Question Number 31336 by NECx last updated on 06/Mar/18
$${Find}\:{the}\:{principal}\:{value}\:{of} \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} .{Hence}\:{find}\:{the} \\ $$$${modulus}\:{of}\:{the}\:{result}. \\ $$
Commented by abdo imad last updated on 06/Mar/18
$${we}\:{have}\:{z}=\:{e}^{\left(\mathrm{1}+{i}\right){ln}\left(\mathrm{1}−{i}\right)} \:{but}\:\mathrm{1}−{i}=\sqrt{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{i}\right)={ln}\left(\sqrt{\mathrm{2}}\right)\:−\frac{{i}\pi}{\mathrm{4}}\:\Rightarrow\left(\mathrm{1}+{i}\right){ln}\left(\mathrm{1}−{i}\right)=\left(\mathrm{1}+{i}\right)\left({ln}\left(\sqrt{\mathrm{2}}\right)−\frac{{i}\pi}{\mathrm{4}}\right) \\ $$$$={ln}\left(\sqrt{\mathrm{2}}\right)−\frac{{i}\pi}{\mathrm{4}}\:+{iln}\left(\sqrt{\mathrm{2}}\right)\:+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}\:+{ln}\left(\sqrt{\mathrm{2}}\right)\:+{i}\left({ln}\left(\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$$${z}=\:{e}^{\frac{\pi}{\mathrm{4}}+{ln}\left(\sqrt{\mathrm{2}}\right)} \left({cos}\left({ln}\left(\sqrt{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)+{isin}\left({ln}\left(\sqrt{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)\:{and}\right.\right.\right. \\ $$$$\mid{z}\mid=\:{e}^{\frac{\pi}{\mathrm{4}}\:+{ln}\left(\sqrt{\mathrm{2}}\right)} \:\:. \\ $$