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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?

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    You should write $z=\rho e^{i\theta}$ and so $z^\pi=\rho^\pi e^{i\pi\theta}$. This implies immediately $\rho=1$ and $\theta = 2k$ being $k\in\mathbb{Z}$.2012-01-13

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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.

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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: $z\in\{1,\exp(2\mathbf i),\exp(-2\mathbf i)\}$.

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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 $q$.

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    It is a bit tricky: note that $z^\pi$ itself is a multivalued function. Typically one defines it via the principle branch of the logarithm as $e^{\pi \log z}$. As |\mathop{\rm Im}\log z | <\pi there are only 3 distinct solutions. (if you take your favorite calculator and evaluate $(e^{4i})^\pi$ you won't get 1!)2012-01-14