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The real part of the principle value of $2^{-i}$ is
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The correct answer is:
$\cos \left[\log \left(\frac{1}{2}\right)\right]$
Let $z=2^{-i}$
Taking $\log$ on both sides, we get $\begin{aligned} & \log z=\log \left(2^{-i}\right) \\ \Rightarrow & \log z=-i \log 2 \\ \Rightarrow & \log z=i \log \left(\frac{1}{2}\right) \\ \Rightarrow & z=e^{i \log (1 / 2)} \quad\left(\because e^{i \theta}=\cos \theta+i \sin \theta\right) \\ \Rightarrow & z=\cos \left(\log \frac{1}{2}\right)+i \sin \left(\log \frac{1}{2}\right) \\ \therefore \text { The real part of } z=2^{-i} \text { is } \cos \left(\log \frac{1}{2}\right) \end{aligned}$
Taking $\log$ on both sides, we get $\begin{aligned} & \log z=\log \left(2^{-i}\right) \\ \Rightarrow & \log z=-i \log 2 \\ \Rightarrow & \log z=i \log \left(\frac{1}{2}\right) \\ \Rightarrow & z=e^{i \log (1 / 2)} \quad\left(\because e^{i \theta}=\cos \theta+i \sin \theta\right) \\ \Rightarrow & z=\cos \left(\log \frac{1}{2}\right)+i \sin \left(\log \frac{1}{2}\right) \\ \therefore \text { The real part of } z=2^{-i} \text { is } \cos \left(\log \frac{1}{2}\right) \end{aligned}$
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