**Conjugate root theorem.** For a polynomial \(p(z)\) with real coefficients, if \(\alpha\) is a zero, then so is \(\overline{\alpha}\).

It’s important to note that the conjugate root theorem only applies to polynomials with *real* coefficients.

**Example. **If we are given that \(1-i\) is a root of the equation \(z^2-2z+2=0\), then we immediately know that \(1+i\) is a root.

**Worked exercise.** Factorise \(p(z)=z^4-z^3-5z^2-z-6\) into real linear quadratic factor, given \(z=i\) is a zero. **Solution**. By the conjugate root theorem, \(z=-i\) is also a zero. By the fundamental theorem of algebra, the polynomial has four complex zeroes, so let the other two roots be \(\alpha\) and \(\beta\). By using Vieta’s formulas for the sum and product of roots, we have $$\begin{align*}

i+(-i)+\alpha+\beta&=1 &\Longrightarrow \alpha+\beta&=1 \\

i(-i)\alpha\beta&=-6 &\Longrightarrow \alpha\beta&=-6

\end{align*} $$ By Vieta’s formulas again, the quadratic expression with zeroes \(\alpha\) and \(\beta\) is \(z^2-z-6 = (z-3)(z+2)\). Hence $$p(z)=(z-i)(z+i)(z-3)(z+2)=(z^2+1)(z-3)(z+2).$$

Because of the conjugate root theorem and the fundamental theorem of algebra, every polynomial with real coefficients can be factorised into real linear and quadratic factors. If \(z-\alpha\) is a factor, then so is \(z-\overline{\alpha}\), and $$(z-\alpha)(z-\overline{\alpha})=z^2-2\text{Re}(\alpha)z+|\alpha|^2$$ is a real quadratic factor.

## Proof of the conjugate root theorem

By looking at the quadratic formula, we can see that the conjugate root theorem is true for quadratic equations with real coefficients. Recall that the solutions to the equation \(az^2+bz+c=0\) are $$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ If \(b^2-4ac \ge 0\), then the roots are real (because \(a\), \(b\) and \(c\) are real) and therefore conjugate root theorem is trivially true because the conjugate of a real number is itself.

If \(b^2-4ac<0\), then \(z= -\frac{b}{2a}\pm i \frac{\sqrt{4ac-b^2}}{2a}\), which are conjugates.

Let’s prove this for polynomials of all degrees.

**Proof of the conjugate root theorem.** Let \(p(z)\) be an arbitrary polynomial of degree \(n\) with real coefficients, that is, $$p(z) = c_0+c_1 z+c_2 z^2 \dots+c_n z^n,$$ where \(c_0, \dots, c_n\) are real (\(c_n\ne 0\)). Let \(\alpha\) be a zero of the polynomial. Therefore $$c_0+c_1 \alpha+c_2 \alpha^2 \dots+c_n \alpha^n = 0.$$ If we take the conjugate of both sides, and apply conjugate properties, we have $$\overline{c_0}+\overline{c_1} \overline{\alpha}+\overline{c_2} \overline{\alpha}^2 \dots+\overline{c_n} \overline{\alpha}^n = 0.$$ However, because all of the \(c_i\) are real, we have \(\overline{c_i} = c_i\). Hence, $$c_0+c_1 \overline{\alpha}+c_2 \overline{\alpha}^2 \dots+c_n \overline{\alpha}^n = 0.$$ This simply means that \(p(\overline{\alpha})=0\), so \(\overline{\alpha}\) is a zero of \(p(z)\).

## Problems

- Write down a polynomial with real integer coefficients with zeroes \(\frac{1}{2}\), \(3-i\) and \(i\).
- Factorise \(8z^4+16z^3+10z^2+2z-6\) into real linear and quadratic factors, given that \(z=\frac{-1+\sqrt{3}i}{2}\).
- Factorise \(z^4-2z^3+6z^2-2z+5\) into real linear and quadratic actors, given that \(z=1-2i\).
*Hard*. You are given the polynomial equation \(z^3+2z^2+2z+1=0\).- Show that the equation has only one real root.
- It is given that this real root is an integer. Find the real root.
- Without finding all the roots, show that all the roots lie on the unit circle \(|z|=1\) (that is, all the roots have modulus 1).