# Convert the complex number $\;z=\large\frac{i-1}{cos \large\frac{\pi}{3} + i sin \large\frac{\pi}{3}}\;$ in the polar form

$(a)\;0\qquad(b)\;\sqrt{2} (cos \large\frac{ \pi}{12} + i sin \large\frac{ \pi}{12})\qquad(c)\;\sqrt{2} (cos \large\frac{5 \pi}{12} + i sin \large\frac{5 \pi}{12})\qquad(d)\; cos \large\frac{5 \pi}{12} + i sin \large\frac{5 \pi}{12}$

Answer : $\;\sqrt{2} (cos \large\frac{5 \pi}{12} + i sin \large\frac{5 \pi}{12})$
Explanation :
We have , $\;z=\large\frac{i-1}{cos \large\frac{\pi}{3} + i sin \large\frac{\pi}{3}}\;$
$\;z=\large\frac{i-1}{\large\frac{1}{2} + i \large\frac{\sqrt{3}}{2}}\;$
$= \large\frac{2 (i-1)}{ 1+\sqrt{3}i}$
$= \large\frac{2 (i-1)}{ 1+\sqrt{3}i} \times \large\frac{1-\sqrt{3}i}{1-\sqrt{3} i}$
$= \large\frac{2(i + \sqrt{3} -1+ \sqrt{3} i)}{1+3}$
$= \large\frac{\sqrt{3}-1}{2} + i \large\frac{\sqrt{3} +1}{2}$
Now , put $\;\large\frac{\sqrt{3} -1}{2}= r cos \theta\;$ and $\; \large\frac{\sqrt{3} +1}{2} = r sin \theta$
Squaring and adding , we obtain
$r^{2} = (\large\frac{\sqrt{3} -1}{2})^{2} + (\large\frac{\sqrt{3} +1}{2})^{2}$
$= \large\frac{2((\sqrt{3})^{2}+1)}{4}$
$= \large\frac{2 \times 4}{4}=2$
$r = \sqrt{2}$
$cos \theta = \large\frac{\sqrt{3} -1}{2\sqrt{2}}\;,sin \theta = \large\frac{\sqrt{3} +1}{2\sqrt{2}}$
Therefore , $\; \theta = \large\frac{\pi}{4} + \large\frac{\pi}{6}= \large\frac{5 \pi}{12}$
Hence the polar form is , $\sqrt{2} (cos \large\frac{5 \pi}{12} + i sin \large\frac{5 \pi}{12})$