Complex numbers polynomial divisibility proof

[U.Renko]

I'm not sure whether this should go in this forum or another. feel free to move it if needed

Homework Statement

Suppose that $z_0 \in \mathbb{C}.$ A polynomial $P(z)$ is said to be dvisible by $z-z_0$ if there is another polynomial $Q(z)$ such that $P(z)=(z-z_0)Q(z).$
Show that for every $c \in\mathbb{C}$ and $k \in\mathbb{N}$, the polynomial $c(z^k - z_0^k )$ is divisible by $z - z_0$

Homework Equations

The Attempt at a Solution

This is just the part a of the problem.
I haven't tried the other parts yet, because I am not sure if my approach was correct.
well here it is:

basically I interpreted the problem as: find $Q(z)$ such that $P(z) = (z-z_0)Q(z)$ and $P(z) = c(z^k - z_0^k )$

then $Q(z) = \Large\frac{P(z)}{z-z_0)} = \Large c\frac{(z^k - z_0^k)}{z-z_0}$
then using complex division (AKA multiply by the conjugate):
$Q(z) = \Large c\frac{z^k(z+z_0) }{z^2 + z_0^2} -\Large c\frac{z_0^k(z+z_0) }{z^2 +z_0^2}$

$Q(z) = \Large\frac{cz^{k+1} +cz^kz_0}{z^2 +z_0^2} - \Large\frac{cz_0^kz+cz_0^{k+1}}{z^2+z_0^2}$

and then: $Q(z) = \Large\frac{c}{(z^2 +z_0^2)}\normalsize(z^{k+1} + z^kz_0 -z_0^kz - z_0^{k+1})$ which is a polynomial.

So: there exists a polynomial satisfying the required conditions and therefore $P(z)$ is divisible by $z-z_0$


it looks okay but I'm not 100% confident on this...

 

[HallsofIvy]

I'm not sure whether this should go in this forum or another. feel free to move it if needed

Homework Statement

Suppose that $z_0 \in \mathbb{C}.$ A polynomial $P(z)$ is said to be dvisible by $z-z_0$ if there is another polynomial $Q(z)$ such that $P(z)=(z-z_0)Q(z).$
Show that for every $c \in\mathbb{C}$ and $k \in\mathbb{N}$, the polynomial $c(z^k - z_0^k )$ is divisible by $z - z_0$

Homework Equations

The Attempt at a Solution

This is just the part a of the problem.
I haven't tried the other parts yet, because I am not sure if my approach was correct.
well here it is:

basically I interpreted the problem as: find $Q(z)$ such that $P(z) = (z-z_0)Q(z)$ and $P(z) = c(z^k - z_0^k )$

then $Q(z) = \Large\frac{P(z)}{z-z_0)} = \Large c\frac{(z^k - z_0^k)}{z-z_0}$
then using complex division (AKA multiply by the conjugate):
$Q(z) = \Large c\frac{z^k(z+z_0) }{z^2 + z_0^2} -\Large c\frac{z_0^k(z+z_0) }{z^2 +z_0^2}$

$Q(z) = \Large\frac{cz^{k+1} +cz^kz_0}{z^2 +z_0^2} - \Large\frac{cz_0^kz+cz_0^{k+1}}{z^2+z_0^2}$

and then: $Q(z) = \Large\frac{c}{(z^2 +z_0^2)}\normalsize(z^{k+1} + z^kz_0 -z_0^kz - z_0^{k+1})$ which is a polynomial.

I think you will need to give an argument that this is a polynomial!

So: there exists a polynomial satisfying the required conditions and therefore $P(z)$ is divisible by $z-z_0$


it looks okay but I'm not 100% confident on this...

 

[U.Renko]

Well...
now that you mention it I'm not quite sure I have an argument for that...