Question

Covert the complex number
$z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$ in the polar form.

Answer 1

Given,

$z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$

$=\frac{i-1}{\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

$=\frac{2(i-1)}{1+\sqrt{3}i}$

$=\frac{2i-2}{1+\sqrt{3}i}\times \frac{1-\sqrt{3}i}{1-\sqrt{3}i}$

$=\frac{(-1+\sqrt{3})+i(1+\sqrt{3})}{2}$

$\therefore z=\frac{-1+\sqrt{3}}{2}+i\frac{1+\sqrt{3}}{2}$

$r=\sqrt{{\left(\frac{-1+\sqrt{3}}{2}\right)}^2+{\left(\frac{1+\sqrt{3}}{2}\right)}^2}$

$r=\sqrt{\frac{8}{4}}=\sqrt{2}$

$\theta ={\tan }^{-1}\left(\frac{\frac{1+\sqrt{3}}{2}}{\frac{-1+\sqrt{3}}{2}}\right)={75}^{\circ }=\frac{5\pi }{12}$

$\therefore z=\sqrt{2}(\cos \frac{5\pi }{12}+\sin \frac{5\pi }{12})$