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What is the value of $1^i$?
I was thinking, what would 1^i be? Then I did: $e^{i\pi}=-1\rightarrow e^{i\pi}\cdot e^{i\pi}=e^{2i\pi}=-1\cdot -1=1$ Now raise to the power i: $1^i=(e^{2i\pi})^i=e^{2i^2\pi}=e^{-2\pi}=\frac{1}{e^{2\pi}}$ Is this correct?
$\endgroup$ 42 Answers
$\begingroup$$$1^i = e^{i\log 1}=e^{i(\log |1|+i\arg 1)}=e^{i(i 2\pi n)}=e^{-2\pi n}$$
Where the principal branch of the logarithm is given by $n=0$.
$\endgroup$ 8 $\begingroup$The power function $(x,y)\mapsto x^y$ is indeed defined via the exponential $x\mapsto e^x$, as $x^y:=e^{\log x\cdot y}$, where $\log$ is the inverse of $\exp$. But, $\exp$ is not injective: it is periodic by $2\pi i$, hence the $\log$ is not unique, only up to $+k2\pi i$ for some $k\in\mathbb Z$. So, $1^i$ has infinite many values: $$1^i=e^{(2k\pi i)i} = e^{-2k\pi} $$ Similarly for $i^i$. Can you find all of its values?
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