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 $|a|-|b|\leq |a-b|$.

 

[Architeuthis Dux]

I would suggest approaching this problem using the fundamental theorem of algebra, which states that a polynomial of degree n has n complex roots (counting multiplicities). In this case, we know that P(z) has at least one complex root, z0.

To show that |z0| > 1, we can use the fact that the modulus of a complex number is related to its conjugate. Let z0 = a + bi, where a and b are real numbers. Since P(z0) = 0, we can rewrite the polynomial as:

P(z0) = a0 + a1(a+bi) + a2(a+bi)^2 + ... + an(a+bi)^n = 0

By taking the modulus of both sides, we get:

|P(z0)| = |a0 + a1(a+bi) + a2(a+bi)^2 + ... + an(a+bi)^n| = |0|

Using the triangle inequality, we can expand the left side to:

|P(z0)| = |a0| + |a1(a+bi)| + |a2(a+bi)^2| + ... + |an(a+bi)^n|

And since we know that all the coefficients are positive (a0 > a1 > a2 > ... > an > 0), we can simplify this further to:

|P(z0)| = a0 + a1|a+bi| + a2|a+bi|^2 + ... + an|a+bi|^n

Now, we can use the fact that |z|^2 = z*z', where z' is the conjugate of z. This means that:

|a+bi|^n = (a+bi)(a-bi)^n = (a+bi)(a^n + b^n + i(a^n-b^n))

Therefore, our equation becomes:

|P(z0)| = a0 + a1|a+bi| + a2|a+bi|^2 + ... + an(a^n + b^n + i(a^n-b^n))

Now, since we are assuming that |z0| < 1, we know that 0 < |a+bi| < 1. This means that all the terms in our equation are positive, except for the last term, which is imaginary. This leads to a contradiction, as the left