# Complex numbers polynomial divisibility proof

• U.Renko
In summary, the problem is to show that for every c in the set of complex numbers and k in the set of natural numbers, the polynomial c(z^k - z_0^k) is divisible by z - z_0. The approach is to find Q(z) such that P(z) = (z-z_0)Q(z) and P(z) = c(z^k - z_0^k) and then show that Q(z) is a polynomial. Using complex division, it is shown that Q(z) can be expressed as a polynomial, satisfying the required conditions and proving that P(z) is divisible by z - z_0. However, an argument needs to be provided for why Q(z)
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 $z - z_0$

## 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...

U.Renko said:
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 $z - z_0$

## 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...

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

## 1. What is a complex number?

A complex number is a number that contains both a real and imaginary component. It is expressed in the form a + bi, where a represents the real part and bi represents the imaginary part (where i is the square root of -1).

## 2. How do you prove polynomial divisibility with complex numbers?

To prove polynomial divisibility with complex numbers, we use the division algorithm. This involves dividing the polynomial by the divisor and checking if the remainder is equal to zero. If the remainder is zero, then the divisor is a factor of the polynomial and therefore it is divisible.

## 3. Can complex numbers be used to divide polynomials in the same way as real numbers?

Yes, complex numbers can be used to divide polynomials in the same way as real numbers. The only difference is that the division may result in a complex number as the quotient or remainder.

## 4. How do you determine if a polynomial is divisible by a complex number?

To determine if a polynomial is divisible by a complex number, you can use the division algorithm or check if all the coefficients of the polynomial are divisible by the real and imaginary parts of the complex number.

## 5. Why is divisibility by complex numbers important in mathematics?

Divisibility by complex numbers is important in mathematics because it allows us to factor polynomials and find their roots, which is crucial in solving equations and understanding the behavior of functions. It also has applications in fields such as engineering and physics.

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