As I understand it, $i^i$ returns more than one real number. Why does this happen? When I raise $i$ to $i$, I'm raising a specific number to itself. So why do I get multiple values of this, as if it was, for lack of a better word, some sort of a "function"?
EDIT: My question wasn't "why is $i^i$ real". I know it's real. Furthermore, I know how to calculate $i^i$. I was just looking for intuition as to why this number has multiple values.
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$\begingroup$There can be multiple values in the same sense that the equation $x^2=1$ yields several square roots ($\pm1$).
But we use the convention that $\sqrt x$ is a positive number, i.e. there is a preferred "branch".
$$\sqrt1=1.$$
Similarly, the complex logarithm has a preferred branch, such that $\log i=i\dfrac\pi2$, allowing us to write
$$i^i:=e^{-\pi/2}.$$
With other branches you would get
$$e^{-\pi/2-2k\pi}.$$
$\endgroup$ 3 $\begingroup$To calculate $i^i$, you need to go via the logarithm. But there’s a problem with the complex logarithm function, as several of the other answerers have pointed out.
The real logarithm is the inverse of the exponential function $x\mapsto e^x$, which is one-to-one, as you see by looking at its graph. To define the complex logarithm, we would want to make it the inverse of the complex exponential, $z\mapsto e^z$. But this function is not one-to-one, as you see from the equation $1=e^0=e^{2\pi i}$, so the complex exponential doesn’t have an inverse. Something radical needs to be done to define a logarithm function: as @YvesDaoust makes clear, a choice must be made to do so.
That is to say that there is no the logarithm, once you move into the complex world. As my grandmother liked to say, You pays your money and takes your choice. Most of us make a choice leading to $i^i=e^{-\pi/2}$, but another may make a choice leading to another value of $i^i$, and none of us can deny the rightness of such choice.
$\endgroup$ $\begingroup$The other answers focus on the calculation, but your problem might lie in the concepts itself. Reflect on what it means to raise a number by a certain power. For natural numbers, this is for a lack of better words a "natural" operation, that is for $n\in \mathbb{N}$, you can just define
$$x^n := \underbrace{x \cdot x \cdot \ldots \cdot x}_{n-\text{times}}$$
For negative integers, you need to exclude $x=0$, but then you can define $x^{-n} := 1/x^n$. However when you try to extend this to more exponents, things get progressively harder. For rational numbers you can define $x^\frac{p}{q} := \sqrt[q]{x}^p$, but then you need to consider what the $q$-th root of a number means. This you would define as the solution $y$ of $y^q = x$, but this already will not be unique, especially in the complex numbers. The convention here is to use the unique positive real solution, for which however $x$ needs also to be a positive real number.
You can extend this continuously to real numbers, but essentially this is the end of the line. If you want to raise a complex number to a complex power, there is no way to multiply $x$ with itself $i$-times. The only reasonable way is to say that the equation $$x^a =\exp(\log(x)a)$$ holds for all $x >0$ and $a \in \mathbb{R}$. So you simply try take it as the definition of exponentiation on complex numbers. However then you exactly arrive at the problem that the logarithm is no longer well defined on the complex numbers, as $\exp(2\pi i) = \exp(0) =1$ and thus there is no inverse of the exponential function. So here you will have to take into account all possible solutions $y$ of $x = \exp(y)$ in place of $\log x$, which then leads to the usual calculation.
$\endgroup$ $\begingroup$First you need to realize that this is a multi-valued function.
Let us choose the principal logarithm for convenience. So our argument $\theta$ will lie between $(-\pi,\pi]$.
We then write $\text{a}+\text{b}\cdot i=\text{R}\cdot\exp\left(\theta\cdot i\right)=\exp\left(\ln\left(\text{r}\right)+\theta\cdot i\right)$.
Now, we get $\left(\text{a}+\text{b}\cdot i\right)^{\text{c}+\text{d}\cdot i}=\exp\left(\left(\ln\left(\text{r}\right)+\theta\cdot i\right)\cdot\left(\text{c}+\text{d}\cdot i\right)\right)$. Do the necessary algebraic manipulations in the exponent to get $\exp\left(\left(\text{c}\cdot\ln\left(text{r}\right)-\text{d}\cdot\theta\right)+\left(\text{d}\cdot\ln\left(\text{r}\right)+\text{c}\cdot\theta\right)\cdot i\right)$. You might also want to take a look at the previous question asked on a similar topic.
$\endgroup$ 4 $\begingroup$The expression $i^i$ is just a short form of $\exp(i\log (i))$. We can compute this as soon as we know what $\log(i)$ is. Note that $\log(i)$ is
$$\text{the answer to the question which $z$ satisfies $\exp(z)=i$.}$$
The problem is that there is no single answer. So $\log(i)$ could be assinged more than one meaningful value. But we do not do this. Instead we just choose one, i.e. $i\pi/2$. Now the value of $i^i$ is unique. But since this process involves a more or less arbitrary choice I usually do not like to write $a^b$ with anything else than positive real exponents (for general basis).
The same problem appears when computing roots, e.g. $\sqrt{1}$. It is the answer for the question which $z$ satisfies $z^2=1$. And there are two values. We just agreed to always use the positive one.
$\endgroup$ $\begingroup$Remeber that $i$ can be written as $e^{i (\pi/2 + 2\pi k)}$ $\forall k \in \mathbb{N}\cup \{0\}$
thus $i^i = (e^{i(\pi/2 + 2 \pi k})^i = e^{-(\pi/2 + 2\pi k)} = \{{k = 0}\} = e^{- \pi/2}$
But of course you can choose other $k$, if you want a deeper explanation look at Jan Eerlands answer.
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