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Question: Answered & Verified by Expert
$$
\left[\sqrt{2}\left(\cos 56^{\circ} 15^{\prime}+i \sin 56^{\circ} 15^{\prime}\right)\right]^8=
$$
MathematicsComplex NumberAP EAMCETAP EAMCET 2023 (15 May Shift 1)
Options:
  • A 1
  • B $\mathrm{i}$
  • C 16
  • D $16 \mathrm{i}$
Solution:
2045 Upvotes Verified Answer
The correct answer is: $16 \mathrm{i}$
$\begin{aligned} & \text { }\left[\sqrt{2}\left(\cos 56^{\circ} 15^{\prime}+i \sin 56^{\circ} 15^{\prime}\right)\right]^8=(\sqrt{2})^8\left(i^\epsilon 6^{\circ} 15^1\right)^8 \\ & =2^4 e^{i\left(450^{\circ}\right)}=2^4 e^{i(360+90)} \quad\left\{\because e^{i \theta}=\cos \theta+i \sin \theta\right\} \\ & =2^4\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)=16 i\end{aligned}$

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