IB Analysis and Approaches HL · Complex numbers · Section 4 of 4

Polynomials with complex roots

3 worked examples · By Álvaro López

Key idea

A polynomial of degree $n$ always has $n$ roots if complex numbers are allowed, and when its coefficients are real, the non-real roots appear in conjugate pairs. So knowing one complex root gives you another one for free, and the sum and product of the roots can be read directly from the coefficients. In the IB exam this is used to find missing roots, rebuild a polynomial and find parameters.

Every polynomial has as many roots as its degree

In topic 1 you saw that a zero $c$ of $P (x)$ is equivalent to the factor $x - c$ (the factor theorem), but only with real numbers. If complex numbers are allowed, none is lost. Here the variable is written $z$ , because the roots may be non-real.

It is not proved in this course. What matters is the consequence: the equation $P (z) = 0$ of degree $n$ has exactly $n$ complex solutions, counting repeated ones. Each time you find one, the factor theorem lets you divide and reduce the degree by one.

For example, $z^{2} + 4 = 0$ has no real roots, but it has two complex ones, $z = 2 i$ and $z = - 2 i$ , and $z^{2} + 4 = (z - 2 i) (z + 2 i)$ .

Conjugate roots

The reason lies in how conjugation behaves with the operations (addition, multiplication and powers, which you already know from the Cartesian form) and in the fact that a real number equals its own conjugate. If $P (w) = 0$ , conjugate both sides:

$0 = 0^{*} = (a_{n} w^{n} + \dots + a_{1} w + a_{0})^{*} = a_{n}^{*} (w^{*})^{n} + \dots + a_{1}^{*} w^{*} + a_{0}^{*} = a_{n} (w^{*})^{n} + \dots + a_{1} w^{*} + a_{0} = P (w^{*})$

The key step is $a_{k}^{*} = a_{k}$ : this only holds if the coefficients are real. If any of them is not, the property fails.

Consequences worth keeping in mind:

Non-real roots come in pairs $a + bi$ and $a - bi$ . If the polynomial is real of odd degree, at least one real root is left without a partner.
Each pair forms a quadratic factor with real coefficients:

$(z - (a + bi)) (z - (a - bi)) = z^{2} - 2 a z + (a^{2} + b^{2})$

The sum of the pair is $2 a$ and its product is $∣ w ∣^{2} = a^{2} + b^{2}$ : both real, as they should be.

Factorising over the reals and over the complex numbers

A real quadratic $z^{2} + p z + q$ has no real roots if its discriminant $p^{2} - 4 q$ is negative; then its roots are a conjugate pair. So, over the reals, a polynomial factorises into linear and irreducible quadratic factors, and over the complex numbers only into linear ones.

Rebuilding a polynomial from its roots

If you know all the roots $w_{1}, \dots, w_{n}$ and the leading coefficient, the fundamental theorem gives the polynomial simply by multiplying the factors. With real coefficients, always add the conjugates of the non-real roots: if you are told it has the root $3 + i$ , the real polynomial of lowest degree with that root also needs $3 - i$ , and so it has degree at least 2.

Sum and product of the roots

When you expand $a_{n} (z - w_{1}) \dots (z - w_{n})$ and compare it with $a_{n} z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{0}$ , the coefficients are linked to the roots. You do not need to know the roots to use these relations, and they also hold with complex coefficients.

Specific cases that are often used:

Degree	Polynomial	Sum of roots	Product of roots
2	$a z^{2} + b z + c$	$- \frac{b}{a}$	$\frac{c}{a}$
3	$a z^{3} + b z^{2} + cz + d$	$- \frac{b}{a}$	$- \frac{d}{a}$
4	$a z^{4} + b z^{3} + c z^{2} + d z + e$	$- \frac{b}{a}$	$\frac{e}{a}$

The check in degree 2 is immediate: $a (z - w_{1}) (z - w_{2}) = a z^{2} - a (w_{1} + w_{2}) z + a w_{1} w_{2}$ . Pay attention to the sign and to the division by the leading coefficient: the polynomial $2 z^{3} - 6 z^{2} + \dots$ has a sum of roots $- \frac{- 6}{2} = 3$ , not $6$ or $- 6$ .

Using these together with conjugation is very powerful: if the roots are $w$ and $w^{*}$ (and others), their sum gives $2 Re (w)$ and their product gives $∣ w ∣^{2}$ , two real numbers that are easy to handle.

Worked examples

Example 1 Basic

One complex root and the others

The equation $z^{3} - 5 z^{2} + 17 z - 13 = 0$ has the root $z = 2 + 3 i$ .

a) Write down another non-real root.

b) Find the real root.

c) Factorise $P (z) = z^{3} - 5 z^{2} + 17 z - 13$ as the product of a linear factor and a quadratic factor, both with real coefficients.

Solution

a) The coefficients of $P$ are real, so the conjugate of $2 + 3 i$ is also a root: $z = 2 - 3 i$ .

b) The two conjugate roots give the real quadratic factor. Their sum is $(2 + 3 i) + (2 - 3 i) = 4$ and their product is $2^{2} + 3^{2} = 13$ , so

$(z - (2 + 3 i)) (z - (2 - 3 i)) = z^{2} - 4 z + 13$

We divide $P (z)$ by $z^{2} - 4 z + 13$ . The quotient has degree 1, $z + m$ , and expanding $(z^{2} - 4 z + 13) (z + m) = z^{3} + (m - 4) z^{2} + (13 - 4 m) z + 13 m$ and comparing with $P$ : the constant term gives $13 m = - 13$ , so $m = - 1$ . Check of the coefficient of $z^{2}$ : $m - 4 = - 5$ . Correct, and also $13 - 4 m = 17$ . The third factor is $z - 1$ , so the real root is $z = 1$ .

Another way: the sum of the three roots is $- \frac{- 5}{1} = 5$ , so $(2 + 3 i) + (2 - 3 i) + r = 5$ and $r = 1$ .

c) $P (z) = (z - 1) (z^{2} - 4 z + 13)$ .

Example 2 Medium

Rebuilding a real polynomial of degree 4

A monic polynomial $P (z)$ of degree $4$ with real coefficients has the roots $2 i$ and $3 + i$ .

a) Write down the other two roots.

b) Find $P (z)$ expanded.

c) Factorise $P (z)$ over $R$ (linear or quadratic factors with real coefficients).

Solution

a) As $P$ has real coefficients, the conjugates $- 2 i$ and $3 - i$ are also roots. With the two given, they make up the four roots of the degree-4 polynomial, which are all of them.

b) We group the conjugate pairs:

Pair $\pm 2 i$ : sum $0$ and product $2^{2} = 4$ , factor $z^{2} + 4$ .
Pair $3 \pm i$ : sum $6$ and product $3^{2} + 1^{2} = 10$ , factor $z^{2} - 6 z + 10$ .

As it is monic,

$P (z) = (z^{2} + 4) (z^{2} - 6 z + 10) = z^{4} - 6 z^{3} + 10 z^{2} + 4 z^{2} - 24 z + 40 = z^{4} - 6 z^{3} + 14 z^{2} - 24 z + 40$

Check with the sum of the roots: $2 i - 2 i + (3 + i) + (3 - i) = 6$ , which matches $- \frac{- 6}{1} = 6$ . The product of the roots is $40 = (- 1)^{4} \cdot \frac{40}{1}$ , and it equals $4 \cdot 10 = 40$ .

c) The two quadratic factors have negative discriminant ( $0 - 16 < 0$ and $36 - 40 < 0$ ), so they cannot be factorised further over the reals:

$P (z) = (z^{2} + 4) (z^{2} - 6 z + 10)$

Example 3 Hard

Parameters from the sum and product of roots

Consider the polynomial $P (z) = z^{4} - 8 z^{3} + b z^{2} + cz + 65$ , with $b, c \in R$ . The four roots of $P (z) = 0$ are not real, and two of them are $1 + 2 i$ and $p + q i$ , with $p, q \in R$ , $q > 0$ , and $p + q i$ different from $1 + 2 i$ .

a) Find the sum and the product of the four roots.

b) Find $p$ and $q$ .

c) Find $b$ and $c$ .

Solution

As the coefficients are real and no root is real, the four roots are two conjugate pairs: $1 + 2 i$ , $1 - 2 i$ , $p + q i$ and $p - q i$ .

a) With $a_{4} = 1$ , $a_{3} = - 8$ and $a_{0} = 65$ :

$sum = - \frac{- 8}{1} = 8, product = (+ 1) \cdot \frac{65}{1} = 65$

b) The sum of the four roots is $2 \cdot 1 + 2 p = 8$ , so $p = 3$ . The product of each pair is the square of its modulus: $(1^{2} + 2^{2}) (p^{2} + q^{2}) = 5 (9 + q^{2}) = 65$ . Then $9 + q^{2} = 13$ , that is $q^{2} = 4$ , and since $q > 0$ , $q = 2$ . The roots are $1 \pm 2 i$ and $3 \pm 2 i$ .

c) We form the real factors of each pair: $1 \pm 2 i$ gives $z^{2} - 2 z + 5$ and $3 \pm 2 i$ gives $z^{2} - 6 z + 13$ . Then

$P (z) = (z^{2} - 2 z + 5) (z^{2} - 6 z + 13) = z^{4} - 8 z^{3} + (13 + 12 + 5) z^{2} + (- 26 - 30) z + 65$

$P (z) = z^{4} - 8 z^{3} + 30 z^{2} - 56 z + 65$

and therefore $b = 30$ and $c = - 56$ . The coefficient of $z^{3}$ and the constant term match those in the statement, which confirms the calculation.

In the exam: justify why the roots form two conjugate pairs (real coefficients and non-real roots) before using the sum and product: that sentence earns the method marks. And write the sum with the correct sign, $- \frac{a _{n - 1}}{a _{n}}$ .

Common mistakes

Summary

Fundamental theorem of algebra: a polynomial of degree $n \geq 1$ has $n$ complex roots (counting repetitions) and $P (z) = a_{n} (z - w_{1}) \dots (z - w_{n})$ .
If the coefficients are real and $w$ is a root, $w^{*}$ is also a root. Non-real roots come in pairs $a \pm bi$ ; a real polynomial of odd degree has at least one real root.
Each pair forms a real factor $z^{2} - 2 a z + (a^{2} + b^{2})$ : sum $2 a$ , product $a^{2} + b^{2}$ . Over $R$ it factorises into linear and quadratic factors with no real roots; over $C$ , only into linear ones.
To rebuild a real polynomial: add the conjugates of the non-real roots, group them into quadratic factors, multiply and adjust the leading coefficient.
Sum of roots $= - \frac{a _{n - 1}}{a _{n}}$ ; product of roots $= (- 1)^{n} \frac{a _{0}}{a _{n}}$ . For degree 2: $- \frac{b}{a}$ and $\frac{c}{a}$ .
With the GDC (Paper 2) you can obtain complex roots in complex mode, but always write down the set-up.

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