Question Number 47389 by gunawan last updated on 09/Nov/18
$$\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{1}+\mathrm{2}{i}} =… \\ $$
Commented by maxmathsup by imad last updated on 09/Nov/18
$${we}\:{have}\:\mid\sqrt{\mathrm{3}}−{i}\mid=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{3}}−{i}\:=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left({cos}\left(−\frac{\pi}{\mathrm{6}}\right)+{isin}\left(−\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \:\Rightarrow\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{1}+\mathrm{2}{i}} \:=\left\{\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right\}^{\mathrm{1}+\mathrm{2}{i}} =\mathrm{2}^{\mathrm{1}+\mathrm{2}{i}} \:{e}^{−\frac{{i}\pi\left(\mathrm{1}+\mathrm{2}{i}\right)}{\mathrm{6}}} \\ $$$$={e}^{−\frac{{i}\pi−\mathrm{2}\pi}{\mathrm{6}}} \:={e}^{\frac{\pi}{\mathrm{3}}} \:{e}^{−\frac{{i}\pi}{\mathrm{6}}} \:\mathrm{2}^{\mathrm{1}+\mathrm{2}{i}} ={e}^{\frac{\pi}{\mathrm{3}}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{2}\left({cos}\mathrm{2}\:+{isin}\mathrm{2}\right) \\ $$$$=\:\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){e}^{\frac{\pi}{\mathrm{3}}} \:\left({cos}\mathrm{2}\:+{isin}\mathrm{2}\right)\:. \\ $$