# Are Complex Roots of a Polynomial Greater Than 1?

• kurnimaha
So, for the first part, we have:|z0P(z0)-P(z0)| \leq |z0P(z0)| + |P(z0)|= |z0||P(z0)| + |P(z0)|= (|z0|+1)|P(z0)|= 0. (since P(z0) = 0)So, we have |z0|+1 = 0, which is a contradiction since |z0| is a positive number. Therefore, |z0| must be greater than 1.
kurnimaha

## Homework Statement

Let a0 > a1 > a2 > ... > an > 0 be coefficients of a polynomial P(z) = a0 + a1z + a2z2 + ... + anzn. 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 + ... + (an-1 - an)z0n + anz0n+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?

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

Last edited:

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

## 1. What does it mean for a polynomial to have complex roots?

A polynomial is an expression containing variables and coefficients, and its roots are the values of the variables that make the expression equal to zero. Complex roots are values that involve the imaginary unit, i, which is the square root of -1. This means that the roots of a polynomial with complex roots involve a combination of real and imaginary numbers.

## 2. How do you find the complex roots of a polynomial?

To find the complex roots of a polynomial, you can use the quadratic formula or other methods such as factoring and graphing. The quadratic formula is particularly useful for polynomials of degree 2, while factoring and graphing can be used for polynomials of higher degrees.

## 3. Can a polynomial have more than two complex roots?

Yes, a polynomial can have multiple complex roots, including both real and imaginary roots. The number of complex roots a polynomial has is equal to its degree, so a polynomial of degree 3 can have up to 3 complex roots.

## 4. What is the significance of complex roots in polynomial equations?

Complex roots are important in polynomial equations because they allow for a complete and accurate solution to the equation. In some cases, the presence of complex roots can also provide insights into the behavior and properties of the polynomial.

## 5. Can complex roots be used to solve real-world problems?

Yes, complex roots can be used to solve real-world problems in various fields such as engineering, physics, and economics. For example, in electrical engineering, complex roots are used to analyze circuits and determine their stability and behavior. In finance, complex roots can be used to model and predict the behavior of stock prices.

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