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$(-1+i \sqrt{3})^{20}$ is equal to
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$\begin{aligned} & \text { Let } z=-1+i \sqrt{3}, r=\sqrt{1+3}=2 \\ & \theta=\tan ^{-1}\left(\frac{\sqrt{3}}{-1}\right)=\frac{2 \pi}{3} \\ & \therefore z=2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \\ & \therefore(z)^{20}=\left[2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)\right]^{20} \\ & =2^{20}\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)^{20}=2^{20}\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{20}\end{aligned}$
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