# Proof that if a polynomial has a complex zero it's conjugate is also a zero

• mindauggas
In summary, this equation states that if z is a zero of a polynomial, then the complex conjugate of z is also a zero of the same polynomial.
mindauggas

## Homework Statement

If $P(x)$ is a polynomial with real coefficients, then if $z$ is a complex zero of $P(x)$, then the complex conjugate $\bar{z}$ is also a zero of $P(x)$.

## Homework Equations

Book provides a hint: assume that $z$ is a zero for $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ and use the fact that $\bar{z}=z$ if $z$ is real (since every real can be written as a complex number with a zero imaginary part) and that $\overline{z+w}=\bar{z}+\bar{w}$ and $\overline{z*w}=\bar{z}*\bar{w}$ for all complex numbers.

## The Attempt at a Solution

Well, it is proven for any real since the complex conjugate of a real number is that same number.

I'm lost from here.

mindauggas said:

## Homework Statement

If $P(x)$ is a polynomial with real coefficients, then if $z$ is a complex zero of $P(x)$, then the complex conjugate $\bar{z}$ is also a zero of $P(x)$.

## Homework Equations

Book provides a hint: assume that $z$ is a zero for $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ and use the fact that $\bar{z}=z$ if $z$ is real (since every real can be written as a complex number with a zero imaginary part) and that $\overline{z+w}=\bar{z}+\bar{w}$ and $\overline{z*w}=\bar{z}*\bar{w}$ for all complex numbers.

## The Attempt at a Solution

Well, it is proven for any real since the complex conjugate of a real number is that same number.

I'm lost from here.

Well, if z is a zero of P(x), then $a_{n}z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+a_{0} = 0$.

Now take the conjugate of both sides.

For the LHS, use the rules they mentioned to systematically simplify the expression till you get it to: $a_{n}{\bar{z}}^{n}+a_{n-1}{\bar{z}}^{n-1}+...+a_{1}{\bar{z}}+a_{0}$. You'll need to apply $\overline{z+w}=\bar{z}+\bar{w}$ first on the entire polynomial, followed by $\overline{z*w}=\bar{z}*\bar{w}$ on each term and finally, "$\bar{z}=z$ if $z$ is real" on the real coefficients.

For the RHS, the conjugate of 0 is of course 0.

You've now established $P(\bar{z}) = 0$ and you're done.

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I deleted a question that was here since I misunderstood the post.

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## 1. What is a complex zero?

A complex zero is a value for a variable in a polynomial equation that, when substituted, results in an output of zero. It is a solution to the equation that involves both real and imaginary numbers.

## 2. What does it mean for a polynomial to have a complex zero?

Having a complex zero means that one or more of the solutions to the polynomial equation involve complex numbers, rather than just real numbers. This can happen when the polynomial has terms with an imaginary component, such as a coefficient with the square root of a negative number.

## 3. What is a conjugate of a complex number?

The conjugate of a complex number is another complex number with the same real part, but the imaginary part has the opposite sign. For example, the conjugate of 3+4i is 3-4i. In general, the conjugate of a+bi is a-bi.

## 4. Why is it important for a polynomial's complex zero to have a conjugate zero?

If a polynomial has complex zeros, it is important for their conjugates to also be zeros because it ensures that the polynomial has only real coefficients. This makes it easier to work with and analyze the polynomial, as well as to graph it.

## 5. How can you prove that a polynomial's complex zero has a conjugate zero?

This can be proven using the fundamental theorem of algebra, which states that a polynomial of degree n has exactly n complex zeros, including repeated zeros. Since the complex zeros occur in conjugate pairs, if one zero is present, its conjugate must also be a zero of the polynomial.

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