solve-1-1-i-i-2-log-1-i-pii-

Question Number 131021 by mohammad17 last updated on 31/Jan/21

s o l v e

(1) : {(1 + i)}^{i}

(2) : l o g {(1 + i)}^{π i}

Answered by Dwaipayan Shikari last updated on 31/Jan/21

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

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

Answered by mr W last updated on 31/Jan/21

(1)

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

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

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

(2)

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

= π i (\ln \sqrt{2} + \frac{π i}{4})

= π (- \frac{π}{4} + i \ln \sqrt{2})

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