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Because $x^2=1$ has 2 solutions. $x^3=1$ has 3 solutions in complex numbers. So should $x^{\sqrt{5}}=1$ have 2 and a bit solutions?!!!

Or does it have infinite solutions? If we say the solution to $x^n=1$ is to go a fraction $m/n$ around a circle for any $m$.

If we go a 'fraction' $m/\sqrt{5}$ around a circle for any $m$ this is an infinite number of solution!

What is wrong here? Is there notation of powers self consistent?

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  • $\begingroup$ $\frac{2\pi i k}{n}$ since $k = n $ if $ k \in \mathbb{Z}$ you get finite number of solutions because they will repeat after $k > n$, i.e $k\mod n$. In this case $n=\sqrt{5}$ so you won't find any cycle for $k$ so you have infinite solutions. $\endgroup$ – Anonymous Oct 30 '16 at 17:28
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Added later: I'm leaving this answer as is, for what it's worth. But Christian Blatter's thorough analysis makes it clear that sometimes what seems safe to say is anything but. I jumped too quickly to formal manipulations, overlooking the fact that one must deal first with what is meant by an expression of the form $x^\alpha$ when the power is not an integer. (End of edit.)

Since $1=e^{2\pi in}$ if and only if $n\in\mathbb{Z}$, we can safely say that

$$x^\sqrt5=1\iff x=e^{2\pi in/\sqrt5}=\cos\left(2\pi n\over \sqrt5\right)+i\sin\left(2\pi n\over\sqrt5\right)\quad\text{with }n\in\mathbb{Z}$$

For example, $x\approx-0.9455+0.3256i$ is a solution (corresponding to $n=1$).

As for $x^\sqrt5=1$ having "$2$ and a bit solutions" in analogy to $x^2$ and $x^3$ having $2$ and $3$ solutions, respectively, would you also say that $x^0=1$ has $0$ solutions, or $x^{-1}=1$ has $-1$ solutions?

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    $\begingroup$ But if put n from 1 to infinity you would get any point on the circle to any accuracy you liked! $\endgroup$ – zooby Oct 30 '16 at 18:46
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You can say it has a real solution, because you take logarithm of both sides and it gives you

$$\sqrt 5\log x=0$$

so

$$x=1$$

is a solution.

You can not really talk about the solutions for $x$ negative or complex if you have not defined how you compute powers for complex numbers.

One way of doing that is defining a complex logarithme (which won't be defined on all $\mathbb C$), but there are several definitions for logarithm, and you have to add some context if you want to study the number of solutions on $\mathbb C$.

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  • $\begingroup$ @OscarLanzi could you explain why? $\endgroup$ – I'mAnAccountantIKnowAlotOfMath Oct 30 '16 at 19:11
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    $\begingroup$ I like this answer best of all, ’cause it makes clear the role of opinion in the formulation of any answer. You may not like my definition of logarithm, after all. $\endgroup$ – Lubin Oct 31 '16 at 19:57
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Before we can look at your equation we have to make sense of the term $x^{\sqrt{5}}$. For rational exponents ${p\over q}$ we can define $x^{p/q}:=\root q\of {x^p}$. As $\sqrt{5}$ is irrational we would have to choose a sequence $(r_n)_{n\geq1}$ of rational numbers that converges to $\sqrt{5}$, and hope that the limit $\lim_{n\to\infty} x^{r_n}=:x^{\sqrt{5}}$ exists, and is independent of the chosen sequence.

Since some limiting procedure is called for anyway one takes a more elegant route to powers with arbitrary real (or complex) exponents. After introducing the exponential function $\exp:\>{\mathbb R}\to{\mathbb R}_{>0}$ and $\log:\>{\mathbb R}_{>0}\to{\mathbb R}$ one defines $$x^\alpha:=\exp(\alpha\log x)\qquad (x>0,\quad\alpha\in{\mathbb R})\ .\tag{1}$$ In this way $(x,\alpha)\mapsto x^\alpha$ is continuous in its variables $x$ and $\alpha$ wherever defined, and obeys the well known rules of "power algebra". If we want $x^\alpha=1$ in this context this implies $\alpha\log x=0$, or $(\alpha=0)\vee(\log x=0)$. As $\alpha=\sqrt{5}$ in the example at hand we are left with $\log x=0$, hence $x=1$, as only solution.

In the complex environment the defining formula $(1)$ is extended to $$z^\alpha:={\rm pv}\bigl(z^\alpha\bigr):=\exp\bigl(\alpha\,{\rm Log}(z) \bigr)\qquad (z\in\Omega,\quad\alpha\in{\mathbb C})\ ,$$ where $\Omega$ denotes the complex plane with the negative real axis removed. Any $z\in\Omega$ can be written in the form $$z=r\,e^{i\phi},\qquad r>0,\quad |\phi|<\pi\ ,$$ and one has $${\rm Log}(z):=\log r+i\phi\ .$$ The given equation then amounts to $$\sqrt{5}(\log r+i\phi)=2k\pi i, \qquad k\in{\mathbb Z},\quad|\phi|<\pi\ .$$ This leads to $\log r =0$, hence $r=1$, and to $$\phi={2k\over\sqrt{5}}\pi\ .$$ The condition $|\phi|<\pi$ then implies that necessarily $k\in\{-1,0,1\}$, so that we obtain three solutions in all.

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  • $\begingroup$ Are you saying that $x=e^{4\pi i/\sqrt5}$ is not a solution to $x^\sqrt5=1$? $\endgroup$ – Barry Cipra Oct 30 '16 at 17:36
  • $\begingroup$ @BarryCipra: Tell me your definition of $x^{\sqrt{5}}$. – If $x^{\sqrt{5}}$ is a set valued function, to be defined properly, then $x^{\sqrt{5}}=1$ makes no sense, only $1\in x^{\sqrt{5}}$ does. In this case the envisaged set is dense in $S^1$. I don't know whether the OP had this in mind. $\endgroup$ – Christian Blatter Oct 30 '16 at 18:38
  • $\begingroup$ You are quite right. I clearly need to revise my answer! (I'll do so when I have a bit more time.) $\endgroup$ – Barry Cipra Oct 30 '16 at 21:36
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In the complex numbers, we define $x^a=e^{a \log x}$ The logarithm is multivalued because $e^{2 \pi i}=1$, so $e^b = e^{2n\pi i+b}$ for any integer $n$. Given that $x=1$ is a solution, then $x=1+\frac {2n\pi i}{ \sqrt 5}$ is also a solution for any integer $n$.. Yes, there are an infinite number of solutions.

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    $\begingroup$ $(1+2n\pi i)^{\sqrt5}=1$???? $\endgroup$ – Barry Cipra Oct 30 '16 at 13:56
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    $\begingroup$ The edit is still wrong. Did you drink decaf by mistake this morning? ;-) $\endgroup$ – Barry Cipra Oct 30 '16 at 15:27

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