Inspired by roots of unity, when we can find the remaining roots of the polynomial $x^n - 1$ using complex exponents, I wonder whether there is any similar way to construct the remaining roots of $x^n$ (that is apart from the trivial $x_0 = 0$) using complex exponentiation.
My first thought was that $e^{i \tau} = 0$, but isn't that just saying the same thing? I mean with that logic, wouldn't $1-1$ also be a distinct root?
TL;DR Find all roots of $x^n$.
Thanks in advance.
There are n roots of this equation but all of them are equal. Same like we have in perfect squares,cubes,etc.
$x^n=(x-0)^n$
$\Rightarrow x=0$