$i$ is an imaginary number,when squared,gives a negative result

In general, it is true that $e^{\log x} = x$ , if by "log" we mean the natural logarithm (also written $\ln x$ ).

In this case, $x$ happens to be $i^i$ .

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Note : The value of $i^i$ is also $i^i = \left(e^{\frac{\pi}2i}\right)^i = e^{-\frac{\pi}{2}}.$