# 2 Complex numbers

## 2.9 Roots of polynomials

We begin by reminding you of what we mean by the word ‘root’. In this unit we use this term in two different, but related, senses, as given below.

### Definition

If p(z) is a polynomial, then the solutions of the polynomial equation p(z) = 0 are called the roots of p(z).

If a is a complex number, then the solutions of the equation zn = a are called the nth roots of a.

The roots of a polynomial are also called its zeros.

Thus the nth roots of a are the roots of the polynomial zna.

In this subsection we explain how to find the nth roots of any complex number and we discuss the roots of polynomial equations more generally.

The formula obtained in the previous section for finding the product of two complex numbers in polar form can be generalised to a product of several complex numbers.

### Strategy 2

To find the product of the complex numbers z1, z2, …, zn given in polar form, multiply their moduli and add their arguments.

### Exercise 18

Let z1 = −1 + i, and z3 = −5.

Use the solution to Exercise 16 to express z1z2z3 and in polar form.

### Solution

From the solution to Exercise 16,

Hence

using the principal argument.

Also

An important special case of the result in Strategy 2 is obtained when z1 = z2 = … = zn = r(cos θ + i sin θ); that is, when the complex numbers z1, z2, …, zn are all equal. In this case, the result gives

Note: nθ may not be the principal argument of (cos θ + isin θ)n.

### Example 5

Find z4, where z = 1 + i.

### Solution

So z4 = −4.

The significant new information obtained from equation (2.4) is that

This result is true for all n and is known as de Moivre's Theorem.

Abraham de Moivre was a French mathematician who worked in England from 1685 after the expulsion of the Huguenots from France.

### Theorem 2: de Moivre's Theorem

If z = cos θ + i sin θ, then, for any n ,

We have seen that de Moivre's Theorem is true when n is a positive integer. We now show that it is true for other integers.

For n = 0,

For n = −m, where m is a positive integer,

Here we use formula (2.3).

One application of de Moivre's Theorem is in finding the nth roots of complex numbers; that is, in solving equations of the form zn = a, where a . Before we illustrate this, we use the theorem to verify some solutions of such an equation.

### Exercise 19

(a)  Express the complex number 1 in polar form.
(b)  Show that each of the three complex numbers with polar forms 1(cos 0 + i sin 0),
and
satisfies the equation z3 = 1.
(c)  Express in Cartesian form the three solutions to the equation z3 = 1 given in part (b).

### Solution

(a)  1 = 1(cos 0 + i sin 0)
(b)

(c)  In Cartesian form,

The solution to Exercise 19 verifies that the three given complex numbers are solutions of the equation z3 = 1. However, what we really want is a method which will enable us to find solutions of such an equation without knowing them in advance. Fortunately, de Moivre's Theorem enables us to do this.

### Example 6

(a)  Express the equation z3= −27 in polar form.
(b)  Find three different solutions of the equation z3= −27, by using the fact that adding any multiple of 2 to an argument of −27 gives another argument of −27.

### Solution

(a)  In polar form, −27 = 27(cos + i sin ). If z = r(cos θ + i sin θ), then the equation z3= −27 can be written as

(b)  From equation (2.5), r3 = 27, so r = 3.
Also from equation (2.5), cos 3θ = cos and sin 3θ = sin , so one solution for θ is obtained by taking 3θ = , so
.
However, we could also take 3, 5, 7, … as arguments of −27, so 3θ = 3, 3θ = 5, 3θ = 7, … also give solutions. Hence we can have
....
But , so taking gives the same complex number as .
Similarly, increasing the value of the argument of −27 by any further multiple of 2 repeats solutions we already have.
So the three different solutions of z3 = −27 are given by

These solutions can also be written in Cartesian form as

They are shown in the diagram below.

The solution z3 could be rewritten in terms of its principal argument as 3(cos(−/3) + i sin(−/3)).

Now we show how we can use the method of Example 6 to find the solutions of any complex equation of the form

where a is a known complex number. We write both z and a in polar form so that, say,

where r and θ are variables whose values we must find, and ρ and φ are known real numbers.

Then zn = a gives, using de Moivre's Theorem,

Hence we must have rn = ρ, so r = ρ1/n. Also nθ must represent the same angle as φ. Now we use the fact that a complex number has many arguments. Since adding on any multiple of 2 to the argument φ of a gives the same complex number a, we can have

that is,

If k = n we have θ = φ/n + 2, which is the same angle as φ/n. So taking k = 0, 1, 2, …, n − 1 will give the n different solutions of the equation zn = a. We thus arrive at the following conclusion.

### Roots of a complex number

Let a = ρ(cos φ + i sin φ) be a complex number in polar form. Then, for any n , the equation zn = a has n solutions, given by

for k = 0, 1, …, n − 1.

### Exercise 20

(a)  Express the equation z6 = 1 in polar form.
(b)  Use the method described above to find the six solutions of z6 = 1 in polar form.
(c)  Sketch the position of each solution in the complex plane.
(d)  Find the Cartesian form of each solution.

### Solution

(a)  Let z = r(cos θ + i sin θ). Then, since

we have

(b)  Hence r = 11/6 = 1 and

for k = 0, 1, … 5, and the six solutions of z6 = 1 are given by

Hence the solutions are

(c)
(d)

In Exercise 20 you found the solutions of the equation z6 = 1. These are known as the sixth roots of unity, and in the complex plane they are equally spaced around the circle of radius 1, centre the origin. More generally, the solutions of the equation zn = 1 are known as the nth roots of unity and in the complex plane they are equally spaced around the circle of radius 1, centre the origin. For any n , one of the nth roots of unity is 1.

The nth roots of any complex number are equally spaced around a circle with centre the origin, but the circle may not have radius 1 and there may not be a root on the real axis.

### Exercise 21

Solve the equation z4 = −4, expressing your answers in Cartesian form. Mark your solutions on a diagram of the complex plane.

### Solution

Let z = r(cos θ + i sin θ). Then, since

we have

Hence r = 41/4 = and for k = 0, 1, 2, 3.

So the solutions are

### Exercise 22

Solve the equation z3 = 8i, expressing your answers in Cartesian form. Mark your solutions on a diagram of the complex plane.

### Solution

Let z = r(cos θ + i sin θ).

Since , we have

Hence r = 81/3 = 2 and for k = 0, 1, 2.

So the solutions are

The result in the box above gives the n solutions of any equation of the form zn = a, where a is a non-zero complex number. Now the equation zn = a, which can be written as zna = 0, is an example of a polynomial equation whose coefficients are complex numbers. Other examples of polynomial equations with complex coefficients are

and

Remarkably, it can be shown that any polynomial equation

where an, an−1, …, a0 and an ≠ 0, has at least one solution in . Moreover, it cannot have more than n solutions.

Note: If we count coincident solutions separately so that, for example, in the equation

the solution 1 is counted three times, the solution −4 is counted twice and the solution 5 is counted once, then a polynomial equation of degree n has exactly n solutions.

The complex numbers are, unlike the reals, an ‘algebraically closed’ system of numbers. By this we mean that any polynomial equation with coefficients in has a solution in . The fact that any polynomial equation with complex coefficients necessarily has a complex solution is called the Fundamental Theorem of Algebra. We do not prove this theorem in this unit.

It is often not easy to find such solutions! However, in some cases we can combine the methods of the Polynomial Factorisation Theorem below and the use of de Moivre's Theorem to find solutions of a polynomial equation.

You met the Polynomial Factorisation Theorem for real roots of a polynomial in OpenLearn unit M208_5 Mathematical language. The theorem is also true in .

### Theorem 3: Polynomial Factorisation Theorem

Let p(z) be a polynomial of degree n with coefficients in and let . Then p() = 0 if and only if

where q(z) is a polynomial of degree n−1 with coefficients in .

In this statement of the theorem, we have used z in place of x, as this is the label usually used for a complex variable. The proof is otherwise exactly the same as the proof for the theorem in , so we do not include it here.

From the Fundamental Theorem of Algebra, mathematical induction and the Polynomial Factorisation Theorem, we can deduce the following important corollary.

### Corollary

Every polynomial p(z) = anzn + an−1zn−1 + ··· + a1z + a0, where n ≥ 1, ai for each i and an ≠ 0, has a factorisation

where the complex numbers 1, 2, …, n are the roots (not necessarily distinct) of p(z).

Proof

Let S(n) be the statement of the corollary for general n ≥ 1.

Note: S(n) is the statement in the box above. We use the notation S(n) instead of the more usual P(n) to avoid confusion with the polynomial p(z).

Then S(1) is true, since the polynomial

(where a1 ≠ 0) has the root 1 = −a0/a1 and has a factorisation

Now suppose that S(k) is true, and consider any polynomial of degree k + 1:

where ak+1 ≠ 0. By the Fundamental Theorem of Algebra, the equation p(z) = 0 has at least one solution, say k+1, in . Then, by the Polynomial Factorisation Theorem,

where q(z) is a polynomial of degree k. Now the coefficient of zk in q(z) must be ak+1. Thus, by S(k), this polynomial has a factorisation

Thus

Therefore S(k) true S(k + 1) true, so by mathematical induction S(n) is true for every positive integer n.

You may have noticed that it follows from the quadratic formula (see Exercise 3) that, for a quadratic equation with real coefficients, the roots are either both real or occur as a complex conjugate pair.

This holds for polynomial equations in general; if p(z) is a polynomial in z with real coefficients, then whenever is a root of p, so is . (We omit the proof of this result.) Moreover, the factors z and can be combined to give a real quadratic factor of p(z), namely

which has real coefficients, since

and .

### Example 7

(a)  Show that z = i is a root of the polynomial

(b)  Hence find all the roots of p(z).

### Solution

(a)
so i is a root of p(z).
(b)  Since p has real coefficients, z = −i is also a root of p(z), so (zi)(z + i) = z2 + 1 is a factor of p(z). By equating coefficients, we obtain

So the remaining two roots of p(z) are given by the solutions of the equation z2 − 3z + 1 = 0.
Using the quadratic formula, we have

Hence the four roots of p(z) are i, −i,
and
.

### Exercise 23

Find, in the form anzn + … + a1z + a0, a polynomial whose roots are 1, −2, 3i and −3i.

### Solution

A suitable polynomial is

that is,

or

Earlier in this subsection we used de Moivre's Theorem to find roots of complex numbers. Another use of de Moivre's Theorem is to find further trigonometric identities similar to those given in OpenLearn unit M208_4 Real functions and graphs.

### Exercise 24

Using de Moivre's Theorem, we have

for all θ .

(a)  Expand the left-hand side of the above expression using the Binomial Theorem. Then express your answer in the form x + iy, where x and y are expressions involving cos θ and sin θ.
(b)  By equating real and imaginary parts, use your answer to part (a) to obtain formulas for cos 3θ and sin 3θ in terms of cos θ and sin θ.
(c)  Use your answer to part (b) and the identity cos2 θ + sin2 θ = 1 to obtain a formula for cos 3θ in terms of cos θ and a formula for sin 3θ in terms of sin θ.

### Solution

(a)

(b)
By part (a),

so

and

(c)  Since sin2 θ = 1 − cos2 θ and cos2 θ = 1 − sin2 θ, we have

so

and

so

The method of the solution to Exercise 24 generalises to produce formulas for cos nθ and sin nθ for all n .

### Exercise 25

(a)  Use de Moivre's Theorem to obtain formulas for cos 5θ and sin 5θ in terms of cos θ and sin θ.
(b)  Use your answer to part (a) and the identity cos2 θ + sin2 θ = 1 to find a formula for cos 5θ in terms of cos θ and a formula for sin 5θ in terms of sin θ.

### Solution

(a)

Equating real and imaginary parts gives

and

(b)
Since sin2 θ = 1 − cos2 θ and sin4 θ = (sin2 θ)2 =(1 − cos2 θ)2, on substituting into the formula for cos 5θ found in part (b) we obtain

so

Similarly,

so