Question

The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :

• A. $\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)$
• B. $\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}$
• C. $\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)$
• D. $\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$

Answer 1

Answer: (D)

$Z =\frac{ i -1}{\cos \frac{\pi}{3}+ i \sin \frac{\pi}{3}}=\frac{ i -1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i }$
$=\frac{ i -1}{\frac{1}{2}+\frac{\sqrt{3}}{2} i } \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2} i }}{\frac{1}{2}-\sqrt{3 / 2} i }=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} i$
Apply polar form,
$ r \cos \theta=\frac{\sqrt{3}-1}{2} $
$ r \sin \theta=\frac{\sqrt{3}+1}{2}$
Now, $\tan \theta=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
So, $\theta=\frac{5 \pi}{12}$