Complex $n^{th}$ roots of a complex number. How would one find all the solutions to the equation
$$a=z^{1/n}$$
where $a, z, n \in \Bbb C$?
Also how many roots does this number have?

Work So Far
If $z$ can be written as
$$r_ze^{i\theta_z}$$
then the $n^{th}$ root is
$$r_z^{1/n}e^{i\theta_z/n}$$
I can't work out how to simplify this any further as I have never dealt with complex radicals before.
EDIT:
Further guessing:
$$n = x + iy$$
$$
\begin{align}
a &= z^{1/n}\\
&= r^{1/n}e^{i\theta/n}\\
&=\exp\left(\frac{\ln r+i\theta}n\right)
\end{align}
$$
 A: To solve $a = z^{1/n}$ where $n, z \in \mathbb{C}$. Note for complex $n$ the natural interpretation of the $1/n$-th power is that $z^{1/n} = \exp \frac{1}{n} \log z$. Suppose $z = re^{i\theta}$ where $r,\theta \in \mathbb{R}$ then $\log (re^{i\theta}) = \ln r+ i(\theta+2\pi k)$ for $k \in \mathbb{Z}$ (it is a set of values, indeed so is $z^{1/n}$) thus,
$$ z^{1/n} = \exp \left( \frac{1}{n} \left[\ln r+ i(\theta+2\pi k) \right]\right) $$
To go further, I need to break $n$ into real and imaginary parts, let $n= \eta+i\xi$ hence $\frac{1}{n} = \frac{\eta-i\xi}{\eta^2+\xi^2}$ then multiply,
$$ \frac{\eta-i\xi}{\eta^2+\xi^2}\left[\ln r+ i(\theta+2\pi k) \right] = 
\underbrace{\frac{\eta \ln r +\xi(\theta+2\pi k)}{\eta^2+\xi^2}}_{A} - i\underbrace{\frac{\xi\ln r+\eta(\theta+2\pi k)}{\eta^2+\xi^2}}_{-B} $$
Thus,
$$ z^{1/n} = \exp(A+iB) = e^A \left( \cos B+i\sin B \right) $$
As $k$ ranges over $\mathbb{Z}$ we could have many values. Clearly, $\xi=0$ will be needed to have a finite set of roots. Otherwise, the $e^A$ term will continue to vary as we range over the integers. If $\xi=0$ and $\eta$ is irrational, then your still looking at infinitely many solutions because $B$ lacks the needed periodicity. I think it is possible that no root exists, I saw another question recently, I'll add a comment when I find it.
A: It seems you got it.
The only thing might be missing is that there are other solutions as well, because the same $z$ can also be written as $z=r\cdot e^{i(\theta+2k\pi)}$ for any $k\in\Bbb Z$.
So, can you find all the $n$ solutions?
A: If you want the exponent really to be a complex number, then you are, so to speak, asking for trouble. Your comments show that you realize that there is a problem, stemming from the fact that the logarithm is a “multi-valued function”. Let me rephrase your question:
You ask to solve $a=z^{1/n}$, and because $n$ is complex, I’m going to write it $\beta$ instead. It seems that you’re aware that solving your equation is the same as finding all values of $a^\beta=z$. But this too you know, because by definition, $a^\beta=\exp(\beta\log a)$. The problem is with the logarithmic term: if $L$ is one complex number such that $\exp L=a$, the others all are of the form $L+2k\pi i$. Now it all falls into place, your solutions all are of the form $\exp(\beta(L+2k\pi i))$, $L$ being any one of the infinitely many numbers such that $\exp(L)=a$.
