_2\right| \left[ \cos\left({\theta_1+\theta_2}\right) + i\sin\left({\theta_1+\theta_2}\right) \right]$

The theorem follows from the definition of $\arg$.