How many solutions does $1=x^π$ have?

Question

I was wondering how many solutions there are to $1 = x^\text{irrational number}$, since the cube root of 1 has 3 solutions and the 4th root has 4 etc and since the number of solutions to $x = x^{a/b}$ is b (where $a$ and $b$ share no factors), how many would $x^π=1$ have? Infinity, none or something else?

Answer 1

By the definition used in the context of complex numbers, the multivaled expression $z^b = \exp(b \log(z))$ where $\log(z)$ is any branch of the logarithm. In this case $\log(1) = 2 n \pi i$ for an integer $i$, so $1^b = \exp(2 n b \pi i)$. If $b$ is irrational, these are all distinct, so there are infinitely many values.

Answer 2

$$1^\pi=e^{\pi(\ln1+2k\pi i)}=e^{i2k\pi^2}\hspace{1cm}k\in\mathbb{Z}$$

Answer 3

We have $$1^\pi = (\mathrm e^{2\pi\mathrm in})^\pi = \mathrm e^{2\pi^2\mathrm in}$$ Now $2\pi^2$ is incommensurable with $2\pi$. Therefore we have a countable infinity of solutions, one for each $n\in\mathbb Z$. In particular, the solutions are dense on the unit circle.