The polar form of complex numbers has a pleasant property because multiplication of complex numbers in this form has an aesthetic aspect as well as a practical geometrical advantage over the Cartesian form.
Recall that multiplication of complex numbers in Cartesian form involves very complicated and unnatural-looking operations:
\[(x+iy)(u+iv) = (xu-yv)+i(xv+yu),\]
where&nbsp;\(x\), \(y\), \(u\) and \(v\) are real numbers.
Let&nbsp;\(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\), where
<ul class="bullet-list"><li><div class="list-item-content">\(r_1, r_2 &gt; 0\) are the respective moduli; and</div></li><li><div class="list-item-content">\(\theta_1\) and \(\theta_2\) are the respective arguments of \(z_1\) and \(z_2\).</div></li></ul>
Then the multiplication of \(z_1\) and \(z_2\) will work out nicely because of the following compound angle (addition) formula which we record here first:
\[\begin{aligned} \cos(\theta_1 + \theta_2) &= \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \\ \sin(\theta_1 + \theta_2) &= \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 \end{aligned}\]
We have:
\[\begin{aligned} &z_1 z_2 \\ &= [r_1(\cos \theta_1 + i \sin \theta_1)][r_2(\cos \theta_2 + i \sin \theta_2)] \\ &= r_1 r_2[(\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i(\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2)] \\ &= r_1 r_2[\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] \end{aligned}\]

The polar form of complex numbers has a pleasant property because multiplication of complex numbers in this form has an aesthetic aspect as well as a practical geometrical advantage over the Cartesian form.

Recall that multiplication of complex numbers in Cartesian form involves very complicated and unnatural-looking operations:

where \(x\), \(y\), \(u\) and \(v\) are real numbers.

Let \(z_1 = r_1(\cos \theta_1 + i \sin \theta_1)\) and \(z_2 = r_2(\cos \theta_2 + i \sin \theta_2)\), where

\(r_1, r_2 > 0\) are the respective moduli; and

\(\theta_1\) and \(\theta_2\) are the respective arguments of \(z_1\) and \(z_2\).

Then the multiplication of \(z_1\) and \(z_2\) will work out nicely because of the following compound angle (addition) formula which we record here first:

\[\begin{aligned} \cos(\theta_1 + \theta_2) &= \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \\ \sin(\theta_1 + \theta_2) &= \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 \end{aligned}\]

\[\begin{aligned} &z_1 z_2 \\ &= [r_1(\cos \theta_1 + i \sin \theta_1)][r_2(\cos \theta_2 + i \sin \theta_2)] \\ &= r_1 r_2[(\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i(\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2)] \\ &= r_1 r_2[\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] \end{aligned}\]

Multiplication via polar form

Multiplication via polar form