TITLE: How many solutions has $z^\pi = 1$?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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: $z\in\{1,\exp(2\mathbf i),\exp(-2\mathbf i)\}$.