How to prove $ z^n - z^n_0 = (z-z_0) \sum_0^{n-1} z^kz_0^{n-1-k} $ I want to prove that with $z_0$ a root of $1+z^n$, I have
$$ z^n - z^n_0 = (z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k}$$
 A: For $z \neq z_0$ and $z_0 \neq 0$:
$$\begin{align}
\sum_{k=0}^{n-1} z^k z_0^{n-1-k} & = z_0^{n-1} \sum_{k=0}^{n-1} \left( \frac{z}{z_0} \right)^k \\
& = z_0^{n-1} \frac{\left(\frac{z}{z_0}\right)^n - 1}{\frac{z}{z_0} - 1} \\
& = \frac{\frac{z^n}{z_0} - z_0^{n-1}}{\frac{z}{z_0} - 1} \\
& = \frac{z^n - z_0^n}{z - z_0}
\end{align}$$
Therefore $z^n - z_0^n = (z - z_0) \sum_{k=0}^{n-1} z^k z_0^{n-1-k}$, whenever $z \neq z_0$ and $z_0 \neq 0$. And the relation is obviously true for $z = z_0$ or $z_0 = 0$.
A: Anytime you have $f(z)=z^n-z_0^n$, you can factor out $z-z_0$ with no remainder, because $f(z_0)=0$, making $z_0$ a root. To find the other factor, you can perform polynomial long division, which will yield $\sum_{k=0}^{n-1}{z^kz_0^{n-1-k}}$.
A: The formula
$z^n - z^n_0 = (z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k} \tag{1}$
is an identity which holds for all $z, z_0 \in \Bbb C$, whether $z_0^n + 1 = 0$ or not; to wit:
$(z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k} = z\sum_0^{n-1} z^kz_0^{n-1-k} - z_0\sum_0^{n-1} z^kz_0^{n-1-k}$
$= \sum_0^{n-1} z^{k + 1}z_0^{n-1-k} - \sum_0^{n-1} z^kz_0^{n-k} =  \sum_1^n z^kz_0^{n- k} - \sum_0^{n-1} z^kz_0^{n-k}$
$= z^n + \sum_1^{n - 1} z^kz_0^{n- k} - \sum_1^{n-1} z^kz_0^{n-k} - z_0^n = z^n - z_0^n, \tag{2}$
since the two "$\Sigma$" expressions cancel one another out.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
