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Question Number 56144 by gunawan last updated on 11/Mar/19

$$\mathrm{calculate}{(i-1)}^{49}{(\cos \frac{\pi }{40}+i\sin \frac{\pi }{40})}^{10}$$

Answered by Smail last updated on 11/Mar/19

$${(i-1)}^{49}={\left[\sqrt{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})\right]}^{49}$$$$=\sqrt{2^{49}}{\left(e^{i\frac{\pi }{4}}\right)}^{49}=2^{24}\sqrt{2}{e}^{i\frac{49\pi }{4}}=2^{24}\sqrt{2}{e}^{i(\frac{\pi }{4}+12\pi )}$$$$=2^{24}\sqrt{2}{e}^{i\frac{\pi }{4}}$$$${(cos\frac{\pi }{40}+isin\frac{\pi }{40})}^{10}={\left(e^{i\frac{\pi }{40}}\right)}^{10}=e^{i\frac{\pi }{4}}$$$${(i-1)}^{49}{(cos\frac{\pi }{40}+isin\frac{\pi }{40})}^{10}=\left(2^{24}\sqrt{2}{e}^{i\frac{\pi }{4}}\right)e^{i\frac{\pi }{4}}$$$$=2^{24}\sqrt{2}{e}^{i\frac{\pi }{2}}$$$$=2^{24}\sqrt{2}i$$

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