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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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    possible duplicate of [How to raise a complex number to the power of another complex number?](http://math.stackexchange.com/questions/9776/how-to-raise-a-complex-number-to-the-power-of-another-complex-number), [Understanding imaginary expoenents](http://math.stackexchange.com/questions/9770/understanding-imaginary-exponents), [Complex exponents](http://math.stackexchange.com/questions/39913/complex-exponents)2012-10-19
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    Why is $e^{\ln 2}$ an integer, when both these numbers are irrational?2012-10-19
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    The important property is that if $x+yi$ is a complex number such that $x^2+y^2=1$ then $(x+yi)^i$ is real for all the values. That's because all the values of the natural logarithm of elements of the unit circle are purely imaginary.2012-10-19
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    @Thomas: Except $1$... :-)2012-10-19
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    Well, depends on whether you consider $0$ to be purely imaginary, I suppose, @AsafKaragila.2012-10-19
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    @Thomas: Well, $0\in\mathbb R$, so no. But I suppose that one can define purely imaginary as $Re(z)=0$, in which case yes.2012-10-19
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    @AsafKaragila Why would you define it any other way? Allowing zero simplifies everything dealing with "purely imaginary" numbers - it makes the set topologically closed and additively closed. You nearly never need to talk about the non-zero imaginary axis as a set.2012-10-19
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    I don't understand how this question got closed because of "exact duplicate": it is not so, at least not of none of the links written by MJD. Sometimes, fro a greenhorn, it's not easy to generalise from some lemma to some particular case.2012-10-19
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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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    Don't forget that $e^{i\pi/2}$ is **not** the only possible representation of $i$: in fact, $i=e^(2k+1)i\pi/2$ for any integer $k$. So in fact, $i^i$ can take the value $e^(2n+1)\pi/2$ for any integer $n$; i.e., it can take infinitely many values, all of which are real. However, if we define the logarithm of $i$ to be the principal value $i\pi/2$, then we do indeed get $e^{-\pi/2}$.2012-10-19
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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