Why Do Complex Roots of a Polynomial With Real Coefficients Occur in Conjugate Pairs?

Complex roots occur in conjugate pairs because complex conjugation leaves every real coefficient unchanged and preserves addition and multiplication. Therefore, if $\alpha$ is a non-real root of a polynomial with real coefficients, its complex conjugate $\overline{\alpha}$ must also be a root.

The Conjugate Root Theorem

Let

$$p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0,$$

where every coefficient $a_k$ is real. If $\alpha$ is a root, then

$$p(\alpha)=0.$$

Taking the complex conjugate of both sides gives

$$\overline{p(\alpha)}=\overline{0}=0.$$

Conjugation distributes over sums, products, and integer powers. Therefore,

$$\overline{p(\alpha)} =\overline{a_n}\,\overline{\alpha}^{\,n}+\cdots+ \overline{a_1}\,\overline{\alpha}+ \overline{a_0}.$$

Since every coefficient is real, $\overline{a_k}=a_k$. Hence,

$$\overline{p(\alpha)} =a_n\overline{\alpha}^{\,n}+\cdots+a_1\overline{\alpha}+a_0 =p(\overline{\alpha})=0.$$

Thus, $\overline{\alpha}$ is also a root. This is the conjugate root theorem, included in IB Mathematics AA HL sub-topic AHL 1.14.

The result can also be understood through factors. If $a+bi$ and (a-bi) are roots, their corresponding factors multiply to

$$(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2,$$

which has real coefficients. Pairing conjugates therefore removes the imaginary terms.

The factor viewpoint also gives a useful degree argument. A polynomial of odd degree with real coefficients must have at least one real root, because all non-real roots can be grouped into pairs, contributing an even number to the total number of roots when multiplicity is counted. Repeated non-real roots are paired with equal multiplicity.

For example, if $2+3i$ is a root of multiplicity two, then $2-3i$ must also occur twice. Together they produce the real quadratic factor $((x-2)^2+9)^2$. This does not mean every polynomial has non-real roots; it only restricts how any non-real roots may appear. This pairing is structural, not a coincidence of the quadratic formula.

Worked Example

Consider

$$p(x)=x^2-2x+5.$$

Using the quadratic formula,

$$x=\frac{2\pm\sqrt{(-2)^2-4(1)(5)}}{2} =\frac{2\pm\sqrt{-16}}{2} =1\pm2i.$$

The roots $1+2i$ and $1-2i$ form a conjugate pair. Their factors give

$$(x-(1+2i))(x-(1-2i))=(x-1)^2+4=x^2-2x+5.$$

A common misconception is that every complex root needs a distinct partner. Real roots are their own conjugates because $\overline{r}=r$. The theorem also requires real coefficients: $p(z)=z-i$ has root $i$, but not (-i).

Exam Technique

For show, write the conjugation steps explicitly and state why $\overline{a_k}=a_k$. Simply quoting the theorem may not earn all reasoning marks on AA HL Paper 1 or Paper 2.