JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 12)
The value of $${\left( {{{1 + \sin {{2\pi } \over 9} + i\cos {{2\pi } \over 9}} \over {1 + \sin {{2\pi } \over 9} - i\cos {{2\pi } \over 9}}}} \right)^3}$$ is
$$ - {1 \over 2}\left( {1 - i\sqrt 3 } \right)$$
$$ - {1 \over 2}\left( {\sqrt 3 - i} \right)$$
$${1 \over 2}\left( {1 - i\sqrt 3 } \right)$$
$${1 \over 2}\left( {\sqrt 3 + i} \right)$$
Explanation
$z=\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3}$
$$ \begin{aligned} & 1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18} \\\\ & =1+2 \cos ^{2} \frac{5 \pi}{36}-1+2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36} \\\\ & =2 \cos \frac{5 \pi}{36}\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)=2 \cos \frac{5 \pi}{36} e^{i \frac{5 \pi}{36}} \\\\ & z=-\frac{\sqrt{3}}{2}+\frac{1}{2} i=\frac{1}{2}(i-\sqrt{3})=-\frac{1}{2}(\sqrt{3}-i) \end{aligned} $$
$$ \begin{aligned} & 1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}=1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18} \\\\ & =1+2 \cos ^{2} \frac{5 \pi}{36}-1+2 i \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36} \\\\ & =2 \cos \frac{5 \pi}{36}\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)=2 \cos \frac{5 \pi}{36} e^{i \frac{5 \pi}{36}} \\\\ & z=-\frac{\sqrt{3}}{2}+\frac{1}{2} i=\frac{1}{2}(i-\sqrt{3})=-\frac{1}{2}(\sqrt{3}-i) \end{aligned} $$
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