Question Number 127979 by bobhans last updated on 03/Jan/21
$$\:\:\left(\mathrm{1}+{i}\right)^{\mathrm{2020}} \:=? \\ $$
Answered by liberty last updated on 03/Jan/21
$$\:\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2020}} \:=\:\left(\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:+\:\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{2020}} \\ $$$$\:=\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2020}} \:\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}\right)+\mathrm{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2020}} \\ $$$$\:=\:\mathrm{2}^{\mathrm{1010}} \:\left(\mathrm{cos}\:\left(\mathrm{505}\pi\right)\:+\:\mathrm{i}\:\mathrm{sin}\:\left(\mathrm{505}\pi\right)\right) \\ $$$$\:=\:\mathrm{2}^{\mathrm{1010}} \:\left(−\mathrm{1}+\mathrm{0}\right)\:=−\:\mathrm{2}^{\mathrm{1010}} \\ $$
Commented by JDamian last updated on 03/Jan/21
$$\boldsymbol{\mathrm{MJS\_new}}\:\mathrm{is}\:\mathrm{right}\:−\:\mathrm{cos}\left(\mathrm{505}\pi\right)\:{is}\:−\mathrm{1} \\ $$$$\mathrm{cos}\left(\mathrm{k}\pi\right)=\left(−\mathrm{1}\right)^{{k}} \:\:\:\forall{k}\in\boldsymbol{\mathrm{Z}} \\ $$
Answered by MJS_new last updated on 03/Jan/21
$$\mathrm{1}+\mathrm{i}=\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2020}} =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2020}} \mathrm{e}^{\mathrm{i}\frac{\mathrm{2020}\pi}{\mathrm{4}}} =\mathrm{2}^{\mathrm{1010}} \mathrm{e}^{\mathrm{505}\pi\mathrm{i}} =−\mathrm{2}^{\mathrm{1010}} \\ $$