Question Number 132655 by mohammad17 last updated on 15/Feb/21
$${write}\:\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{23}} }{\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{13}} }\:{by}\:{re}^{{i}\theta} \:{such}\:{that}\:{r}>\mathrm{0},−\pi\leqslant\theta<\pi \\ $$
Answered by physicstutes last updated on 15/Feb/21
$$\mathrm{1}\:−{i}\:=\:\sqrt{\mathrm{2}}\:{e}^{−\frac{\pi}{\mathrm{4}}{i}} \\ $$$$\Rightarrow\:\left(\mathrm{1}−{i}\right)^{\mathrm{23}} \:=\:\sqrt{\mathrm{2}}\:^{\mathrm{23}} {e}^{−\frac{\mathrm{23}\pi}{\mathrm{4}}{i}} \\ $$$$\:\sqrt{\mathrm{3}}\:−\:{i}\:=\:\mathrm{2}{e}^{−\frac{\pi}{\mathrm{6}}{i}} \:\Rightarrow\:\left(\sqrt{\mathrm{3}}\:−{i}\right)^{\mathrm{13}} \:=\:\mathrm{2}^{\mathrm{13}} {e}^{−\frac{\mathrm{13}\pi}{\mathrm{6}}{i}} \\ $$$$\:{z}\:=\:\frac{\mathrm{2048}\sqrt{\mathrm{2}}\:{e}^{−\frac{\mathrm{23}\pi}{\mathrm{4}}{i}} }{\mathrm{2048}\:×\mathrm{4}{e}^{−\frac{\mathrm{13}\pi}{\mathrm{6}}{i}} }\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{e}^{−\frac{\mathrm{43}\pi}{\mathrm{12}}{i}} \\ $$$$\: \\ $$
Commented by MJS_new last updated on 15/Feb/21
$$\mathrm{wrong}\:\mathrm{format}.\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{e}^{\frac{\mathrm{5}\pi}{\mathrm{12}}\mathrm{i}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer} \\ $$
Commented by mohammad17 last updated on 16/Feb/21
$${can}\:{you}\:{solve}\:{this}\:{sir} \\ $$
Commented by physicstutes last updated on 16/Feb/21
$$\mathrm{yes}\:\mathrm{yes}\:\mathrm{sir},\:\mathrm{thank}\:\mathrm{you},\:\mathrm{i}\:\mathrm{just}\:\mathrm{saw}\:\mathrm{that} \\ $$