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According to WolframAlpha, $i^i=e^{-\pi/2}$ but I don't know how I can prove it.

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To add to the answer below, the morals of the story is that once you represent $i$ as a point on the unit circle, then you can add $\pm 2\pi$ as many times as you like, and still get the same point, (i.e. revolve around as many complete cycles as you like). –  user2468 Sep 5 '12 at 18:12
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Do you know the definition of $i^i$? –  Jack Sep 5 '12 at 18:22
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A pedantic point: is a complex number with a 0 imaginary part the same as a real number? –  James Sep 6 '12 at 12:11
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@James: Unless you know some secret that I don't, yes it is. –  Cameron Buie Sep 6 '12 at 20:58
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@CameronBuie I agree for most practical purposes you don't need to distinguish but formally, since complex numbers and reals have different properties, do you have to do an intermediate conversion? For instance, can you assert $1+0i < 2+0i$ in the same way you can assert $1 < 2$? –  James Sep 7 '12 at 7:42

4 Answers 4

up vote 22 down vote accepted

Write $i=e^{\frac{\pi}{2}i}$, then $i^i=(e^{\frac{\pi}{2}i})^i = e^{-\frac{\pi}{2}} \in \mathbb{R}$. Be careful though, taking complex powers is more... complex... than it may appear on first sight $-$ see here for more info.

In particular, it's not well-defined (until we make some choice that makes it well-defined); we could just have well written $i=e^{\frac{5\pi}{2}i}$ and obtained $i^i=e^{-\frac{5\pi}{2}}$. But $i^i$ can't be equal to both $e^{-\frac{\pi}{2}}$ and $e^{-\frac{5\pi}{2}}$ can it?

Despite the lack-of-well-defined-ness, though, $i^i$ is always real, no matter which '$i^{\text{th}}$ power of $i$' we decide to take.


More depth: If $z,\alpha \in \mathbb{C}$ then we can define $$z^{\alpha} = \exp(\alpha \log z)$$ where $\exp w$ is defined in some independent manner, e.g. by its power series. The complex logarithm is defined by $$\log z = \log \left| z \right| + i\arg z$$ and therefore depends on our choice of range of argument. If we fix a range of argument, though, then $z^{\alpha}$ becomes well-defined.

Now, here, $z=i$ and so $\log i = i\arg i$, so $$i^i = \exp (i \cdot i\arg i) = \exp (-\arg i)$$ so no matter what we choose for our range of argument, we always have $i^i \in \mathbb{R}$.

Fun stuff, eh?

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I am going to choose this answer as accepted answer, but there are some other answers that are very insightful and worth to take a look. –  Isaac Sep 12 '13 at 19:48

Using the representation that $i = e^{i \pi/2}$, we have $i^i = \left(e^{i\pi/2}\right)^i = e^{i^2\pi/2} = e^{-\pi/2}$.

$i = e^{i\pi/2}$ comes from the representation that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, which for $\theta = \pi/2$ gives us $e^{i\pi/2} = \cos \pi/2 + i \sin \pi/2 = 0+i\cdot 1 = i$.

Edit: To add to the other fantastic answers/comments, this is the result on the principal branch. Others have commented that you can equivalently represent $i = e^{i(2k+1/2)\pi}$ and obtain other real-valued answers for $i^i$. Wolfram Alpha gives you $e^{-\pi/2}$ because its default setting is to return the principal value.

Edit again:

It may seem weird that we resort to this "out of nowhere" polar representation of complex numbers, but it is a powerful tool.

Over the reals, the concept that "exponentiation = repeated multiplication" breaks down when you have non-integer exponents, so you have to start defining exponentiation using suprema of sets, which exploits the ordered field nature of the reals.

The complex field is not an ordered field, so the equivalent notion of a supremum doesn't exist. So how do we take any number to the power $i$, let alone a complex number? The polar representation allows us to deal with this issue in a rather clever fashion.

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$i^i$ takes infinitely many values.

$$ i^i = e^{i \log i} = e^{i(i\pi/2 + 2 \pi i m)} = e^{-\pi/2}e^{-2 \pi m}, $$ where $m$ is an integer.

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All of those are real –  Henry Sep 5 '12 at 22:26
    
No, it can't possibly take on multiple values, its not a multi-function (its merely A number). You are introducing a "magical" step much in the way one may prove that 1=-1. –  Squirtle Sep 12 '12 at 6:27
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@GiovanniDeGaetano Then please tell me what is wrong. This is straight out of "Complex Analysis" by Gamelin (page 24). Are you thinking about the principal value? –  N.U. Oct 24 '12 at 15:32
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@GiovanniDeGaetano, dustanalysis, actually, this is the best answer out of the lot. It's somewhat embarrassing for the site that it has so few votes. –  S123 Apr 1 '13 at 12:56
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Wooops! You are perfectly right. I cannot justify my comment in any way, it would have been enough to stop and reflect before downvoting. I can only apologize with the community and specifically with N.U. for it, and thank Steve for tagging me here. Unfortunately now it's too late to remove the downvote. –  Giovanni De Gaetano Apr 5 '13 at 13:53

Here's a proof that I absolutely do not believe: take its complex conjugate, which is $\bigl({\bar i}\bigr)^{\bar i}=(1/i)^{-i}=i^i$. Since complex conjugation leaves it fixed, it’s real!

EDIT: In answer to @Isaac’s comment, I think that to justify the formula above, you have to go through exactly the same arguments that most of the other answerers did. For complex numbers $u$ and $v$, we define $u^v=\exp(v\log u)$. Now, the exponential and the logarithm are defined by series with all real coefficients; alternatively you can say that they are analytic, sending reals to reals. Thus $\overline{\exp u}=exp(\bar u)$ and $\overline{\log(u)}=\log\bar u$. The result follows, always sweeping under the rug the fact that the logarithm is not well defined.

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$+1 {}{}{}{}{}$ –  user2468 Sep 5 '12 at 23:18
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Its a proof that belongs in "the book". –  Squirtle Sep 6 '12 at 1:57
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It's... it's beautiful! Just a question: Does complex conjugate of $(a+ib)^{(c+id)}$ equal $(a-ib)^{(c-id)}$? –  Isaac Sep 6 '12 at 16:28

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