JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 8)
The complex number $z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$ is equal to :
$\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}$
$\sqrt{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)$
$\sqrt{2} i\left(\cos \frac{5 \pi}{12}-i \sin \frac{5 \pi}{12}\right)$
$\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$
Explanation
$\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}$
$=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} \mathrm{i}$
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}$
$=\frac{i-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}} \times \frac{\frac{1}{2}-\sqrt{\frac{3}{2}} \mathrm{i}}{\frac{1}{2}-\sqrt{3 / 2} \mathrm{i}}=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} \mathrm{i}$
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}$
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