The imaginary part of i^{i} is | vedclass

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Question

(A) Let $z = i^{i}$.Taking the natural logarithm on both sides,we get $\ln(z) = i \ln(i)$.We know that $i = e^{i(\pi/2 + 2n\pi)}$ for any integer $n$.

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $z = i^{i}$.Taking the natural logarithm on both sides,we get $\ln(z) = i \ln(i)$.We know that $i = e^{i(\pi/2 + 2n\pi)}$ for any integer $n$. Taking the principal value where $n=0$,we have $\ln(i) = i\pi/2$.Substituting this into the equation,we get $\ln(z) = i(i\pi/2) = i^{2}(\pi/2) = -\pi/2$.Therefore,$z = e^{-\pi/2}$.Since $e^{-\pi/2}$ is a purely real number,it can be written as $e^{-\pi/2} + 0i$.Thus,the imaginary part of $i^{i}$ is $0$.

The imaginary part of $i^{i}$ is

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