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Possible Duplicate:
How to raise a complex number to the power of another complex number?

My calculator (as well as WolframAlpha) gives me the approximation:

$0.2078795763507619085469...$

But I don't understand how exponentiating two purely imaginary constructs yields a real (albeit irrational) number. When I do $i^{i+1}$ it gives me an imaginary number as well as $(i+1)^i$. So why does $i^i$ fall into that precise point where it is real and no longer imaginary? What is happening? I understand that exponentiation is not repeated multiplication, and it wouldn't make sense to multiply $i$ by itself $i$ times (because it would only yield $i$, $-i$, $1$, or $-1$). So what are we doing behind the scenes to get such a number?

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    This question is an exact duplicate of http://math.stackexchange.com/questions/191572/prove-that-ii-is-a-real-number2012-10-19

1 Answers 1

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Using Euler's formula:

$ i = e^{i\pi / 2} $

So:

$ i^i = (e^{i\pi / 2})^i = e^{i^2\pi/2} = e^{-\pi/2} = 0.207... $

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    It's worth noting that $i = e^{i\pi(2kn+\frac{1}{2})}$ for any $n$, so really, it depends on which branch of the natural logarithm you use.2012-10-19