Find the principal value of z=(1−i)^(1+i) .Hence find the modulus of the result.


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Question Number 31336 by NECx last updated on 06/Mar/18

$F i n d t h e p r i n c i p a l v a l u e o f$ $z = {(1 - i)}^{1 + i} . H e n c e f i n d t h e$ $m o d u l u s o f t h e r e s u l t .$
Commented by abdo imad last updated on 06/Mar/18

$w e h a v e z = e^{(1 + i) l n (1 - i)} b u t 1 - i = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow$ $l n (1 - i) = l n (\sqrt{2}) - \frac{i π}{4} \Rightarrow (1 + i) l n (1 - i) = (1 + i) (l n (\sqrt{2}) - \frac{i π}{4})$ $= l n (\sqrt{2}) - \frac{i π}{4} + i l n (\sqrt{2}) + \frac{π}{4} = \frac{π}{4} + l n (\sqrt{2}) + i (l n (\sqrt{2}) - \frac{π}{4})$ $z = e^{\frac{π}{4} + l n (\sqrt{2})} (c o s (l n (\sqrt{2} - \frac{π}{4}) + i s i n (l n (\sqrt{2} - \frac{π}{4}) a n d$ $∣ z ∣= e^{\frac{π}{4} + l n (\sqrt{2})} .$
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