- #1

kurnimaha

- 4

- 0

## Homework Statement

Let a

_{0}> a

_{1}> a

_{2}> ... > a

_{n}> 0 be coefficients of a polynomial P(z) = a

_{0}+ a

_{1}z + a

_{2}z

^{2}+ ... + a

_{n}z

^{n}. Let z

_{0}be complex number such that P(z

_{0}) = 0. Show that |z

_{0}| > 1.

## Homework Equations

Triangle inequality? Not sure if that's enough.

## The Attempt at a Solution

I started with asumption that P(z

_{0}) = 0 with some |z

_{0}| < 1. Then I made a pair of equations:

z

_{0}P(z

_{0}) = 0

P(z

_{0}) = 0

Subtraction gives a new equation

z

_{0}P(z

_{0}) - P(z

_{0}) = 0

I rearranged the terms

-a

_{0}+ (a

_{0}- a

_{1})z

_{0}+ ... + (a

_{n-1}- a

_{n})z

_{0}

^{n}+ a

_{n}z

_{0}

^{n+1}= 0

and took a modulus of it. Then by using triangle inequality (and the asumption that |z

_{0}| < 1) I got a contradiction 0 > 0. So the modulus of z

_{0}must be greater or equal to 1.

For the second part of the task, I also tried a similar approach (assumed that |z

_{0}| = 1 and tried to make it lead to a contradiction) but couldn't get anything out of it. In the first part I needed the asumption that |z

_{0}| < 1 to get the contradiction 0 < 0. Now the same approach leads to inequality 0 [tex]\leq[/tex] 0 which is true.

Any tips?