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

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})} .