Find-the-pricipal-value-of-1-i-1-i-

Question Number 19936 by ajfour last updated on 18/Aug/17

F i n d t h e p r i c i p a l v a l u e o f

{(1 - i)}^{1 + i} .

Commented by prof Abdo imad last updated on 22/Jun/18

{(1 - i)}^{1 + i} = {(\sqrt{2} e^{- \frac{i π}{4}})}^{1 + i}

= {(\sqrt{2})}^{1 + i} e^{- \frac{i π}{4} (1 + i)}

= \sqrt{2} {(\sqrt{2})}^{i} e^{- \frac{i π}{4} + \frac{π}{4}}

= \sqrt{2} e^{i l n (\sqrt{2}) - \frac{i π}{4}} e^{\frac{π}{4}}

= \sqrt{2} e^{\frac{π}{4}} e^{i {l n (\sqrt{} 2) - \frac{π}{4}}}

= \sqrt{2} e^{\frac{π}{4}} {c o s (l n (\sqrt{2}) - \frac{π}{4}) + i s i n (l n (2) - \frac{π}{4})} .

Answered by sma3l2996 last updated on 18/Aug/17

{(1 - i)}^{1 + i} = {[\sqrt{2} (\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2})]}^{1 + i} = {(\sqrt{2} e^{- i \frac{π}{4}})}^{1 + i}

= 2^{\frac{1 + i}{2}} \times e^{\frac{π}{4} (1 - i)} = 2^{\frac{1}{2}} \times 2^{\frac{i}{2}} \times e^{\frac{π}{4}} \times e^{- i \frac{π}{4}}

= \sqrt{2} \times e^{i \frac{l n 2}{2}} \times e^{π / 4} \times e^{- i π / 4}

{(1 - i)}^{1 + i} = \sqrt{2} e^{π / 4} \times e^{i (\frac{2 l n 2 - π}{4})}

Commented by ajfour last updated on 18/Aug/17

t h a n k s s i r, c o r r e c t a n s w e r .

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