Question Number 56144 by gunawan last updated on 11/Mar/19
$$\mathrm{calculate}\:\left({i}−\mathrm{1}\right)^{\mathrm{49}} \left(\mathrm{cos}\:\frac{\pi}{\mathrm{40}}+{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{40}}\right)^{\mathrm{10}} \\ $$
Answered by Smail last updated on 11/Mar/19
![(i−1)^(49) =[(√2)(((√2)/2)−i((√2)/2))]^(49) =(√2^(49) )(e^(i(π/4)) )^(49) =2^(24) (√2)e^(i((49π)/4)) =2^(24) (√2)e^(i((π/4)+12π)) =2^(24) (√2)e^(i(π/4)) (cos(π/(40))+isin(π/(40)))^(10) =(e^(i(π/(40))) )^(10) =e^(i(π/4)) (i−1)^(49) (cos(π/(40))+isin(π/(40)))^(10) =(2^(24) (√2)e^(i(π/4)) )e^(i(π/4)) =2^(24) (√2)e^(i(π/2)) =2^(24) (√2)i](https://www.tinkutara.com/question/Q56153.png)
$$\left({i}−\mathrm{1}\right)^{\mathrm{49}} =\left[\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right]^{\mathrm{49}} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{49}} }\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{49}} =\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{49}\pi}{\mathrm{4}}} =\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{12}\pi\right)} \\ $$$$=\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\left({cos}\frac{\pi}{\mathrm{40}}+{isin}\frac{\pi}{\mathrm{40}}\right)^{\mathrm{10}} =\left({e}^{{i}\frac{\pi}{\mathrm{40}}} \right)^{\mathrm{10}} ={e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\left({i}−\mathrm{1}\right)^{\mathrm{49}} \left({cos}\frac{\pi}{\mathrm{40}}+{isin}\frac{\pi}{\mathrm{40}}\right)^{\mathrm{10}} =\left(\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right){e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$=\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{2}^{\mathrm{24}} \sqrt{\mathrm{2}}{i} \\ $$