Show that $$\sqrt[n]{z}=\{ |\sqrt[n]{z}| \cdot\left(\cos\left(\frac{\theta}{n}\right) +i \sin\left( \frac{\theta}{n} \right) \right) \zeta^{k} \mid k= 0,1,\ldots,n−1\}.\tag{1}$$
My attempt:
Let $z=r(\cos{ \theta }+i\sin\theta )$ and let $w = \rho \left( cos{ \phi } + i \sin{ \phi } \right)$ and $\theta = arg\left(z\right)$.
Note by De Moivre's Formula for the polar representation of powers of
complex numbers,
$z^{n} = r^{n}\left( \cos{\theta n} + i\sin{\theta n} \right)$
Now to find $z^{n} = w$, we have,
$z^{n} = r^{n}\left( \cos{\theta n} + i\sin{\theta n} \right) = \rho \left( cos{ \phi} + i \sin{ \phi} \right) = w$.
Thus, we have $r^{n} = \rho$ and $\theta n = \phi +2 \pi k$ for some $k \in \mathbf{Z}$.
When we solve for $r$ and $\theta$, $r= \sqrt[n]{\rho}$ and $\theta =\frac{\phi + 2\pi k}{n}$.
Therefore, $z= \sqrt[n]{\rho}\left( \cos{\frac{\phi + 2\pi k}{n}} + i\sin{\frac{\phi + 2\pi k}{n}} \right)$,
which is satisfied by all $k = \{0,1,2,...,n-1 \}$ and equivalent to (1).
Is there anyway that this can be improved?
