Core Pure · Complex Numbers · Year 1

Complex Roots of Polynomials

Prove that non-real roots of a real-coefficient polynomial always come in conjugate pairs, then use that fact to crack open cubics and quartics.

15 min read 4 worked examples

The conjugate root theorem

Theorem 4.1 — conjugate root theorem

If $p(z)$ is a polynomial with real coefficients and $z_0$ is a root of $p(z)=0$, then $z_0^{*}$ (the conjugate of $z_0$) is also a root of $p(z)=0$.

This relies on two small conjugation facts, both provable directly from the definition $z^{*}=a-bi$ for $z=a+bi$: conjugation respects sums, $(u+v)^{*}=u^{*}+v^{*}$, and respects products, $(uv)^{*}=u^{*}v^{*}$. It also fixes real numbers: if $c\in\mathbb{R}$ then $c^{*}=c$.

ProofConjugating a root of a real-coefficient polynomial gives another root

Let $p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$, with every $a_k\in\mathbb{R}$, and suppose $p(z_0)=0$.

General real-coefficient polynomial with a known root z₀.

Take the conjugate of both sides: $\big(a_nz_0^n+\cdots+a_1z_0+a_0\big)^{*} = 0^{*}=0$.

Conjugating 0 gives 0.

$(a_nz_0^n)^{*}+\cdots+(a_1z_0)^{*}+a_0^{*} = 0$

Conjugation distributes over addition.

$a_n^{*}(z_0^{*})^n+\cdots+a_1^{*}z_0^{*}+a_0^{*} = 0$

Conjugation distributes over products (and powers, applying it repeatedly).

Since every $a_k$ is real, $a_k^{*}=a_k$, giving $a_n(z_0^{*})^n+\cdots+a_1z_0^{*}+a_0 = 0$.

Real coefficients are unchanged by conjugation.

This last line says exactly $p(z_0^{*})=0$.

Substituting z₀* into p gives zero, so z₀* is also a root.

▪ QED

Common mistake

This theorem needs real coefficients — it is false in general if any coefficient of the polynomial is itself non-real. Always check (or quote) that the coefficients are real before invoking the theorem; exam questions occasionally set a trap by using a complex coefficient specifically to block this shortcut.

Turning a conjugate pair into a real quadratic factor

If $\alpha$ and $\alpha^{*}$ are both roots, then $(z-\alpha)(z-\alpha^{*})$ is a factor of the polynomial. Expanding, with $\alpha=a+bi$:

$$(z-\alpha)(z-\alpha^{*}) = z^2 - (\alpha+\alpha^{*})z+\alpha\alpha^{*} = z^2-2az+(a^2+b^2)$$

using $\alpha+\alpha^{*}=2a$ (twice the real part) and $\alpha\alpha^{*}=|\alpha|^2=a^2+b^2$. Crucially, this quadratic factor has entirely real coefficients — which means the remaining factor, found by polynomial division, also has real coefficients, and the whole process can be repeated on it if it is degree $\geq2$.

Example 1 · Solving a cubic given one complex root

Given that $z=2+3i$ is a root of $z^3-3z^2+9z+13=0$, find the other two roots.

The coefficients are real, so by Theorem 4.1, $z=2-3i$ is also a root.

The corresponding real quadratic factor is $(z-(2+3i))(z-(2-3i)) = z^2-4z+13$ (using $2a=4$, $a^2+b^2=4+9=13$).

Dividing the cubic by this quadratic:

$$z^3-3z^2+9z+13 = (z^2-4z+13)(z+1)$$

which can be checked by expanding the right-hand side. So the third root is $z=-1$. The full solution set is $\{2+3i,\ 2-3i,\ -1\}$.

Exam technique — checking a division

After dividing, always expand your quotient back out (or compare a couple of coefficients) to confirm it reproduces the original polynomial — a small arithmetic slip in long division is easy to make and easy to catch this way before it costs the final accuracy mark.

Example 2 · A quartic with two conjugate pairs

Given that $z=1+2i$ is a root of $z^4-8z^3+27z^2-50z+50=0$, and that the equation has another non-real root with real part $3$, find all four roots.

Since coefficients are real, $z=1-2i$ is also a root, giving the real quadratic factor $z^2-2z+5$ (using $2a=2$, $a^2+b^2=1+4=5$).

The remaining two roots are non-real with real part $3$, so by Theorem 4.1 they too form a conjugate pair: $3+ki$ and $3-ki$ for some real $k$, with quadratic factor $z^2-6z+(9+k^2)$.

Dividing the quartic by $z^2-2z+5$ gives the quotient $z^2-6z+10$ (comparing coefficients, or by long division) — so $9+k^2=10 \Rightarrow k^2=1 \Rightarrow k=\pm1$, giving roots $3+i$ and $3-i$.

The full solution set is $\{1+2i,\ 1-2i,\ 3+i,\ 3-i\}$.

Forming a polynomial from given roots

The process also runs in reverse: given a set of roots including a conjugate pair, multiply out $(z-\alpha)(z-\alpha^{*})$ and any real roots' factors together to reconstruct the original polynomial.

Example 3 · Constructing a cubic from its roots

Find, with integer coefficients, a cubic equation with roots $2$ and $-1+i$.

Since one root is $-1+i$ and the coefficients must be real, $-1-i$ is also a root. Real quadratic factor: $(z-(-1+i))(z-(-1-i)) = z^2+2z+2$ (using $2a=-2$, $a^2+b^2=1+1=2$).

Multiplying by the factor for the real root $2$: $(z-2)(z^2+2z+2) = z^3+2z^2+2z-2z^2-4z-4 = z^3-2z-4$.

So $z^3-2z-4=0$ is a cubic equation with the required roots.

Exam-style questions

Question 1 6 marks

(a) Show that $z=3+2i$ is a root of $z^3-4z^2+z+26=0$. (b) Find the other two roots of the equation.

(a) $(3+2i)^2=9+12i-4=5+12i$; $(3+2i)^3=(3+2i)(5+12i)=15+36i+10i-24=-9+46i$ (M1).

$z^3-4z^2+z+26 = (-9+46i)-4(5+12i)+(3+2i)+26 = (-9-20+3+26)+(46-48+2)i = 0+0i=0$ ✓ (A1)

(b) Real coefficients $\Rightarrow$ $3-2i$ is also a root (B1). Quadratic factor: $z^2-6z+13$ (M1). Dividing: $z^3-4z^2+z+26=(z^2-6z+13)(z+2)$, so the third root is $z=-2$ (M1, A1).

Question 2 5 marks

Find, with integer coefficients, a cubic equation with roots $2$ and $-1+i$.
(Hint: you'll need the conjugate root theorem before you can multiply anything out.)

Real coefficients required $\Rightarrow$ $-1-i$ is also a root (B1).

$(z-(-1+i))(z-(-1-i)) = z^2+2z+2$ (M1, A1)

$(z-2)(z^2+2z+2) = z^3+2z^2+2z-2z^2-4z-4 = z^3-2z-4$ (M1, A1)

So $z^3-2z-4=0$.

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Loci in the Complex Plane

De Moivre's Theorem