Verified Solution • 2025–26 Session moderate QJEEM23Ma032e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma032e Solution The value of (1+sin2π9+icos2π91+sin2π9−icos2π9)3{\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}(1+sin92π−icos92π1+sin92π+icos92π)3 is Multiple Choice Options A −12(1−i3)- {1 \over 2}\left( {1 - i\sqrt 3 } \right)−21(1−i3) B −12(3−i)- {1 \over 2}\left( {\sqrt 3 - i} \right)−21(3−i) C 12(1−i3){1 \over 2}\left( {1 - i\sqrt 3 } \right)21(1−i3) D 12(3+i){1 \over 2}\left( {\sqrt 3 + i} \right)21(3+i) Step-by-Step Verified Solution z=(1+sin2π9+icos2π91+sin2π9−icos2π9)3z=\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}z=(1+sin92π−icos92π1+sin92π+icos92π)3 amp;1+sin2π9+icos2π9=1+cos5π18+isin5π18amp;=1+2cos25π36−1+2isin5π36cos5π36amp;=2cos5π36(cos5π36+isin5π36)=2cos5π36ei5π36amp;z=−32+12i=12(i−3)=−12(3−i)\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}amp;1+sin92π+icos92π=1+cos185π+isin185πamp;=1+2cos2365π−1+2isin365πcos365πamp;=2cos365π(cos365π+isin365π)=2cos365πei365πamp;z=−23+21i=21(i−3)=−21(3−i) Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring