# How many solutions does $1=x^π$ have? [duplicate]

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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?

## marked as duplicate by Jack D'Aurizio, Lucian, Dietrich Burde, Community♦Jan 26 '17 at 22:53

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• See the comment here. – Dietrich Burde Jan 26 '17 at 20:59
• $1^\pi$ has no solutions, since $1^\pi$ is a number, not an equation. – Jack D'Aurizio Jan 26 '17 at 20:59
• I edited the question since obviously the intention was to write $x^\pi = 1$ – ThomasR Jan 26 '17 at 21:02
• Yes, of course, my title was badly worded, sorry about that – Simon Goodwin Jan 26 '17 at 21:08

## 3 Answers

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.

• OK, thanks, as a follow up question are any of these solutions -1, i or -i? And if I was instead to adjust this to (-1)^n = x, are there any purely real solutions? (Where n is irrational) – Simon Goodwin Jan 26 '17 at 21:12

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

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.