<p>A direct consequence of the above development would be to understand division of complex numbers involves subtraction of arguments.</p>
<p>First of all,</p>
<p>\[
\begin{align*}
\frac{1}{z_2} 
& = \frac{1}{r_2(\cos \theta_2 + i \sin \theta_2)} \\
& = \frac{1}{r_2} \cdot \frac{\cos \theta_2 - i \sin \theta_2}{\cos^2 \theta_2 + \sin^2 \theta_2} \\
& = \frac{1}{r_2}(\cos \theta_2 - i \sin \theta_2) \\
& = \frac{1}{r_2}[\cos(-\theta_2) + i \sin(-\theta_2)].
\end{align*}
\]</p>
<p>The effect of taking the reciprocal of a complex number</p>
<ul class="bullet-list"><li><div class="list-item-content"><span class="list-item-text">reciprocates its length; and</span></div></li><li><div class="list-item-content"><span class="list-item-text">reflects it about the positive real axis.</span></div></li></ul>
<p>In particular, when&nbsp;\(r_2=1\), i.e., when&nbsp;\(z_2\) is a complex number of unit length, then reciprocating it is equivalent to conjugating it.</p>
<p>Next, we proceed to divide \(z_1\) by \(z_2\).</p>
<p>\[\frac{z_1}{z_2} = z_1 \cdot \frac{1}{z_2} = [r_1(\cos \theta_1 + i \sin \theta_1)][\frac{1}{r_2}(\cos(-\theta_2) + i \sin(-\theta_2))] = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)].\]</p>

A direct consequence of the above development would be to understand division of complex numbers involves subtraction of arguments.

\[
\begin{align*}
\frac{1}{z_2} 
& = \frac{1}{r_2(\cos \theta_2 + i \sin \theta_2)} \\
& = \frac{1}{r_2} \cdot \frac{\cos \theta_2 - i \sin \theta_2}{\cos^2 \theta_2 + \sin^2 \theta_2} \\
& = \frac{1}{r_2}(\cos \theta_2 - i \sin \theta_2) \\
& = \frac{1}{r_2}[\cos(-\theta_2) + i \sin(-\theta_2)].
\end{align*}
\]

The effect of taking the reciprocal of a complex number

reflects it about the positive real axis.

In particular, when \(r_2=1\), i.e., when \(z_2\) is a complex number of unit length, then reciprocating it is equivalent to conjugating it.

Next, we proceed to divide \(z_1\) by \(z_2\).

\[\frac{z_1}{z_2} = z_1 \cdot \frac{1}{z_2} = [r_1(\cos \theta_1 + i \sin \theta_1)][\frac{1}{r_2}(\cos(-\theta_2) + i \sin(-\theta_2))] = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)].\]

Division via polar form

Division via polar form