Complex Roots of Polynomials

A2 · P3 · 17 min

A polynomial with real coefficients can have non-real roots, but they never come alone: if $a + bi$ is a root then so is $a - bi$ . This one fact is what the syllabus means by "use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs". In the exam it turns up as a cubic or quartic where you are given one complex root (or asked to verify it) and must find all the others, sometimes with unknown coefficients to determine first. The skills are the ones from the algebra unit (division and the factor theorem) combined with complex arithmetic.

The conjugate root theorem

Look back at a real quadratic with negative discriminant: its roots are $-\dfrac{b}{2a} \pm \dfrac{\sqrt{4ac - b^2}}{2a}i$ , a conjugate pair. That was not a coincidence of the quadratic formula. It holds for polynomials of every degree.

Conjugate pairs

If $p(z)$ is a polynomial with real coefficients and $\alpha$ is a root, then $\alpha^*$ is also a root.

So non-real roots of real polynomials occur in conjugate pairs, and each pair $a \pm bi$ gives a real quadratic factor

(z - (a + bi))(z - (a - bi)) = z^2 - 2az + (a^2 + b^2).

Proof

Let $p(z) = c_nz^n + \cdots + c_1z + c_0$ with every $c_k$ real, and suppose $p(\alpha) = 0$ . Take the conjugate of both sides. The conjugate of a sum is the sum of conjugates, the conjugate of a product is the product of conjugates, so $(\alpha^k)^* = (\alpha^*)^k$ , and $c_k^* = c_k$ because $c_k$ is real. Hence

0 = 0^* = \left(p(\alpha)\right)^* = c_n(\alpha^*)^n + \cdots + c_1\alpha^* + c_0 = p(\alpha^*).

So $\alpha^*$ is a root.

The proof shows exactly where "real coefficients" is used: in $c_k^* = c_k$ . If any coefficient is complex, the result fails.

Consequences for cubics and quartics

A polynomial of degree $n$ has exactly $n$ roots, counted with repetition (you may use this fact; it is not proved at A Level). Combined with conjugate pairs:

Degree (real coefficients)	Possible numbers of real roots	Possible numbers of non-real roots
quadratic	$2$ or $0$	$0$ or $2$
cubic	$3$ or $1$	$0$ or $2$
quartic	$4$ , $2$ or $0$	$0$ , $2$ or $4$

In particular every real cubic has at least one real root, because the non-real roots pair off and three is odd. You can see this on a graph: a cubic $y = p(x)$ goes from $-\infty$ to $+\infty$ (or the reverse), so it must cross the $x$ -axis.

y = x^3 - 5x^2 + 9x - 5 (1, 0)

The curve $y = x^3 - 5x^2 + 9x - 5$ crosses the $x$ -axis only once, at $x = 1$ . Its other two roots, $2 \pm i$ , are non-real and do not appear on the graph.

The sum and product shortcut

For the conjugate pair $\alpha = a + bi$ and $\alpha^* = a - bi$ :

the sum is $\alpha + \alpha^* = 2a$ , which is $2\operatorname{Re}\alpha$ ;
the product is $\alpha\alpha^* = a^2 + b^2$ .

So the quadratic factor is $z^2 - (\text{sum})z + (\text{product})$ . For $2 + i$ : sum $4$ , product $5$ , factor $z^2 - 4z + 5$ . Do this in your head and write it down; there is no need to expand $(z - 2 - i)(z - 2 + i)$ each time, though it is a good check.

The three standard question types

Type 1: one complex root given, all coefficients known

Write down the conjugate, stating that the coefficients are real.
Form the real quadratic factor $z^2 - 2az + (a^2 + b^2)$ .
Divide the polynomial by it (long division, or compare coefficients in $p(z) = (\text{quadratic})(\text{other factor})$ ).
Solve the other factor for the remaining roots.

Type 2: one complex root given, coefficients unknown

Either substitute the root, expand the powers, and equate real and imaginary parts to zero (two equations for two unknowns);

or form the quadratic factor and compare coefficients in $p(z) = (\text{quadratic})(\text{other factor with unknowns})$ .

Then finish as in Type 1.

Type 3: real root findable, complex roots wanted

Use the factor theorem to find a real root $r$ (try $\pm 1, \pm 2, \ldots$ , divisors of the constant term over divisors of the leading coefficient).
Divide by $(z - r)$ , or by $(kz - m)$ if the root is $\tfrac{m}{k}$ , to get a quadratic.
Solve the quadratic, using $i$ when the discriminant is negative.

Comparing coefficients is usually quicker than long division for a cubic. For example, if $z^3 - 5z^2 + 9z - 5 = (z^2 - 4z + 5)(z + k)$ , the constant terms give $5k = -5$ immediately.

Worked examples

Type 1: a cubic with one complex root given

Given that $2 + i$ is a root of $z^3 - 5z^2 + 9z - 5 = 0$ , find the other two roots.

Solution

The coefficients are real, so $2 - i$ is also a root. The sum of this pair is $4$ and the product is $4 + 1 = 5$ , so $z^2 - 4z + 5$ is a factor.

Write $z^3 - 5z^2 + 9z - 5 = (z^2 - 4z + 5)(z + k)$ .

Constant terms: $5k = -5$ , so $k = -1$ .

Check another coefficient, $z^2$ : $k - 4 = -5$ . Correct.

The third factor is $z - 1$ , so the roots are $2 + i$ , $2 - i$ and $1$ .

Type 3: find the real root first

Solve the equation $z^3 - 4z^2 + 14z - 20 = 0$ .

Solution

Try small integers: $p(2) = 8 - 16 + 28 - 20 = 0$ , so $(z - 2)$ is a factor.

Divide (by comparing coefficients): $z^3 - 4z^2 + 14z - 20 = (z - 2)(z^2 + bz + 10)$ , where the constant $10$ comes from $-2 \times 10 = -20$ . The $z^2$ coefficient gives $b - 2 = -4$ , so $b = -2$ . Check $z$ : $10 - 2b = 14$ .

Solve $z^2 - 2z + 10 = 0$ :

z = \frac{2 \pm \sqrt{4 - 40}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i.

The roots are $2$ , $1 + 3i$ and $1 - 3i$ .

Type 2: unknown coefficients by substitution

The equation $z^3 + az^2 + bz - 10 = 0$ , where $a$ and $b$ are real, has a root $1 + 2i$ .

(a) Find $a$ and $b$ .

(b) Find the other two roots.

Solution

(a) Work out the powers:

z^2 = (1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i,

z^3 = (-3 + 4i)(1 + 2i) = -3 - 6i + 4i + 8i^2 = -11 - 2i.

Substitute into the equation:

(-11 - 2i) + a(-3 + 4i) + b(1 + 2i) - 10 = 0.

Real parts: $-21 - 3a + b = 0$ . Imaginary parts: $-2 + 4a + 2b = 0$ , i.e. $b = 1 - 2a$ .

Substitute: $-21 - 3a + 1 - 2a = 0$ , so $-5a = 20$ , $a = -4$ and $b = 9$ .

(b) The equation is $z^3 - 4z^2 + 9z - 10 = 0$ . The coefficients are real, so $1 - 2i$ is a root and $z^2 - 2z + 5$ is a factor. Then $z^3 - 4z^2 + 9z - 10 = (z^2 - 2z + 5)(z - 2)$ (constant terms: $5 \times (-2) = -10$ ).

The other roots are $1 - 2i$ and $2$ .

Type 2: a quartic by comparing coefficients

The equation $z^4 + az^3 + bz^2 - 8z + 20 = 0$ , with $a$ and $b$ real, has a root $1 - 2i$ . Find $a$ and $b$ , and solve the equation.

Solution

The conjugate $1 + 2i$ is also a root. Sum $2$ , product $5$ : the factor is $z^2 - 2z + 5$ .

Write

z^4 + az^3 + bz^2 - 8z + 20 = (z^2 - 2z + 5)(z^2 + pz + q).

Compare coefficients, starting with the ones that involve only known numbers:

constant: $5q = 20$ , so $q = 4$ ;
$z$ : $5p - 2q = -8$ , so $5p = 0$ and $p = 0$ ;
$z^3$ : $a = p - 2 = -2$ ;
$z^2$ : $b = q - 2p + 5 = 9$ .

The other factor is $z^2 + 4$ , so $z = \pm 2i$ .

$a = -2$ , $b = 9$ , and the roots are $1 + 2i$ , $1 - 2i$ , $2i$ , $-2i$ .

Building a polynomial from its roots

Find the cubic equation with real coefficients and leading coefficient $1$ that has roots $3$ and $-1 + i\sqrt{2}$ .

Solution

Real coefficients means $-1 - i\sqrt{2}$ is also a root. That pair has sum $-2$ and product $(-1)^2 + (\sqrt{2})^2 = 3$ , so its factor is $z^2 + 2z + 3$ .

(z - 3)(z^2 + 2z + 3) = z^3 + 2z^2 + 3z - 3z^2 - 6z - 9 = z^3 - z^2 - 3z - 9.

The equation is $z^3 - z^2 - 3z - 9 = 0$ .

Exam-hard: verify a root, then factorise a quartic

(a) Show that $z = 1 + i$ is a root of the equation $z^4 + 3z^2 - 6z + 10 = 0$ .

(b) Hence find the other three roots.

Solution

(a) $z^2 = (1 + i)^2 = 2i$ and $z^4 = (2i)^2 = -4$ . Then

z^4 + 3z^2 - 6z + 10 = -4 + 6i - 6 - 6i + 10 = 0.

(b) The coefficients are real, so $1 - i$ is also a root. Sum $2$ , product $2$ : factor $z^2 - 2z + 2$ .

z^4 + 0z^3 + 3z^2 - 6z + 10 = (z^2 - 2z + 2)(z^2 + pz + q).

Constant: $2q = 10$ , so $q = 5$ . $z^3$ : $p - 2 = 0$ , so $p = 2$ . Check $z^2$ : $q - 2p + 2 = 5 - 4 + 2 = 3$ . Check $z$ : $2p - 2q = 4 - 10 = -6$ .

Solve $z^2 + 2z + 5 = 0$ : $(z + 1)^2 = -4$ , so $z = -1 \pm 2i$ .

The other roots are $1 - i$ , $-1 + 2i$ and $-1 - 2i$ .

Exam-hard: a non-monic cubic

Solve the equation $2z^3 - z^2 + 5z + 3 = 0$ , giving the complex roots in exact form.

Solution

The constant is $3$ and the leading coefficient is $2$ , so try $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$ .

$p\left(-\tfrac{1}{2}\right) = 2\left(-\tfrac{1}{8}\right) - \tfrac{1}{4} - \tfrac{5}{2} + 3 = -\tfrac{1}{4} - \tfrac{1}{4} - \tfrac{5}{2} + 3 = 0$ .

So $(2z + 1)$ is a factor. Write $2z^3 - z^2 + 5z + 3 = (2z + 1)(z^2 + bz + 3)$ . The $z^2$ coefficient gives $2b + 1 = -1$ , so $b = -1$ . Check $z$ : $b + 6 = 5$ .

Solve $z^2 - z + 3 = 0$ :

z = \frac{1 \pm \sqrt{1 - 12}}{2} = \frac{1 \pm i\sqrt{11}}{2}.

The roots are $-\tfrac{1}{2}$ , $\tfrac{1}{2} + \tfrac{\sqrt{11}}{2}i$ and $\tfrac{1}{2} - \tfrac{\sqrt{11}}{2}i$ .

On an Argand diagram the roots of a real polynomial are always symmetric about the real axis: each non-real root has its mirror image, and the real roots sit on the axis.

Common mistakes

Applying the rule to complex coefficients

The conjugate-pair result needs real coefficients. $z^2 - (3 + i)z + (4 + 3i) = 0$ has roots $2 - i$ and $1 + 2i$ , which are not conjugates. Always check the coefficients before writing "the conjugate is also a root".

Wrong signs in the quadratic factor

The factor from $a \pm bi$ is $z^2 - 2az + (a^2 + b^2)$ . For $-3 + 2i$ it is $z^2 + 6z + 13$ , not $z^2 - 6z + 13$ . The middle coefficient is minus the sum; the sum here is $-6$ .

Equating only one part

When substituting a root, the result $X + Yi = 0$ means $X = 0$ and $Y = 0$ . You need both equations to find two unknowns.

Sloppy powers

Build powers step by step: $z^2$ , then $z^3 = z^2 \cdot z$ , then $z^4 = (z^2)^2$ . Expanding $(1 + 2i)^3$ in one go with the binomial theorem and powers of $i$ is where most sign errors happen.

Exam technique

Exam tip

State the reason when you use the conjugate: "since the coefficients are real, $2 - i$ is also a root". This sentence is often worth a mark.
"Verify that $\alpha$ is a root" or "show that $\alpha$ is a root" needs the substitution with each power shown and a final line equal to $0$ . Showing $X + Yi$ with $X = 0$ and $Y = 0$ is ideal.
When comparing coefficients, use the highest and lowest powers first (they involve the fewest unknowns), then use a remaining coefficient as a check. Say "check" and show it.
Give all roots at the end, as a list. Do not leave a factorised polynomial as the answer to "solve".
If a question says "find the roots" and you have found a real root from a graph or calculator, you still need the factor theorem working to justify it.

Summary

For a polynomial with real coefficients, non-real roots come in conjugate pairs $a \pm bi$ .
Each pair gives the real quadratic factor $z^2 - 2az + (a^2 + b^2)$ : minus the sum, plus the product.
A real cubic always has at least one real root; a real quartic has $0$ , $2$ or $4$ real roots.
Given one complex root: write the conjugate, form the quadratic factor, divide or compare coefficients.
Unknown coefficients: substitute the root and equate real and imaginary parts, or compare coefficients.
A real root can often be found by the factor theorem first, leaving a quadratic.
On an Argand diagram the roots of a real polynomial are symmetric about the real axis.

Practice

Question

Given that $1 + 3i$ is a root of $z^3 - 4z^2 + 14z - 20 = 0$ , find the other roots without using the factor theorem.
Find the quartic equation with real coefficients and leading coefficient $1$ whose roots include $2i$ and $1 + i$ .
The equation $z^3 + pz^2 + qz + 26 = 0$ , with $p$ and $q$ real, has a root $-3 + 2i$ . Find $p$ , $q$ and the third root.
Show that $z = 2i$ is a root of $z^3 - 3z^2 + 4z - 12 = 0$ and find the other roots.
Solve $z^3 - z^2 + z + 3 = 0$ .
The equation $z^3 + az^2 + bz + 6 = 0$ , with $a$ and $b$ real, has a root $1 - i$ . Find $a$ and $b$ by substituting the root, and find the other roots.
Verify that $2i$ is a root of $z^4 - 4z^3 + 9z^2 - 16z + 20 = 0$ and find all the roots.
Find the quartic equation with real coefficients and leading coefficient $1$ that has roots $1 + i$ and $2 - i$ , giving it in expanded form.
The equation $z^3 + kz + 20 = 0$ , where $k$ is real, has a root $1 + 3i$ . Find $k$ and the other two roots.

Answers

Real coefficients, so $1 - 3i$ is a root. Sum $2$ , product $10$ : factor $z^2 - 2z + 10$ . Then $z^3 - 4z^2 + 14z - 20 = (z^2 - 2z + 10)(z + k)$ with $10k = -20$ , $k = -2$ (check $z^2$ : $k - 2 = -4$ ). Roots $1 \pm 3i$ and $2$ .
The roots must include the conjugates $-2i$ and $1 - i$ . Factors $z^2 + 4$ and $z^2 - 2z + 2$ : $(z^2 + 4)(z^2 - 2z + 2) = z^4 - 2z^3 + 2z^2 + 4z^2 - 8z + 8 = z^4 - 2z^3 + 6z^2 - 8z + 8$ . So $z^4 - 2z^3 + 6z^2 - 8z + 8 = 0$ .
$-3 - 2i$ is a root; sum $-6$ , product $13$ ; factor $z^2 + 6z + 13$ . $(z^2 + 6z + 13)(z + k)$ with $13k = 26$ , $k = 2$ . Expanding, $z^3 + 8z^2 + 25z + 26$ , so $p = 8$ , $q = 25$ , and the third root is $-2$ .
$z^2 = -4$ , $z^3 = -8i$ : $-8i - 3(-4) + 4(2i) - 12 = -8i + 12 + 8i - 12 = 0$ . With $-2i$ also a root, $z^2 + 4$ is a factor: $z^3 - 3z^2 + 4z - 12 = (z^2 + 4)(z - 3)$ . Roots $2i$ , $-2i$ , $3$ .
$p(-1) = -1 - 1 - 1 + 3 = 0$ , so $(z + 1)$ is a factor: $z^3 - z^2 + z + 3 = (z + 1)(z^2 - 2z + 3)$ . Then $z = \dfrac{2 \pm \sqrt{4 - 12}}{2} = 1 \pm i\sqrt{2}$ . Roots $-1$ , $1 + i\sqrt{2}$ , $1 - i\sqrt{2}$ .
$z^2 = (1 - i)^2 = -2i$ , $z^3 = -2i(1 - i) = -2 - 2i$ . Substitute: $(-2 - 2i) + a(-2i) + b(1 - i) + 6 = 0$ . Real: $4 + b = 0$ , $b = -4$ . Imaginary: $-2 - 2a - b = 0$ , so $-2a = -2$ , $a = 1$ . The equation is $z^3 + z^2 - 4z + 6 = 0 = (z^2 - 2z + 2)(z + 3)$ . Other roots $1 + i$ and $-3$ .
$z^2 = -4$ , $z^3 = -8i$ , $z^4 = 16$ : $16 + 32i - 36 - 32i + 20 = 0$ . With $-2i$ , the factor $z^2 + 4$ : $z^4 - 4z^3 + 9z^2 - 16z + 20 = (z^2 + 4)(z^2 - 4z + 5)$ (constant $4 \times 5 = 20$ ; $z^3$ : $-4$ ; check $z^2$ : $5 + 4 = 9$ ; $z$ : $-16$ ). Then $z^2 - 4z + 5 = 0$ gives $z = 2 \pm i$ . Roots $\pm 2i$ , $2 \pm i$ .
Conjugates $1 - i$ and $2 + i$ are also roots. Factors $z^2 - 2z + 2$ and $z^2 - 4z + 5$ : $(z^2 - 2z + 2)(z^2 - 4z + 5) = z^4 - 4z^3 + 5z^2 - 2z^3 + 8z^2 - 10z + 2z^2 - 8z + 10 = z^4 - 6z^3 + 15z^2 - 18z + 10$ . So $z^4 - 6z^3 + 15z^2 - 18z + 10 = 0$ .
$1 - 3i$ is also a root, giving the factor $z^2 - 2z + 10$ . Write $z^3 + 0z^2 + kz + 20 = (z^2 - 2z + 10)(z + c)$ . The $z^2$ coefficient: $c - 2 = 0$ , so $c = 2$ ; check constant: $10 \times 2 = 20$ . The $z$ coefficient: $k = 10 - 2c = 6$ . So $k = 6$ and the roots are $1 + 3i$ , $1 - 3i$ , $-2$ .

The conjugate root theorem

Consequences for cubics and quartics

The sum and product shortcut

The three standard question types

Worked examples

Common mistakes

Exam technique

Summary

Practice

Console