The real part of (1 - i)^{-i} is | vedclass

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(A) Let $z = (1 - i)^{-i}$. Taking $\log$ on both sides,$\log z = -i \log(1 - i) = -i \log \left( \sqrt{2} e^{-i\pi/4} \right)$$= -i \left[ \log \sqrt

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$e^{-\pi/4} \cos \left( \frac{1}{2} \log 2 \right)$

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(A) Let $z = (1 - i)^{-i}$. Taking $\log$ on both sides,$\log z = -i \log(1 - i) = -i \log \left( \sqrt{2} e^{-i\pi/4} ight)$$= -i \left[ \log \sqrt{2} + \log(e^{-i\pi/4}) ight]$$= -i \left[ \frac{1}{2} \log 2 - \frac{i\pi}{4} ight]$$= -\frac{i}{2} \log 2 - \frac{\pi}{4}$Therefore,$z = e^{-\pi/4} e^{-i(\frac{1}{2} \log 2)}$.Using Euler's formula $e^{i	heta} = \cos 	heta + i \sin 	heta$,we have $z = e^{-\pi/4} \left[ \cos \left( \frac{1}{2} \log 2 ight) - i \sin \left( \frac{1}{2} \log 2 ight) ight]$.The real part is $	ext{Re}(z) = e^{-\pi/4} \cos \left( \frac{1}{2} \log 2 ight)$.

The real part of $(1 - i)^{-i}$ is

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