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Question Number 31335 by NECx last updated on 06/Mar/18

Find the pricipal value of z=(1−i)^i

$${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$${z}=\left(\mathrm{1}−{i}\right)^{{i}} \\ $$

Commented by abdo imad last updated on 06/Mar/18

we have z=e^(iln(1−i)) and ln here mesnd the complexe log but 1−i=(√2)( (1/(√2)) −(i/(√2)) )=e^(−i(π/4)) ⇒ln(1−i)=ln((√2) )−((iπ)/4)⇒ iln(1−i)=iln((√2))+(π/4) ⇒e^(iln(1−i)) =e^(π/4) (cos(ln((√2))+isin(ln((√2))))⇒ (1−i)^i =e^(π/4) cos(ln((√2))) +i e^(π/4) sin(ln((√2))) .

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

Commented by abdo imad last updated on 06/Mar/18

∣(1−i)^i ∣=e^(π/4) .

$$\mid\left(\mathrm{1}−{i}\right)^{{i}} \mid={e}^{\frac{\pi}{\mathrm{4}}} \:. \\ $$

Commented by NECx last updated on 06/Mar/18

thank you so much.

$${thank}\:{you}\:{so}\:{much}. \\ $$

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