Core Pure · Complex Numbers · Year 1

Solving Equations with Complex Roots

Real-coefficient polynomials can have non-real roots, but never alone — the conjugate root theorem guarantees they always arrive in pairs.

17 min read 4 worked examples

Complex roots of quadratic equations

Method 3.1 — completing the square / the quadratic formula

Every quadratic equation $ax^2+bx+c=0$ ($a,b,c$ real, $a\neq0$) can be solved with the quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ When the discriminant $b^2-4ac<0$, the square root is imaginary, and the equation has a pair of complex roots instead of real ones.

Theorem 3.2 — complex roots occur in conjugate pairs

If a polynomial equation has real coefficients, any non-real complex root $z$ must be accompanied by its conjugate $z^*$, which is also a root.

ProofConjugate root theorem (quadratic case)

Suppose $z=p+qi$ ($q\neq0$) is a root of $ax^2+bx+c=0$ with $a,b,c$ real.

Setup.

Then $az^2+bz+c=0$. Take the complex conjugate of both sides.

If two complex numbers are equal, so are their conjugates.

$a(z^2)^*+b z^*+c^*=0^* \Rightarrow a(z^*)^2+bz^*+c=0$

Conjugation distributes over sums and products, and $a,b,c$ are real so equal their own conjugates.

This shows $z^*=p-qi$ also satisfies the equation.

Direct substitution result of step 3.

∎

Example 1 · Solving a quadratic with complex roots

Solve $x^2-6x+13=0$, giving your answers in the form $a+bi$.

$x=\dfrac{6\pm\sqrt{36-52}}{2}=\dfrac{6\pm\sqrt{-16}}{2}=\dfrac{6\pm4i}{2}=3\pm2i$.

The roots are $3+2i$ and $3-2i$ — a conjugate pair, as expected since all coefficients are real.

Example 2 · Forming a quadratic from a given complex root

One root of a quadratic equation with real coefficients is $4-3i$. Find the equation in the form $x^2+bx+c=0$.

By Theorem 3.2, the other root is $4+3i$. Sum of roots $=(4-3i)+(4+3i)=8$; product of roots $=(4-3i)(4+3i)=16+9=25$.

Using $x^2-(\text{sum})x+(\text{product})=0$: $x^2-8x+25=0$.

Complex roots of cubic and quartic equations

Method 3.3 — cubics and quartics with complex roots

For polynomials of degree 3 or more with real coefficients, complex roots still occur in conjugate pairs, but real roots can also be present. To solve, use the factor theorem to find (or be given) one root, divide out the corresponding factor, then solve what's left — repeating the process until every factor is linear or a solvable quadratic.

The possible root structures

A cubic equation with real coefficients always has three roots, and either:

all three roots are real, or
one root is real and the other two form a complex conjugate pair.

A quartic equation with real coefficients always has four roots, and either:

all four roots are real, or
two roots are real and the other two form a complex conjugate pair, or
two roots form one complex conjugate pair and the remaining two roots form a second, different conjugate pair (Example 4 below is exactly this case — no real roots at all).

What's never possible: an odd number of non-real roots. Non-real roots only ever leave or enter in matched conjugate pairs.

Example 3 · Solving a cubic with one real root given

Given that $x=2$ is a root of $x^3-2x^2+9x-18=0$, find the other roots.

Dividing by $(x-2)$: $x^3-2x^2+9x-18=(x-2)(x^2+9)$.

Solving $x^2+9=0 \Rightarrow x^2=-9 \Rightarrow x=\pm3i$.

The three roots are $x=2$, $x=3i$, $x=-3i$.

Example 4 · Solving a quartic given one complex root

Given that $2+i$ is a root of $x^4-4x^3+9x^2-16x+20=0$, find all four roots.

By Theorem 3.2, $2-i$ is also a root. The quadratic factor formed from these is $x^2-(\text{sum})x+(\text{product})=x^2-4x+5$ (sum $=4$, product $=(2+i)(2-i)=4+1=5$).

Dividing the quartic by $x^2-4x+5$: $x^4-4x^3+9x^2-16x+20=(x^2-4x+5)(x^2+4)$.

Solving $x^2+4=0 \Rightarrow x=\pm2i$.

All four roots: $2+i,\ 2-i,\ 2i,\ -2i$.

Exam technique

Once you know a complex root $p+qi$ of a real-coefficient polynomial, immediately form the quadratic factor $x^2-2px+(p^2+q^2)$ (sum of the conjugate pair is $2p$, product is $p^2+q^2$) and use polynomial division to peel it off — this is faster and less error-prone than substituting the complex root directly into long division.