Properties of Principal Argument of Complex Numbers

Angles Add in Products and Subtract in Quotients

Multiplication combines directions by adding angles, while division compares directions by subtracting them. Let $z_1$ and $z_2$ be nonzero complex numbers in polar form:

Where $\theta_1$ is one argument of $z_1$ and $\theta_2$ is one argument of $z_2$.

Argument of Product

The argument of the product of two complex numbers ($z_1 \times z_2$) is the sum of the arguments of the individual complex numbers.

The equality is understood modulo one full turn:

This means if $\theta_1$ is an argument of $z_1$ and $\theta_2$ is an argument of $z_2$,

then $\theta_1 + \theta_2$ is one of the arguments of $z_1 z_2$.

To find the Principal Argument $\text{Arg}(z_1 z_2)$:

1. Calculate $\text{Arg}(z_1) + \text{Arg}(z_2)$.
2. If the result is already within $(-180^\circ, 180^\circ]$ or $(-\pi, \pi]$, it is the Principal Argument.
3. Otherwise, add or subtract full turns until it lies in that interval.

Argument of Quotient

The argument of the quotient of two complex numbers ($\frac{z_1}{z_2}$, with $z_2 \neq 0$) is the difference between the argument of the numerator complex number ($z_1$) and the argument of the denominator complex number ($z_2$).

Again, the equality is understood modulo one full turn:

This means if $\theta_1$ is an argument of $z_1$ and $\theta_2$ is an argument of $z_2$,

then $\theta_1 - \theta_2$ is one of the arguments of $\frac{z_1}{z_2}$.

To find the Principal Argument $\text{Arg}\left(\frac{z_1}{z_2}\right)$:

1. Calculate $\text{Arg}(z_1) - \text{Arg}(z_2)$.
2. If the result is already within $(-180^\circ, 180^\circ]$ or $(-\pi, \pi]$, it is the Principal Argument.
3. Otherwise, bring it into that interval with the same full-turn step as for the product.

Using Argument Properties

Given two complex numbers:

Find the Principal Argument of $z_1 \times z_2$ and $\frac{z_1}{z_2}$.

Solution:

We know the Principal Arguments are:

1. Argument of Product ($z_1 \times z_2$):

   Sum of Principal Arguments:

   $$\begin{aligned} \text{Arg}(z_1) + \text{Arg}(z_2) &= 45^\circ + 95^\circ \\ &= 140^\circ \end{aligned}$$

   Since $140^\circ$ is already within $(-180^\circ, 180^\circ]$:

   $$\text{Arg}(z_1 \times z_2) = 140^\circ$$

   The set of all arguments is $\{140^\circ + k \cdot 360^\circ : k \in \mathbb{Z}\}$

2. Argument of Quotient ($\frac{z_1}{z_2}$):

   Difference of Principal Arguments:

   $$\begin{aligned} \text{Arg}(z_1) - \text{Arg}(z_2) &= 45^\circ - 95^\circ \\ &= -50^\circ \end{aligned}$$

   Since $-50^\circ$ already lies in $(-180^\circ, 180^\circ]$:

   $$\text{Arg}\left(\frac{z_1}{z_2}\right) = -50^\circ$$

   The set of all arguments is $\{-50^\circ + k \cdot 360^\circ : k \in \mathbb{Z}\}$.

Exercise

Given $z_a = 4(\cos 120^\circ + i\sin 120^\circ)$ and $z_b = 2(\cos 50^\circ + i\sin 50^\circ)$. Both numbers are already in polar form, so the two arguments combine directly, and the principal range tells you whether the result needs one more full turn. Find:

1. $\text{Arg}(z_a \times z_b)$
2. $\text{Arg}\left(\frac{z_a}{z_b}\right)$

Worked Solution

Given $\text{Arg}(z_a) = 120^\circ$ and $\text{Arg}(z_b) = 50^\circ$.

The positive moduli do not change the direction of either complex number, so only the two arguments are needed.

1. Argument of Product:

   $$\begin{aligned} \text{Arg}(z_a) + \text{Arg}(z_b) &= 120^\circ + 50^\circ \\ &= 170^\circ \end{aligned}$$

   Since $170^\circ \in (-180^\circ, 180^\circ]$, $\text{Arg}(z_a z_b) = 170^\circ$.

2. Argument of Quotient:

   $$\begin{aligned} \text{Arg}(z_a) - \text{Arg}(z_b) &= 120^\circ - 50^\circ \\ &= 70^\circ \end{aligned}$$

   Since $70^\circ \in (-180^\circ, 180^\circ]$, $\text{Arg}\left(\frac{z_a}{z_b}\right) = 70^\circ$.