# The Complex Conjugate Root Theorem

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So far, we've focused exclusively on real solutions but, sometimes, we need to consider complex solutions to get a complete list of all the possible results. Finding complex roots usually requires locating a quadratic factor then using the Quadratic Formula. There is, however, one rule that tells us what we can expect in terms of complex solutions.

## The Complex Conjugate Root Theorem

If z is a root of a polynomial with real coefficients then $\bar{z}$ is also a root.

# Example 1

Solve the equation $x^3-7x^2+17x-15 = 0$.

The Rational Root Theorem tells us that the possible rational roots are $\pm15, \pm5, \pm3, \pm1$. If you evaluate the polynomial at all eight of those numbers, 3 is the only one that makes it 0 so $x - 3$ is the only factor with rational coefficients.

If we divide the original polynomial by that factor we get:

$$\frac{x^3-7x^2+17x-15}{x-3} = x^2-4x+5$$

So the equation becomes

$$(x^2-4x+5)(x-3)=0$$

The second factor gives us $x=3$ as one solution. The first factor doesn't factor so we have to fall back on the Quadratic Formula.

$$x=\frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2 \cdot 1}$$ $$x=\frac{4 \pm \sqrt{-4}}{2}$$ $$x=\frac{4 \pm 2i}{2}$$ $$x=2 \pm i$$

So the solutions are $x=3, 2+i, 2-i$. Notice how the two imaginary roots, $2 + i$ and $2 - i$, are conjugates as the Complex Conjugate Root Theorem predicted.

# Example 1

Find a fourth degree polynomial that has 2 and 3 + i as two of its roots.

This isn't a particularly practical question but it's one that students often get asked. Based on what we're given in the question, we know a few things:

• The Complex Conjugate Root Theorem says that if $3 + i$ is a factor then $3 - i$ must also be a factor.
• If $2$, $3 + i$ and $3 - i$ are roots then the Factor Theorem says that $x - 2$, $x - (3 + i)$ and $x - (3 - i)$ must be factors.
• This gives us three factors but, to get a fourth degree polynomial, we need a fourth factor that doesn't add a new root. We can't duplicate either of the complex factors since the Complex Conjugate Root Theorem woud require us to also add a factor for its conjugate which would make the degree too high. We can, however, add a second $x - 2$ without adding a new root.

Now that we know the polynomial's factors, we can get the required polynomial by multiplying them together.

$$(x - 2)(x - 2)(x - (3 + i))(x - (3 - i))$$ $$(x^2 - 4x + 4)(x - 3 - i)(x - 3 + i))$$ $$(x^2 - 4x + 4)((x - 3) - i)((x - 3) + i))$$ $$(x^2 - 4x + 4)((x - 3)^2 - i^2)$$ $$(x^2 - 4x + 4)(x^2 - 6x + 9 + 1)$$ $$(x^2 - 4x + 4)(x^2 - 6x + 10)$$ $$x^4-10x^3+38x^2-64x+40$$

# Video Lectures

The Complex Conjugate Root Theorem says that if z is a complex root of a polynomial then the conjugate of z is also a root. This video walks through the proof by first showing four smaller statements that demonstrate some basic methods for proving statements about complex variables.