Q49 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12}=$

  • A
    -1
  • Correct answer
    B
    i
  • C
    $-i$
  • D
    2

Solution

We have, $$ \begin{aligned} & \left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12} \\ & =\left[\frac{2 \cos ^2 \frac{\pi}{16}-2 \sin \frac{\pi}{16} \cos \frac{\pi}{16} i}{2 \cos ^2 \frac{\pi}{16}+2 \sin \frac{\pi}{16} \cos \frac{\pi}{16} i}\right]^{12} \\ & =\left[\frac{\cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}\right)}{\cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}\right)}\right]^{12} \\ & =\left(\frac{\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}}\right)^{12}=\left(\frac{e^{-i \pi / 16}}{e^{i \pi / 16}}\right)^{12} \\ & {\left[\because \cos \theta+i \sin \theta=e^{i \theta}\right]} \\ & =\left(e^{-\frac{i \pi}{8}}\right)^{12}=e^{-i 3 \pi / 2}=\cos \frac{3 \pi}{2}-i \sin \frac{3 \pi}{2} \\ & =0-i(-1)=i \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

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