How many roots has $x^a=b$ where $a$ is a real number? I know that $x^a=b$ has $a$ complex roots if $a$ is an entire number. What can I say about $a$ real?
I've tried some Wolfram Calculations with exponent $n \pi$ and had $[n\pi]+1$ complex roots... So I've thought about $[a]+1$, but cannot prove it.
Thank you so much!

if $x=re^{i\phi}$
then:
$$r^ae^{i\phi a}=b$$
since:
$$|e^{i\phi a}|=1$$ be must have that:
$$|r^a|=|b|$$
or:
$$r=|b|^{1/a}$$
now roots will repeat at a point since $e^{iz}$ has a period of $2\pi$ and so you want to find all $\phi$ which satisfy the equation:
$$e^{i\phi a}=1=e^{2\pi i}=e^{0i}$$
in other words, find all unique $\phi$ for integer $k$ such that:
$$\phi a=2k\pi$$

From all of this we get:
$$\phi=\frac{2k\pi}{a},\qquad 0\le\phi\le2\pi$$
Hopefully you can see where this is going
