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I am confused as to what $$i^{i}$$ looks like. I followed the procedure our prof described and got $$e^{i\log i}=e^{\log i+2ki\pi }=e^{-2k\pi (\cos(\log i)+i\sin(\log i))}.$$

However, $$i^{i}=e^{-\pi/2}$$

Is it the same thing or am I doing something wrong?

Thanks!

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    Use $i = e^{i \pi / 2}$ to evaluate the logarithm in the middle. – 2017-02-25
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    You might wish to define $a^b=e^{b\operatorname{Log}(a)}, where $\operatorname{Log}$ denotes the principal log. – 2017-08-25

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There's no unambiguous definition for complex logarithm, hence several answers are legitimate.