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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!

Tig la Pomme
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JMK
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1 Answers1

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

Tig la Pomme
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