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What is the value of 1^i?

Note that I am absolutely not a mathematician, so this may be silly, but I saw this on Wikipedia's page about $i$:

One definition of $i^i$ is : $i^i = \left( e^{i (2k \pi + \pi/2)} \right)^i = e^{i^2 (2k \pi + \pi/2)} = e^{- (2k \pi + \pi/2)}$ where $k \in \mathbb{Z}$.

The principal value (for $k=0$) is $e^{- \pi/2} $ or approximately $0.207879576350761908546955...$

But if $i^i = i^i$, and I assume it is, then isn't, for example, the following true: $$e^{-(2\times0 \pi + \pi/2)} = e ^{-(2\times4 \pi + \pi/2)}$$

And doesn't than mean that: $$0.207879576350761908546955 \approx 2.5281392565177714\times10^{-12}$$

And I don't think that last one is really true.

Or does the problem lie in my assumption that $i^i = i^i$ is always true?

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...and math.stackexchange.com/questions/3668/what-is-the-value-of-1i looks like exactly what you need –  Grigory M Jun 29 '11 at 11:32
    
Okay, so I have to look at that complex logarithm branch thing. Some day. In the meantime, to compensate for this (fun) mathematical paradox, I am going to amuse myself for hours by bouncing a ball off a wall. Ha ha! Fun! –  eje211 Jun 29 '11 at 11:44
    
If you really want to understand, what's going on, then yes, I'm afraid. But short version is, there is no such thing as the value of $i^i$, it has many different values. –  Grigory M Jun 29 '11 at 13:14
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marked as duplicate by Grigory M, Chandru, t.b., ShreevatsaR, Asaf Karagila Jun 29 '11 at 12:02

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1 Answer

For those who don't like to follow the links: For real numbers x and y, the standard definition is $$ x^y := \exp(y \; \log(x)) $$ For complex numbers one has to be careful because the logarithm is not defined on $\mathbb{C}$, so one has to specify which branch one is using. First, we get $$ i^{i} = \exp(i \log(i)) $$ Since we have $$ e^{(2 \pi k + \pi /2) i} = i $$ picking the main branch value for $log(i)$ we get $$ i^{i} = \exp(i (\pi /2) i) = \exp(- \pi / 2) $$ which should also suffice to point out where the error in the wikipedia article is.

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