How many solutions has $z^\pi = 1$? I know that for $z \in \mathbb C$ and some natural $n\geq 1$, the equation $z^n = 1$ has exactly $n$ solutions. But what if I say $n$ need not be natural, e.g. 
$$ z^\pi = 1.$$ I mean the equation can't have 3.14-something solutions, can't it? 
 A: The number of solutions is always infinity or a nonnegative integer.  Let $z = re^{i \theta}$, to see that
$$
z^{\pi} =  r^{\pi}e^{i \pi\theta} = 1.
$$
The solutions are then $r = 1$ and $\theta = \dots , -4 , -2, 0, 2, 4 \dots$.   
In particular, $z^{\pi} = 1$ has infinitely many solutions.  More generally, $z^a = 1$ has $m$ solutions if $a = \frac{m}{n}$ and $\gcd(m,n) = 1$, and $z^a = 1$ has infinitely many solutions if $a$ is irrational.
A: For non-integral exponent $\alpha$, the function $x\mapsto x^\alpha$ is only well defined for $x$ restricted to being a positive real number (in which case it is given by $x\mapsto \exp(\alpha\ln x)$). On that domain there is only one pre-image of $1$, namely $1$. The reason this function is generally not considered on the complex numbers, is that different branches of the complex logarithm give different values to the expression $\exp(\alpha\ln z)$, unlike the case where $\alpha\in\mathbf Z$. For irrational $\alpha$ one even has that all branches give different values. If you want to talk about solving $z^\pi=1$, then at least you should consider $z\mapsto z^\pi$ to be a true, single-valued, function, otherwise your "solutions" do not make $1$ the value of $z^\pi$, but just one of the (infinitely many) possible values of $z^\pi$.
So to answer the question, $z^\pi=1$ has only one true solution, namely $z=1$. If you choose the principal branch of the complex logarithm to define $z^\pi=\exp(\pi\ln z)$ on $\mathbf C\setminus\mathbf R_{\leq0}$, then the equation has $3$ complex solutions: $z\in\{1,\exp(2\mathbf i),\exp(-2\mathbf i)\}$.
A: There are (countably) infinitely many solutions. Writing $z=e^{i\theta}$, we get $e^{i \pi \theta} = 1$. This is satisfied whenever $\theta = 2n$ for $n \in \mathbb{Z}$, and since $e^{im} \ne e^{in}$ for $m \ne n \in \mathbb{Z}$, these solutions are all distinct.
The only time you get a finite set of solutions is when you're solving $z^q=1$ for rational values of $q$.
