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

$$solve$$$$$$$$\left(1\right):{(1+i)}^i$$$$$$$$\left(2\right):log{(1+i)}^{\pi i}$$

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

$${(1+i)}^i=\sqrt{2}{\left(e^{\frac{\pi }{4}i}\right)}^i=(cos\left(log\left(\sqrt{2}\right)\right)+isin\left(log\left(\sqrt{2}\right)\right))e^{-\frac{\pi }{4}}$$$$\pi ilog(1+i)=\pi ilog\left(\sqrt{2}\right)-\frac{{\pi }^2}{4}$$

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

$$\left(1\right)$$$$(1+i)=\sqrt{2}{e}^{\frac{\pi i}{4}}=e^{\ln \sqrt{2}+\frac{\pi i}{4}}$$$${(1+i)}^i=e^{(\ln \sqrt{2}+\frac{\pi i}{4})i}=e^{-\frac{\pi }{4}}{e}^{i\ln \sqrt{2}}$$$$=e^{-\frac{\pi }{4}}[\cos \left(\ln \sqrt{2}\right)+i\sin \left(\ln \sqrt{2}\right)]$$$$$$$$\left(2\right)$$$$\ln {(1+i)}^{\pi i}=\pi i\ln \left[e^{\ln \sqrt{2}+\frac{\pi i}{4}}\right]$$$$=\pi i(\ln \sqrt{2}+\frac{\pi i}{4})$$$$=\pi (-\frac{\pi }{4}+i\ln \sqrt{2})$$

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