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The complex number $\mathrm{z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}$ is equal to [JEE (Main) 31 Jan 2023 S II]
(A) $\sqrt{2} \mathrm{i}\left(\cos \frac{5 \pi}{12}-\mathrm{i} \sin \frac{5 \pi}{12}\right)$
(B) $\sqrt{2}\left(\cos \frac{5 \pi}{12}+\mathrm{i} \sin \frac{5 \pi}{12}\right)$
(C) $\sqrt{2}\left(\cos \frac{\pi}{12}+\mathrm{i} \sin \frac{\pi}{12}\right)$
(D) $\cos \frac{\pi}{12}-\mathrm{i} \sin \frac{\pi}{12}$

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Solution

$\mathrm{Z}=\frac{\mathrm{i}-1}{\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}}=\frac{\mathrm{i}-1}{\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}}=\frac{2(\mathrm{i}-1)(1-\sqrt{3} \mathrm{i})}{(1+\sqrt{3} \mathrm{i})(1-\sqrt{3} \mathrm{i})}=\frac{2(\mathrm{i}+\sqrt{3} \mathrm{i}-1+\sqrt{3})}{1+3}$

\[ \begin{aligned} & z=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2} i \Rightarrow|z|=\frac{1}{2} \sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}}=\sqrt{2} \\ & \arg (z)=\tan ^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)=75^{\circ}=\frac{5 \pi}{12} \Rightarrow z=\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right) \end{aligned} \]

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Answer

B

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