What does $\sqrt[n]{z}$ mean, when $z$ is an arbitrary complex number? What does $\sqrt[n]{z}$ mean, when $z$ is an arbitrary complex number?  Is it a single complex number, or the set of $n$-th roots of $z$?
 A: When I see that symbol, I expect it to mean a single $n^{th}$ root of $z$. The problem is, without additional context, I don't know which one is meant! The symbol is not well-defined, since there are indeed $n$ distinct $n^{th}$ roots of $z$ if $z$ is not $0$. Usually we define a function sending complex numbers to one of their $n^{th}$ roots, for example by sending $re^{i\theta}$ to $\root n \of re^{i\theta/n}$, where $\theta \in [0,2\pi)$. But notice that this function is not continuous on the positive real line. (The positive real axis is said to be a branch cut for this function. We could in fact use any ray from the origin as a branch cut to define an unambiguous $n^{th}$ root function.) After explaining how the function is defined, one is then free to (ab)use the notation $\root n \of z$.
A: This symbol is for the set  of all $n-th$ root of $z$.
$$z= r(\cos\theta+i\sin\theta)$$
For $k= 0,1,,,n-1$, we have:
$$z^\frac{1}{n}=\{ r^{\frac{1}{n}}[\cos(2\pi k +\theta)+i\sin(2\pi k+\theta)]^\frac{1}{n}| k=0,1,..n-1\}$$
that is 
$$z^\frac{1}{n}=\{ r^{\frac{1}{n}}[\cos\frac{(2\pi k +\theta)}{n}+i\sin\frac{(2\pi k+\theta)}{n}]| k=0,1,..n-1\}$$
