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:
 * $\arg \left({z_1 z_2}\right) = \arg \left({z_1}\right) + \arg \left({z_2}\right) + 2 k \pi$

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

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 definition of complex multiplication, factoring $\left|{z_1}\right|\left|{z_2}\right|$ from all terms, we have:
 * $z_1 z_2 = \left|{z_1}\right| \left|{z_2}\right| \left({\left({\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2}\right) + i \left({\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2}\right)}\right)$

Using Sine of Sum and Cosine of Sum, we have:
 * $z_1 z_2 = \left|{z_1}\right| \left|{z_2}\right| \left({\cos \left({\theta_1 + \theta_2}\right) + i \sin \left({\theta_1 + \theta_2}\right)}\right)$

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


 * $(1): \quad \dfrac{\left|{z_1}\right| \left|{z_2}\right| \cos \left({\theta_1 + \theta_2}\right)} {\left|{z_1}\right| \left|{z_2}\right|} = \cos \left({\arg \left({z_1 z_2}\right)}\right)$


 * $(2): \quad \dfrac{\left|{z_1}\right| \left|{z_2}\right| \sin \left({\theta_1 + \theta_2}\right)} {\left|{z_1}\right| \left|{z_2}\right|} = \sin \left({\arg \left({z_1 z_2}\right)}\right)$

which in turn means that:
 * $\cos \left({\theta_1 + \theta_2}\right) = \cos \left({\arg \left({z_1 z_2}\right)}\right)$
 * $\sin \left({\theta_1 + \theta_2}\right) = \sin \left({\arg \left({z_1 z_2}\right)}\right)$

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:
 * $\arg \left({z_1 z_2}\right) = \theta_1 + \theta_2 = \arg \left({z_1}\right) + \arg \left({z_2}\right) + 2 k \pi$

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