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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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  • $\begingroup$ 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) $\endgroup$ – Simon Goodwin Jan 26 '17 at 21:12
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$$1^\pi=e^{\pi(\ln1+2k\pi i)}=e^{i2k\pi^2}\hspace{1cm}k\in\mathbb{Z}$$

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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.

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