Why does $z^{1/n}$ have exactly $n$ distinct values for $z \in \mathbb{C}$?

Question

In complex analysis, why the set of distinct values of $z^{1/n}$ contains exactly $n$ different values? For example why is $z^{1/3}$ have exactly 3 different roots? I thought that since $$ z^{1/n} = r^{1/n}e^{\frac{i \theta}{n}}$$ implied that there are infinitely many values for different $\theta$.

Answer 1

$$e^{i\theta/n} = \cos(\theta/n) + i \sin(\theta/n)$$ is a periodic function with period $2\pi n$.

For a particular input value $\theta$, one gets $n$-distinct output values of that function by replacing $\theta$ with $\theta + 2\pi k$ for $k=0,...,n-1$.

Answer 2

For fixed z, it has exactly n values, because $\theta$ only changes modulo $2 \pi$(and then you divide by n so only for n distinct multiples of $2\pi$ will you get values of $\frac{\theta}{n}$ inside $[0,2\pi)$

Answer 3

We may assume without loss of generality that $z=1$, i.e. we are asking for the $n$-th roots of unity. These complex numbers are the roots of the polynomial $X^n-1$. A polynomial of degree $n$ over the complex numbers has exactly $n$ roots. They are different, because the roots of unity form a finite subgroup of $(\Bbb{C},\cdot)$, which has to be cyclic then. So the roots of unity are given by $1,\zeta,\zeta^2,\cdots \zeta^{n-1}$, which are pairwise distinct.