The real part of the principal value of 2^{-i | vedclass

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(C) Let $z = 2^{-i}$.Taking the natural logarithm on both sides,we get:$\log z = \log (2^{-i})$$\Rightarrow \log z = -i \log 2$Since $z = e^{\log z}$,

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$\cos \left[\log \left(\frac{1}{2}\right)\right]$

Vedclass · Answer

(C) Let $z = 2^{-i}$.Taking the natural logarithm on both sides,we get:$\log z = \log (2^{-i})$$\Rightarrow \log z = -i \log 2$Since $z = e^{\log z}$,we have:$z = e^{-i \log 2} = e^{i \log(1/2)}$Using Euler's formula,$e^{i 	heta} = \cos 	heta + i \sin 	heta$,where $	heta = \log(1/2)$:$z = \cos(\log(1/2)) + i \sin(\log(1/2))$The real part of $z$ is $\cos(\log(1/2))$.

The real part of the principal value of $2^{-i}$ is

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