Argument of Quotient equals Difference of Arguments

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

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

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.
 * $z_1 = \left\langle{r_1, \theta_1}\right\rangle$
 * $z_2 = \left\langle{r_2, \theta_2}\right\rangle$

From Division of Complex Numbers in Polar Form:
 * $\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \left({\cos \left({\theta_1 - \theta_2}\right) + i \sin \left({\theta_1 - \theta_2}\right)}\right)$

By the definition of argument:
 * $\arg \left({z_1}\right) = \theta_1$
 * $\arg \left({z_2}\right) = \theta_2$
 * $\arg \left({\dfrac {z_1} {z_2} }\right) = \theta_1 - \theta_2$

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 $\dfrac {z_1} {z_2}$.