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What is the value of $1^i$?

I read that $$ (-1)^{2i}=\exp(2i\log -1)=\exp(-2\pi-4\pi k) $$ for $k\in\mathbb{Z}$. How does the second equality follow?

I calculate $\cos(2i\log(-1))=1$ and $\sin(2i\log(-1))=0$, but I don't see how a $-2\pi-4\pi k$ appears like this.

Thanks for clarifying.

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marked as duplicate by pedja, William, Noah Snyder, whuber, draks ... Oct 5 '12 at 6:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 1 down vote accepted

The logarithm can be considered a multi-valued function. Backstory: since $\displaystyle e^{\pi i}+1=0$, we may conclude that $\displaystyle e^{2\pi i}=1$ and so $\displaystyle e^{\pi i+2k\pi i}=-1$ for any integer $k$. So which of these $\pi i+2k\pi i$ is the value of $\log(-1)$ then? Generally a particular branch cut is chosen so that the logarithm only obtains one of these values, but clearly here the formula is valid for integers $k$, only for different choices of branches with each $k$, and this what might be intended in the purported equality.

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