Are Complex Roots of a Polynomial Greater Than 1?

[kurnimaha]

Homework Statement

Let a₀ > a₁ > a₂ > ... > a_n > 0 be coefficients of a polynomial P(z) = a₀ + a₁z + a₂z² + ... + a_nzⁿ. Let z₀ be complex number such that P(z₀) = 0. Show that |z₀| > 1.

Homework Equations

Triangle inequality? Not sure if that's enough.

The Attempt at a Solution

I started with asumption that P(z₀) = 0 with some |z₀| < 1. Then I made a pair of equations:

z₀P(z₀) = 0
P(z₀) = 0

Subtraction gives a new equation

z₀P(z₀) - P(z₀) = 0

I rearranged the terms

-a₀ + (a₀ - a₁)z₀ + ... + (a_n-1 - a_n)z₀ⁿ + a_nz₀ⁿ⁺¹ = 0

and took a modulus of it. Then by using triangle inequality (and the asumption that |z₀| < 1) I got a contradiction 0 > 0. So the modulus of z₀ must be greater or equal to 1.

For the second part of the task, I also tried a similar approach (assumed that |z₀| = 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 |z₀| < 1 to get the contradiction 0 < 0. Now the same approach leads to inequality 0 \leq 0 which is true.

Any tips?

 

[quantumdude]

Not sure if it will work, but I would try estimating it from below. That is, use the triangle inequality |a|-|b|\leq |a-b|.