Question Number 132834 by mohammad17 last updated on 16/Feb/21
$${Solve}\:{e}^{\mathrm{2}{z}} =\sqrt{\mathrm{3}}−{i}\:\:\:,{z}={x}+{iy} \\ $$
Answered by mathmax by abdo last updated on 20/Feb/21
$$\mid\sqrt{\mathrm{3}}−\mathrm{i}\mid=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{3}}−\mathrm{i}=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}\right)\:=\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{6}}} =\mathrm{2e}^{\left(−\frac{\mathrm{i}\pi}{\mathrm{6}}+\mathrm{2ik}\pi\right)} \\ $$$$\mathrm{e}^{\mathrm{2z}} =\mathrm{2}\:\mathrm{e}^{\mathrm{i}\left(−\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\right)} \:\Rightarrow\mathrm{e}^{\mathrm{2z}} \:=\mathrm{e}^{\mathrm{ln2}+\mathrm{i}\left(−\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\right)\:} \:\Rightarrow\mathrm{2z}=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{i}\left(−\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\right)\:\Rightarrow \\ $$$$\mathrm{z}_{\mathrm{k}} =\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\mathrm{i}\left(−\frac{\pi}{\mathrm{12}}+\mathrm{k}\pi\right)\:\:\mathrm{k}\in\mathrm{Z} \\ $$$$\mathrm{ddtermination}\:\mathrm{principale}\:\mathrm{is}\:\mathrm{z}_{\mathrm{0}} =\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{i}\pi}{\mathrm{12}} \\ $$