Question Number 207220 by hardmath last updated on 09/May/24
$$\mathrm{Find}:\:\:\:\left(\boldsymbol{\mathrm{i}}\:−\:\mathrm{1}\right)^{−\mathrm{100}} \:\:=\:\:? \\ $$
Answered by Frix last updated on 10/May/24
$$\left(−\mathrm{1}+\mathrm{i}\right)^{−\mathrm{100}} =\left(\left(−\mathrm{1}+\mathrm{i}\right)^{−\mathrm{1}} \right)^{\mathrm{100}} = \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\right)^{\mathrm{100}} =\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{3}\pi}{\mathrm{4}}\mathrm{i}} \right)^{\mathrm{100}} = \\ $$$$=\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{100}} \mathrm{e}^{−\frac{\mathrm{300}\pi}{\mathrm{4}}\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} }\mathrm{e}^{\mathrm{75}\pi\mathrm{i}} =−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} }=−\mathrm{2}^{−\mathrm{50}} \\ $$
Commented by hardmath last updated on 10/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 10/May/24
$$=\left(\left({i}−\mathrm{1}\right)^{\mathrm{2}} \right)^{−\mathrm{50}} =\left(−\mathrm{2}{i}\right)^{−\mathrm{50}} \\ $$$$=\left\{\left(−\mathrm{2}{i}\right)^{\mathrm{2}} \right\}^{−\mathrm{25}} \\ $$$$=\left(−\mathrm{4}\right)^{−\mathrm{25}} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{4}^{\mathrm{25}} }=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{50}} } \\ $$