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Possible Duplicate:
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?

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    Both answers are correct; there are infinitely many different values for $1^i$ that are all correct.2012-10-02

2 Answers 2

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$1^i = e^{i\log 1}=e^{i(\log |1|+i\arg 1)}=e^{i(i 2\pi n)}=e^{-2\pi n}$

Where the principal branch of the logarithm is given by $n=0$.

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    I edited as I was receiving down votes for, I assume, not including other branches.2012-10-02
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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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    @Badshah: in the complex case, $a^b$ is *defined* as $e^{b\log a}$. The issue is that $\log 1$ is not in general $0$ when we talk about the complex logarithm. Silently assuming that $\log 1 = 0$ can lead to many errors in the complex setting.2012-10-02