Argument of Product equals Sum of Arguments

Theorem
Let $z_1, z_2 \in \C$ be complex numbers.

Let $\arg$ be the argument operator.

Then:
 * $\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$

where $k$ can be $0$, $1$ or $-1$.

Proof
Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.

Then the polar forms of $z_1, z_2$ are:

By the definition of complex multiplication, factoring $\cmod {z_1} \cmod {z_2}$ from all terms, we have:
 * $z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }$

Using Sine of Sum and Cosine of Sum, we have:
 * $z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$

The theorem follows from the definition of $\map \arg z$, which says that $\map \arg {z_1 z_2}$ satisfies the equations:

which in turn means that:

There are $3$ possibilities for the size of $\theta_1 + \theta_2$:


 * $(1): \quad \theta_1 + \theta_2 > \pi$

Then:
 * $-\pi < \theta_1 + \theta_2 - 2 \pi \le \pi$

and we have:

and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $z_1 z_2$ within its principal range.


 * $(2): \quad \theta_1 + \theta_2 \le -\pi$

Then:
 * $-\pi < \theta_1 + \theta_2 + 2 \pi \le \pi$

and we have:

and so $\theta_1 + \theta_2 + 2 \pi$ is within the principal range of $z_1 z_2$.


 * $(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$

Then $\theta_1 + \theta_2$ is already within the principal range of $z_1 z_2$.

Therefore:
 * $\map \arg {z_1 z_2} = \theta_1 + \theta_2 = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$

where $k$ can be $0$, $1$ or $-1$.