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Here's my (obviously flawed) proof that $1=e^{-2 \pi}$: $$ 1^i=1\\ e^{2 \pi i} = 1\\ \left(e^{2\pi i}\right)^i = 1^i\\ e^{-2 \pi} = 1 $$

What's the issue? I understand that exponentiation is not injective (and thus $-1 \neq 1$ even though $(-1)^2 = 1^2$), but I don't think that's an issue here: I'm only raising things to the power of $i$, which I don't think is multi-valued.

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    $\begingroup$ Raising things to the power $i$ is multi-valued. The only single-valued powers are integer powers. In general, $(a^b)^c = a^{bc}$ does not hold. $\endgroup$ Commented Aug 9, 2013 at 21:03
  • $\begingroup$ @DanielFischer Well that was quick. Could you explain why it is multivalued? $\endgroup$ Commented Aug 9, 2013 at 21:05
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    $\begingroup$ It's because exponentiation is built using the exponential function and the natural log: $a^b = e^{b \log(a)}$ and the natural log is multivalued. $\endgroup$
    – Bill Cook
    Commented Aug 9, 2013 at 21:10
  • $\begingroup$ @BillCook Thank you, this has cleared a lot up for me. $\endgroup$ Commented Aug 9, 2013 at 21:13

2 Answers 2

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In complex numbers, either $x^y$ is a multivalued function, or you have to give up the notion that $(x^y)^z = x^{yz}$.

If you allow $x^y$ to be multi-valued, then one of the values for $1^i$ is $e^{-2\pi}$.

If $x^y$ is not multivalued, then you have to pick a single value for $\log 1$ to define $1^y$. We usually pick $\log 1 = 0$, for some reason. :)

The multivalued nature makes sense when you consider that $\sqrt{1}=1^{1/2}$ can be thought of as having two values, $-1$ and $1$. In general, though, when $y$ is irrational, you get $1^y$ (or more generally, $x^y$) can take infinitely many values.

The only time $x^y$ is naturally single-valued is when $y$ is an integer.

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  • $\begingroup$ This makes lots of sense. Thank you! $\endgroup$ Commented Aug 9, 2013 at 21:12
  • $\begingroup$ Nice answer :). $\endgroup$ Commented Aug 11, 2013 at 21:06
  • $\begingroup$ You say "the only time $x^y$ is naturally single-valued is when $y$ is an integer". But what about $2^{\sqrt{2}}$ ? Is that not single-valued? $\endgroup$
    – BIRA
    Commented Aug 28, 2020 at 22:11
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    $\begingroup$ You can define it so it is single-valued on positive reals $x,y,$ but over the complex number $$2^{\sqrt 2}=e^{(2\pi ki+\ln 2)\sqrt 2}$$ where $k$ can be any integer. @bira $\endgroup$ Commented Aug 28, 2020 at 22:20
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The law of exponents $x^{ab} = (x^a)^b$ does not hold in general.

Raising numbers to complex powers is multivalued.

For example, $1^i = e^{\log(1^i)} = e^{i (\ln(1)+2\pi k i)} = e^{i(2\pi k i)} = e^{-2\pi k}$ for any integer $k$. Choosing $k=-1$ matches your left hand side, $k=0$ matches the right.

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