Argument of Quotient equals Difference of Arguments

Theorem
Let $z_1$ and $z_2$ be complex numbers.

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

where:
 * $\arg$ denotes the argument of a complex number
 * $k$ can be $0$, $1$ or $-1$.

Proof
Let $z_1$ and $z_2$ be expressed in polar form.
 * $z_1 = \polar {r_1, \theta_1}$
 * $z_2 = \polar {r_2, \theta_2}$

From Division of Complex Numbers in Polar Form:
 * $\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$

By the definition of argument:
 * $\map \arg {z_1} = \theta_1$
 * $\map \arg {z_2} = \theta_2$
 * $\map \arg {\dfrac {z_1} {z_2} } = \theta_1 - \theta_2$

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 $\dfrac {z_1} {z_2}$ 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 $\dfrac {z_1} {z_2}$.


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

Then $\theta_1 - \theta_2$ is already within the principal range of $\dfrac {z_1} {z_2}$.