Q48 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

$$ \left[\sqrt{2}\left(\cos 56^{\circ} 15^{\prime}+i \sin 56^{\circ} 15^{\prime}\right)\right]^8= $$

  • A
    1
  • B
    $\mathrm{i}$
  • C
    16
  • Correct answer
    D
    $16 \mathrm{i}$

Solution

$\begin{aligned} & \text { }\left[\sqrt{2}\left(\cos 56^{\circ} 15^{\prime}+i \sin 56^{\circ} 15^{\prime}\right)\right]^8=(\sqrt{2})^8\left(i^\epsilon 6^{\circ} 15^1\right)^8 \\ & =2^4 e^{i\left(450^{\circ}\right)}=2^4 e^{i(360+90)} \quad\left\{\because e^{i \theta}=\cos \theta+i \sin \theta\right\} \\ & =2^4\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)=16 i\end{aligned}$

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

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