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Q.
The imaginary part of $i^i $ is
KCETKCET 2007Complex Numbers and Quadratic Equations
Solution:
Let $(a + ib) = i^i$
Taking log on both sides, we get
$\log\left(a +ib\right) = i \,\log \,i $
$ \Rightarrow \log\left(a+ib\right) = i \left(i \,\pi /2\right) $
$ \Rightarrow \log\left(a +ib\right) = - \frac{\pi}{2} $
$ \Rightarrow \left(a +ib\right) = e^{-\pi/2} $
On comparing imaginary part of $i^i$ is 0.