I'd like to show that the equation $z^n = w$ for $z,w \in \mathbb{C}$ and $w \not = 0$.
Let $w = r \cdot e^{ix_2} = r_2 \cdot (cos[x_2] + i\cdot sin[x_2]) $ and $z = r_1\cdot e^{ix_1}$
So we have $z^n = w$ if $r_1^n \cdot e^{ix_1n} = r_2 \cdot e^{ix_2}$
But this is true for $r_1^n = r_2$ AND $nx_1 = x_2 + 2\pi k$
With this I should get the n distinct solutions...?
We will show that $a$ has exactly $n$ different roots. Write $z = r' ( \cos \theta + i \sin \theta ) $. Writing in polar form the equation $z^n = a $ gives:
$$ r'^n e^{i \theta n} = r e^{i \varphi} $$
Solving this equation gives $r'^n = r $ and $ i \theta n = i \varphi + i2 \pi k $ for $ k \in \mathbb{Z}_{n-1}$. In particular,
$$ r' = \sqrt[n]{r}, \; \; \; \; \theta = \frac{ \varphi + 2 \pi k }{n} \; \; \; ( 0 \leq k \leq n-1 )$$
Hence, we have found $n$ distinct roots of a complex number $a$ and we can write an explicit formula for these roots:
\begin{equation} \zeta_k = \sqrt[n]{r} \left( \cos \frac{ \varphi + 2 \pi k}{n} + i \sin \frac{ \varphi + 2 \pi k}{n} \right), 0 \leq k \leq n-1 \end{equation}