How to prove $z^n - z^n_0 = (z-z_0) \sum_0^{n-1} z^kz_0^{n-1-k}$ [duplicate]

I want to prove that with $z_0$ a root of $1+z^n$, I have

$$z^n - z^n_0 = (z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k}$$

• Try multiplying the RHS by $z-z_{0}.$ Commented Apr 13, 2014 at 19:18
• Oops I forgot the $(z-z_0)$ factor indeed Commented Apr 13, 2014 at 19:20

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 $z_0 = 0$.

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 $\sum_{k=0}^{n-1}{z^kz_0^{n-1-k}}$.

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!!!