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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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    $\begingroup$ Do you know the definition of $i^i$? $\endgroup$ – Jack Sep 5 '12 at 18:22
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    $\begingroup$ A pedantic point: is a complex number with a 0 imaginary part the same as a real number? $\endgroup$ – James Sep 6 '12 at 12:11
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    $\begingroup$ @James: Unless you know some secret that I don't, yes it is. $\endgroup$ – Cameron Buie Sep 6 '12 at 20:58
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    $\begingroup$ @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 $1 < 2$? $\endgroup$ – James Sep 7 '12 at 7:42
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    $\begingroup$ @James Yes, if you interpret $<$ to be by lexicographical order. It suffices to be an order relation on the set $\{a+bi \mid a \in \Bbb R, b = 0\} \subset \Bbb C$. $\endgroup$ – Emily Sep 7 '12 at 13:41
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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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    $\begingroup$ 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. $\endgroup$ – Isaac Sep 12 '13 at 19:48
  • $\begingroup$ Relevant: tauday.com/tau-manifesto $\endgroup$ – user117644 Feb 22 '16 at 7:48
  • $\begingroup$ For your shorter version, how can you be sure that exponentiation rules apply for complex powers? Specifically, how do you know that (e^a)^i=e^(ia)? $\endgroup$ – Patrick Cook Apr 21 '18 at 22:23
  • $\begingroup$ @PatrickCook: You prove it ;) It's an immediate consequence of the more general fact that $\exp(zw) = \exp(z)^w$, which isn't too difficult to prove using exponential series. $\endgroup$ – Clive Newstead Apr 21 '18 at 22:30
  • $\begingroup$ @CliveNewstead For z,w in the real numbers sure. But is it really true that (z^i)(w^i)=(zw)^i? Which would be a consequence of what you're saying here. The reason I ask is because I am dealing with (i^i)^4 = i^(4i) = (i^(2))^(2i) = ((-1)^(2))^i = 1^i = 1. So the problem here is then if i^i = e^(-pi/2) and (i^i)^4=1, then e^(-2pi)=1 which is obviously untrue. $\endgroup$ – Patrick Cook Apr 22 '18 at 0:50
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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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    $\begingroup$ It's... it's beautiful! Just a question: Does complex conjugate of $(a+ib)^{(c+id)}$ equal $(a-ib)^{(c-id)}$? $\endgroup$ – Isaac Sep 6 '12 at 16:28
  • $\begingroup$ excellent answer. I just loved it $\endgroup$ – M. A. SARKAR Feb 2 at 14:01
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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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  • $\begingroup$ How is it determined that $(e^{i\pi/2})^i=e^{i^2\pi/2}$? $\endgroup$ – Jonas Meyer Feb 12 '17 at 5:29
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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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    $\begingroup$ All of those are real $\endgroup$ – Henry Sep 5 '12 at 22:26
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    $\begingroup$ How is it possible that a wrong answer got 7 (now 6) upvotes?! I think this is quite insulting for all the high-quality answers to "low-profile" questions that can hope to reach 2 or 3 upvotes at most. $\endgroup$ – Giovanni De Gaetano Oct 24 '12 at 11:24
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    $\begingroup$ @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? $\endgroup$ – N.U. Oct 24 '12 at 15:32
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    $\begingroup$ @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. $\endgroup$ – Stephen Apr 1 '13 at 12:56
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    $\begingroup$ 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. $\endgroup$ – Giovanni De Gaetano Apr 5 '13 at 13:53
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This would come right from Euler's formula. Let's derive it first. There are many ways to derive it though, the Taylor series method being the most popular; here I’ll go through a different proof. Let the polar form of the complex number be equal to $z$ . $$\implies z = \cos x + i\sin x$$ Differentiating on both sides we get,

$$\implies \dfrac{dz}{dx} = -\sin x + i\cos x$$

$$\implies dz = (-\sin x + i\cos x)dx$$ Integrating on both sides,

$$\implies \displaystyle \int \frac{dz}{z} = i \int dx$$ $$\implies \log_e z = ix + K$$ Since $K = 0$, (Set $x = 0$ in the equation), we have, $$\implies z = e^{ix}$$ $$\implies e^{ix} = \cos x + i\sin x$$ The most famous example of a completely real number being equal to a real raised to an imaginary is $$\implies e^{i\pi} = -1$$ which is Euler’s identity. To find $i$ to the power $i$ we would have to put $ x = \frac{\pi}2$ in Euler's formula. We would get $$e^{i\frac{\pi}2} = \cos \frac{\pi}2 + i\sin \frac{\pi}2$$ $$e^{i\frac{\pi}2} = i$$ $${(e^{i\frac{\pi}2})}^{i} = i^{i}$$ $$i^{i} = {e^{i^{2}\frac{\pi}2}} = {e^{-\frac{\pi}2}}$$ $$i^{i} = {e^{-\frac{\pi}2}} = 0.20787957635$$ This value of $i$ to the power $i$ is derived from the principal values of $\sin$ and $\cos$ which would satisfy this equation. There are infinite angles through which this can be evaluated; since $\sin$ and $\cos$ are periodic functions.

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Let say: $f(z) = z^\theta $
We know Euler's formula: $e^{i \theta} = \cos(\theta) + i \sin(\theta)$
Using it we will get: $$z^\theta = e^{\theta \ln(z)} = e^{\theta (\ln\|z\| + i\arg (z))} = e^{\theta \ln\|z\| }e^{ i \theta \arg (z))}= e^{\theta \ln\|z\| }{(\cos(\theta\arg (z)) + i \sin(\theta\arg (z)))} z \in C$$

So if $$z = i \wedge \theta = i \implies$$ $$ z^\theta = i^i = e^{i \ln(i)} = e^{i (\ln\|i\| + i\arg (i))} = e^{i \ln\|i\| }e^{ i i \arg (i))}= e^{i \ln(1) }e^{- \frac{\pi}{2}+2\pi k}= e^0 e^{- \frac{\pi}{2}+2\pi k}=e^{- \frac{\pi}{2}+2\pi k} $$ which is a bunch of Real numbers depending $k \in \mathbb Z$

So it is already proved that $i^i$ is a real number.

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$i^i=x$ so $i^{(i^2)}=x^i$ So $\frac{1}{i} = x^i$ So $-i=x^i$ so $i = -(x^i)$ So $i^i = -x^{(-ix)^i}$

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