# 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.

$z_1 = \polar {r_1, \theta_1}$
$z_2 = \polar {r_2, \theta_2}$
$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$

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:

 $\ds \map \cos {\theta_1 - \theta_2}$ $=$ $\ds \map \cos {\theta_1 - \theta_2 - 2 \pi}$ Cosine of Angle plus Full Angle $\ds \map \sin {\theta_1 - \theta_2}$ $=$ $\ds \map \sin {\theta_1 - \theta_2 - 2 \pi}$ Sine of Angle plus Full Angle

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:

 $\ds \map \cos {\theta_1 - \theta_2}$ $=$ $\ds \map \cos {\theta_1 - \theta_2 + 2 \pi}$ Cosine of Angle plus Full Angle $\ds \map \sin {\theta_1 - \theta_2}$ $=$ $\ds \map \sin {\theta_1 - \theta_2 + 2 \pi}$ Sine of Angle plus Full Angle

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 $\dfrac {z_1} {z_2}$.

$\blacksquare$