Prove that $i^i$ is a real number

Question

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

Answer 1

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.

Answer 2

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.

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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?

Answer 3

$i^i$ 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.

Answer 4

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.

Answer 5

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.

Answer 6

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.