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The imaginary part of $i^{i}$ is
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0
Let $(a+i b)=i^{i}$
Taking log on both sides, we get
$\log (a+i b)=i \log i$
$\Rightarrow \quad \log (a+i b)=i(i \pi / 2)$
$\Rightarrow \quad \log (a+i b)=-\frac{\pi}{2}$
$\Rightarrow \quad(a+i b)=e^{-\pi / 2}$
On comparing imaginary part of $i^{i}$ is 0 .
Taking log on both sides, we get
$\log (a+i b)=i \log i$
$\Rightarrow \quad \log (a+i b)=i(i \pi / 2)$
$\Rightarrow \quad \log (a+i b)=-\frac{\pi}{2}$
$\Rightarrow \quad(a+i b)=e^{-\pi / 2}$
On comparing imaginary part of $i^{i}$ is 0 .
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