Find-the-principal-value-of-z-1-i-1-i-Hence-find-the-modulus-of-the-result-

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^((1+i)ln(1−i)) but 1−i=(√2) e^(−((iπ)/4)) ⇒ ln(1−i)=ln((√2)) −((iπ)/4) ⇒(1+i)ln(1−i)=(1+i)(ln((√2))−((iπ)/4)) =ln((√2))−((iπ)/4) +iln((√2)) +(π/4)=(π/4) +ln((√2)) +i(ln((√2))−(π/4)) z= e^((π/4)+ln((√2))) (cos(ln((√2) −(π/4))+isin(ln((√2) −(π/4)) and ∣z∣= e^((π/4) +ln((√2))) .

$${we}\:{have}\:{z}=\:{e}^{\left(\mathrm{1}+{i}\right){ln}\left(\mathrm{1}−{i}\right)} \:{but}\:\mathrm{1}−{i}=\sqrt{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{i}\right)={ln}\left(\sqrt{\mathrm{2}}\right)\:−\frac{{i}\pi}{\mathrm{4}}\:\Rightarrow\left(\mathrm{1}+{i}\right){ln}\left(\mathrm{1}−{i}\right)=\left(\mathrm{1}+{i}\right)\left({ln}\left(\sqrt{\mathrm{2}}\right)−\frac{{i}\pi}{\mathrm{4}}\right) \\ $$$$={ln}\left(\sqrt{\mathrm{2}}\right)−\frac{{i}\pi}{\mathrm{4}}\:+{iln}\left(\sqrt{\mathrm{2}}\right)\:+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}\:+{ln}\left(\sqrt{\mathrm{2}}\right)\:+{i}\left({ln}\left(\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$$${z}=\:{e}^{\frac{\pi}{\mathrm{4}}+{ln}\left(\sqrt{\mathrm{2}}\right)} \left({cos}\left({ln}\left(\sqrt{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)+{isin}\left({ln}\left(\sqrt{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)\:{and}\right.\right.\right. \\ $$$$\mid{z}\mid=\:{e}^{\frac{\pi}{\mathrm{4}}\:+{ln}\left(\sqrt{\mathrm{2}}\right)} \:\:. \\ $$

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