- #1

U.Renko

- 57

- 1

## Homework Statement

Suppose that [itex] z_0 \in \mathbb{C}. [/itex] A polynomial [itex] P(z)[/itex] is said to be dvisible by [itex] z-z_0 [/itex] if there is another polynomial [itex] Q(z)[/itex] such that [itex] P(z)=(z-z_0)Q(z). [/itex]

Show that for every [itex] c \in\mathbb{C} [/itex] and [itex] k \in\mathbb{N} [/itex], the polynomial [itex] c(z^k - z_0^k ) [/itex] is divisible by [itex] z - z_0 [/itex]

## 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 [itex] Q(z) [/itex] such that [itex] P(z) = (z-z_0)Q(z)[/itex] and [itex] P(z) = c(z^k - z_0^k ) [/itex]

then [itex] Q(z) = \Large\frac{P(z)}{z-z_0)} = \Large c\frac{(z^k - z_0^k)}{z-z_0} [/itex]

then using complex division (AKA multiply by the conjugate):

[itex] 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} [/itex]

[itex] 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} [/itex]

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

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

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