# Complex Conjugate Zeros

## Key Questions

• The complex conjugate of a complex number $a + b i$ can be found by negating the imaginary part, that is, $a - b i$.

For example, the complex conjugate of $2 - 3 i$ is $2 + 3 i$.

If a polynomial has Real coefficients, then any Complex zeros will occur in Complex conjugate pairs.

That is, if $z = a + b i$ is a zero then $\overline{z} = a - b i$ is also a zero.

#### Explanation:

Actually a similar theorem holds for square roots and polynomials with rational coefficients:

If $f \left(x\right)$ is a polynomial with rational coefficients and a zero expressible in the form $a + b \sqrt{c}$ where $a , b , c$ are rational and $\sqrt{c}$ is irrational, then $a - b \sqrt{c}$ is also a zero.

• As a student, if a teacher tells you that a polynomial with real coefficients has $3 i$ for one of its zeros, that you can reason:

With real corrifients, if $3 i$ is a zero, then so is $- 3 i$. So I know that $\left(x - 3 i\right)$ and $\left(x - \left(- 3 i\right)\right) = \left(x + 3 i\right)$ are both factors.

So I know that $\left(x - 3 i\right) \left(x + 3 i\right) = {x}^{2} + 9$ is a factor. That might help me factor the polynomial. (If I've learned division of polynomials.)

I saw this interesting bit of reasoning recently here on Socratic.

We know that a polynomial of degree $n$ has $n$ zeros (counting multiplicity). So a polynomial of odd degree has an odd number of zeros.
We also know that if a polynomial has real coefficients, the any imaginary zeros occur in conjugate pairs.

So we can conclude that a polynomial with real coefficients and odd degee must have at least one real zero. (An odd number of zeros cannot all be imaginary.)

• A complex conjugate is the number to which you have to multiply a complex number in order to make it real.
By using the identity (x+y) . (x-y) = x²-y², we see that, to every complex, there is another to which we can multiply it in order to get a new number that will not depend on $i$.
If $\left(a + b i\right) . \left(c + \mathrm{di}\right)$ is real, ($c + \mathrm{di}$) is ($a + b i$)'s conjugate and it equals ($a - b i$).