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I came across this 'paradox' - $$1=e^{2\pi i}\Rightarrow 1=(e^{2\pi i})^{2\pi i}=e^{2\pi i \cdot 2\pi i}=e^{-4\pi^2}$$

I realized the fallacy lies in the fact that in general $(x^y)^z\ne x^{yz}$. Why doesn't it work with complex numbers even though it is valid in real case? Is it related to the fact that logarithm of complex number is not unique?

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Yes, it is precisely the complex logarithm that causes these problems. This is discussed in a previous question: math.stackexchange.com/questions/1211/… –  Qiaochu Yuan Aug 21 '10 at 0:27
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(Maybe I should summarize my comments from the other question. The point is that requiring (x^y)^z = x^{yz} is equivalent to forcing different branches of the complex logarithm to agree, which... they don't.) –  Qiaochu Yuan Aug 21 '10 at 1:05
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@Qiaochu if instead of taking branches you could interpret $a^b$ as a set (generally countably-infinite), you could ask when the sets $(a^b)^c$ and $a^{bc}$ are equal. I sometimes give this as an exercise to my complex variables classes. The answer is kind of pleasant to work out. –  Ryan Budney Aug 22 '10 at 4:12

2 Answers 2

up vote 15 down vote accepted

Even without any complex numbers: $-1=(-1)^{2\cdot\frac12}\ne((-1)^2)^{\frac12}=1^{\frac12}=1$.

But you're right, the problem is that raising to a (non-integer) power is essentially a multivalued function.

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Did not notice that. So it looks like the only place where that can be applied is when we are dealing with only positive numbers! –  KalEl Aug 21 '10 at 7:02
    
So, I guess it would be more precise to write $1\in e^{2\pi i}$. Isn't it? –  Pantelis Sopasakis Mar 28 '11 at 23:35
    
The "only place where that can be applied is when we are dealing with only positive" bases. $\hspace{.89 in}$ The exponents don't even need to be real for the usual rules to hold. $\:$ –  Ricky Demer Jun 15 '13 at 9:05

This is related to a note by Euler, maybe he was the first to realize that $i^i$ is real. Actually, $$i^i = (e^{i\pi/2})^i = e^{-\pi/2}$$ on the other hand $$i^i = (e^{i(\pi/2+2\pi n})^i = e^{-\pi/2 -2\pi n},\ n\in\mathbb{Z}.$$ So maybe it is better to say that $i^i$ is a subset of the $\mathbb{R}$ and that certain equality signs are to be understood as congruences.

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