Thank you for reporting, we will resolve it shortly
Q.
Complex number z = $ \frac {i-1}{cos (\pi/3)+i sin (\pi/3)} $ in polar form is
AMUAMU 2010Complex Numbers and Quadratic Equations
Solution:
$z=\frac{i-1}{cos\left(\frac{\pi}{3}\right)+i\,sin\left(\frac{\pi}{3}\right)}$
$=\frac{i-1}{\left(\frac{1+i\sqrt{3}}{2}\right)}=\frac{2\left(i-1\right)\left(1-i\sqrt{3}\right)}{\left(1-i^{2}\cdot3\right)}$
$=\frac{2}{4}\left(i-1-i^{2}\sqrt{3}+i\sqrt{3}\right)$
$=\frac{1}{2}\left(\sqrt{3}-1+i\sqrt{3}+i\right)$
$=\frac{\left(\sqrt{3}-1\right)}{2}+\frac{i\left(\sqrt{3}+1\right)}{2} \ldots\left(i\right)$
Let $r\,cos\,\theta=\frac{\sqrt{3}-1}{2} \ldots\left(ii\right)$
$r\,sin\,\theta=\frac{\sqrt{3}+1}{2} \ldots\left(iii\right)$
On squaring and adding Eqs. $\left(ii\right)$ and $\left(iii\right)$
$r^{2}=\frac{1}{4}\left\{3+1-2\sqrt{3}+3+1+2\sqrt{3}\right\}$
$r^{2}=2$
$\Rightarrow r=\sqrt{2}$
From on dividing Eq. $\left(iii\right)$ from Eq. $\left(ii\right)$
$tan\,\theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$tan\,\theta=\frac{\left(\sqrt{3}+1\right)^{2}}{3-1}=\frac{3+1+2\sqrt{3}}{2}$
$tan\,\theta=2+\sqrt{3}$
$\theta=\frac{7\pi}{12}$
Principal value of $\theta=\pi-\frac{7\pi}{12}$
$=\frac{5\pi}{12}$
Hence, the polar form is
$r=\sqrt{2}\left(cos\frac{5\pi}{12}+i\,sin \frac{5\pi}{12}\right)$