Question Number 31335 by NECx last updated on 06/Mar/18
$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{{i}} \\ $$
Commented by abdo imad last updated on 06/Mar/18
$${we}\:{have}\:{z}={e}^{{iln}\left(\mathrm{1}−{i}\right)} \:\:{and}\:{ln}\:{here}\:{mesnd}\:{the}\:{complexe}\:{log} \\ $$$${but}\:\mathrm{1}−{i}=\sqrt{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{{i}}{\:\sqrt{\mathrm{2}}}\:\right)={e}^{−{i}\frac{\pi}{\mathrm{4}}} \Rightarrow{ln}\left(\mathrm{1}−{i}\right)={ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\Rightarrow \\ $$$${iln}\left(\mathrm{1}−{i}\right)={iln}\left(\sqrt{\mathrm{2}}\right)+\frac{\pi}{\mathrm{4}}\:\Rightarrow{e}^{{iln}\left(\mathrm{1}−{i}\right)} ={e}^{\frac{\pi}{\mathrm{4}}} \left({cos}\left({ln}\left(\sqrt{\mathrm{2}}\right)+{isin}\left({ln}\left(\sqrt{\mathrm{2}}\right)\right)\right)\Rightarrow\right. \\ $$$$\left(\mathrm{1}−{i}\right)^{{i}} ={e}^{\frac{\pi}{\mathrm{4}}} {cos}\left({ln}\left(\sqrt{\mathrm{2}}\right)\right)\:+{i}\:{e}^{\frac{\pi}{\mathrm{4}}} {sin}\left({ln}\left(\sqrt{\mathrm{2}}\right)\right)\:. \\ $$
Commented by abdo imad last updated on 06/Mar/18
$$\mid\left(\mathrm{1}−{i}\right)^{{i}} \mid={e}^{\frac{\pi}{\mathrm{4}}} \:. \\ $$
Commented by NECx last updated on 06/Mar/18
$${thank}\:{you}\:{so}\:{much}. \\ $$