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I'm not sure how to begin since this is not in the form $re^{i \theta}$.

Adam Hughes
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Edi Madi
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1 Answers1

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Write $i=e^{i\pi/2}$ then $i^{2i}=e^{2i^2\pi/2}=e^{-\pi}$

Note that $e^{i\theta}$ representations are only defined up to mutliples of $2\pi$ in $\theta$, so really this is only one of a myriad of answers, as

$$i=e^{2\pi i + i\pi/2}=i\implies i^{2i}=e^{4\pi i^2 +2i^2\pi/2} = e^{-5\pi}$$

and so on. So the right answer is that this is not even well-defined.

Adam Hughes
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    Please at least take the time to explain why $i^{2i}$ is not well defined. This is what might be most useful to the OP. – Did Apr 29 '16 at 06:17
  • @Did presumably this is from a basic complex variables class, but I take your point, and it's a good one. I've edited. – Adam Hughes Apr 29 '16 at 06:20