University of Sheffield

School of Mathematics and Statistics

MAS140: Mathematics (Chemical)

MAS152: Civil Engineering Mathematics

MAS152: Essential Mathematical Skills & Techniques

MAS156: Mathematics (Electrical and Aerospace)

MAS161: General Engineering Mathematics

Semester 1 2017–18

Outline Syllabus

• Functions of a real variable. The concept of a function; odd, even and periodic

functions; continuity. Binomial theorem.

• Elementary functions. Circular functions and their inverses. Polynomials.

Exponential, logarithmic and hyperbolic functions.

• Differentiation. Basic rules of differentiation: maxima, minima and curve sketching.

• Partial differentiation. First and second derivatives, geometrical interpretation.

• Series. Taylor and Maclaurin series, L’Hopital’s rule.

• Complex numbers. basic manipulation, Argand diagram, de Moivre’s theorem,

Euler’s relation.

• Vectors. Vector algebra, dot and cross products, differentiation.

Module Materials

These notes supplement the video lectures. All course materials, including examples sheets

(with worked solutions), are available on the course webpage,

http://engmaths.group.shef.ac.uk/mas140/

http://engmaths.group.shef.ac.uk/mas151/

http://engmaths.group.shef.ac.uk/mas152/

http://engmaths.group.shef.ac.uk/mas156/

http://engmaths.group.shef.ac.uk/mas161/

which can also be accessed through MOLE.

1

1 Complex Numbers

The equation

x2 − 1 = 0

has the solutions x = ±1 and the equation

x2 − 6x+ 5 = 0

⇒ (x− 5)(x− 1) = 0

has the roots 5 and 1. However, the equation

x2 + 1 = 0

is not satisfied by any (real) number x (indeed, x2 ≥ 0 for all x ∈ R).

We define the quantity i by i =√−1 so that i2 = −1. The last equation now has the two

roots i and −i. (Do not be tempted to think that one of them is positive and one negative.

The quantities i and −i are neither positive nor negative!)

A quantity z of the form

z = x+ iy,

where x and y are real numbers, is called a complex number. They occur naturally when

solving quadratic equations. Thus if

25z2 − 20z + 13 = 0

then (using the standard formula)

z =20±

√202 − 4× 25× 13

2× 25

=2±√−9

5

=2

5± i3

5

For any real numbers x and y, z = x+ iy is a complex number. In particular, choosing y to

be zero shows that all real numbers are complex. In the same way, all integers are rational.

C ⊃ R ⊃ Q ⊃ Z

We do not distinguish between any of the following:

x+ iy, x+ yi, iy + x, yi+ x.

The symbol j is often used for√−1, especially in electrical engineering.

In what follows, x and y will always be real numbers and z a complex number.

2

If z = x+ iy then

x is called the real part of z

y is called the imaginary part of z

and we write

x = Re(z) and y = Im(z).

Note. The imaginary part of a complex number is real .

Example

If z = 4− 3i then Re(z) = 4 and Im(z) = −3.

(<(z) and =(z) are also used.)

A complex number of the form iy is called pure-imaginary . Its real part is zero.

The complex conjugate of the complex number z = x+ iy is denoted by z and is defined by

z = x− iy.

Example z = 4− 3i⇒ z = 4 + 3i.

Taking the complex conjugate twice gives the original number. i.e. if z1 = z, then z1 = z.

If z = z then Im(z) = 0 and z is a real number.

1.1 Algebraic Operations

Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers.

• Equality

z1 = z2 ⇔ x1 = x2 and y1 = y2

• Additionz1 + z2 = (x1 + iy1) + (x2 + iy2)

= (x1 + x2) + i(y1 + y2)

• Subtractionz1 − z2 = (x1 + iy1)− (x2 + iy2)

= (x1 − x2) + i(y1 − y2)

• Multiplication

z1z2 = (x1 + iy1)(x2 + iy2)

= (x1x2 − y1y2) + i(x1y2 + x2y1)

• Divisionz1z2

=x1 + iy1x2 + iy2

=(x1 + iy1)

(x2 + iy2)

(x2 − iy2)(x2 − iy2)

=(x1x2 + y1y2)

(x22 + y22)+ i

(−x1y2 + y1x2)

(x22 + y22)

3

Example: Find the real and imaginary parts of z =3− 4i

5 + 2i

Solution3− 4i

5 + 2i=

(3− 4i)

(5 + 2i)

(5− 2i)

(5− 2i)

=(3× 5− 4× 2) + i(−3× 2− 4× 5)

(25 + 4)

=7− 26i

29

The real part is Re(z) = 7/29 and the imaginary part is Im(z) = −26/29.

For any complex numbers z1, z2 and z3 we have

z1 + z2 = z2 + z1 (commutative)

z1z2 = z2z1

z1 + (z2 + z3) = (z1 + z2) + z3 (associative)

z1(z2z3) = (z1z2)z3

Note

• z + z = 2Re(z) and z − z = 2iIm(z)

• z1 + z2 = z1 + z2, z1 − z2 = z1 − z2

• z1z2 = (z1)(z2), z1/z2 = z1/z2

1.2 Geometrical Representation: The Argand Diagram

The complex number z = x+iy can be represented by the point

P with coordinates (x, y) in a plane. In this plane, known as

the Argand Diagram , the x-axis is called the real axis and the

y-axis is the imaginary axis.

θ

y

x P

O

1.2.1 Modulus and Argument

Instead of using Cartesian coordinates (i.e. x and y) to specify the position of P in the Argand

diagram, we may use polar coordinates r and θ. Then

x = r cos θ and y = r sin θ

4

and so

z = x+ iy = r(cos θ + i sin θ).

This is called the polar form of the complex number z. The non-negative real number r is

the modulus of z and θ is the argument of z.

We use the notation

|z| = r and arg z = θ.

Clearly

r =√x2 + y2, sin θ = y/r and cos θ = x/r.

The angle θ is measured positive in the anti-clockwise direction from the real axis. Values of θ

which differ by 2π (or 4π etc) correspond to the same direction in the Argand diagram. The

unique angle θ such that −π < θ ≤ π is called the principal value of the argument and we

shall denote it by Argz.

Example: Find the modulus and principal argument of

(a) 1 + i√

3 (b) −1 + i

Solution

(a) Let z = 1 + i√

3, then

|z| =√

12 + (√

3)2 = 2 and θ is such that

cos θ =1

2, sin θ =

√3

2and − π < θ ≤ π.

Hence the principal value of the argument is π/3.

(b) | − 1 + i| =√

(−1)2 + (1)2 =√

2 and θ is such that

cos θ = − 1√2, sin θ =

1√2

and − π < θ ≤ π.

Hence the principal value of the argument is 3π/4.

Note

(a) |z| = |z|

(b) z = x+ iy ⇒ zz = |z|2 = x2 + y2

(c) Re(z) ≤ |z| and Im(z) ≤ |z|

It follows from (b) that1

z=

z

|z|2.

5

In particular, if |z| = 1 then 1/z = z.

If the complex number z has modulus r and argument θ, then this is sometimes written as

z = 〈r, θ〉. Also

z = 〈r, θ〉 ⇒ z = 〈r,−θ〉

1.2.2 Addition and the Argand Diagram

If P represents z1 and Q represents z2 then R will

represent their sum z1 + z2. Note that OPRQ is a

parallelogram. Also the triangle inequality gives

OP + PR ≥ OR

and hence

OP +OQ ≥ OR.

This implies that

|z1 + z2| ≤ |z1|+ |z2|.

y

xP

QR

O

1.3 Multiplication of Complex Numbers in Polar Form

Letz1 = r1(cos θ1 + i sin θ1)

and z2 = r2(cos θ2 + i sin θ2)

thenz1z2 = r1(cos θ1 + i sin θ1)r2(cos θ2 + i sin θ2)

= r1r2[(cos θ1 cos θ2 − sin θ1 sin θ2)+

i(sin θ1 cos θ2 + cos θ1 sin θ2)]

= r1r2[(cos(θ1 + θ2) + i sin(θ1 + θ2)]

Or

〈r1, θ1〉〈r2, θ2〉 = 〈r1r2, θ1 + θ2〉

Alsoz1z2

=r1(cos θ1 + i sin θ1)

r2(cos θ2 + i sin θ2)

=r1r2

(cos θ1 + i sin θ1)(cos θ2 − i sin θ2)

=r1r2

[(cos θ1 cos θ2 + sin θ1 sin θ2)+

i(sin θ1 cos θ2 − cos θ1 sin θ2)]

=r1r2

[(cos(θ1 − θ2) + i sin(θ1 − θ2)]

6

Or〈r1, θ1〉〈r2, θ2〉

=

⟨r1r2, θ1 − θ2

⟩Hence we have the following results for multiplication and division:

(a) |z1z2| = |z1||z2|

(b) |z1/z2| = |z1|/|z2|

(c) arg(z1z2) = arg z1 + arg z2

(d) arg(z1/z2) = arg z1 − arg z2

Note that it is not possible to use Arg in results (c) and (d).

1.4 Loci

If the complex number z satisfies the equation

|z| = 1

then its distance from the origin is 1. So z lies on the circle of radius 1 centred on the origin.

This circle is often referred to as the unit circle .

Alternatively,

|z| = 1 ⇒√x2 + y2 = 1

⇒ x2 + y2 = 1

which is the equation of a circle, radius 1 and centre at the origin.

If |z| satisfies

|z − z0| = r

(where z0 is a fixed complex number and r is a fixed positive number) then z lies on the circle

of radius r with centre at z0.

To see this, let

z = x+ iy and z0 = x0 + iy0

then

z − z0 = (x− x0) + i(y − y0)

and so

|z − z0| = r ⇒√

(x− x0)2 + (y − y0)2 = r

⇒ (x− x0)2 + (y − y0)2 = r2

7

which is the equation of a circle, centre (x0, y0) and radius r.

Example: If z satisfies the equation

|z − i| = |z + i| (?)

what is the locus of z in the Argand diagram?

Solution 1: Geometry

The given equation implies that the distance between the point P representing z and the

point i is the same as that between P and −i. It follows that P lies on the perpendicular

bisector of the points i and −i.So P lies on the real axis.

i

- i

- 2 2

Solution 2: Analysis

Set z = x+ iy so that the equation (?) becomes

|x+ i(y − 1)| = |x+ i(y + 1)|⇒√x2 + (y − 1)2 =

√x2 + (y + 1)2

⇒ x2 + (y − 1)2 = x2 + (y + 1)2

⇒ x2 + y2 − 2y + 1 = x2 + y2 + 2y + 1

⇒ 4y = 0

⇒ y = 0

Thus the imaginary part of z is zero and so z is a real number and hence lies on the real axis.

Example: If |z + 1| =√

2|z − 1|, what is the locus of z?

Solution

Let z = x+ iy, then

|z + 1| = |(x+ 1) + iy| =√

(x+ 1)2 + y2

and |z − 1| = |(x− 1) + iy| =√

(x− 1)2 + y2

8

Thus

(x+ 1)2 + y2 = 2[(x− 1)2 + y2]

x2 + 2x+ 1 + y2 = 2[x2 − 2x+ 1 + y2]

= 2x2 − 4x+ 2 + 2y2

x2 + y2 − 6x+ 1 = 0

(x− 3)2 − 9 + y2 + 1 = 0

⇒ (x− 3)2 + y2 = 8

Thus the point representing z in the Argand diagram lies on a circle with radius√

8 and

centred on (3, 0). This result is known as the circle of Apollonius.

Example

Show that multiplying any complex number by i has the same effect as an anticlockwise

rotation of π/2.

Solution

If z has modulus r and argumentθ then

z = 〈r, θ〉 and i = 〈1, π/2〉

and so

iz = 〈r, θ〉〈1, π/2〉= 〈r × 1, θ + π/2〉= 〈r, θ + π/2〉

So iz has the same modulus as z but the argument is increased by π/2.

z

iz

9

1.5 De Moivre’s Theorem

Let z1 have modulus r1 and argument θ1. Also let z2 have modulus r2 and argument θ2. Then

z1z2 = 〈r1, θ1〉〈r2, θ2〉= 〈r1r2, θ1 + θ2〉

If θ1 = θ2 = θ and r1 = r2 = r then we see that

〈r, θ〉2 = 〈r2, 2θ〉

and so

(cos θ + i sin θ)2 = cos 2θ + i sin 2θ

Also

〈r, θ〉3 = 〈r2, 2θ〉〈r, θ〉= 〈r3, 3θ〉

and

(cos θ + i sin θ)n = cosnθ + i sinnθ

for any natural number n.

The generalization of this result is de Moivre’s theorem which states that

(cos θ + i sin θ)q = cos qθ + i sin qθ

for any rational number q.

More accurately, the r.h.s. gives one of the values of the l.h.s. If q = m/n (where m and n are

integers, n positive) then the l.h.s. has n possible values whereas the r.h.s. has just one value.

1.6 Euler’s Relation

If x is any real number, then

ex = 1 + x+x2

2!+x3

3!+x4

4!+ . . .

sinx = x− x3

3!+x5

5!− x7

7!+ . . .

cosx = 1− x2

2!+x4

4!− x6

6!+ . . .

10

These expansions are now taken to apply to all complex numbers. Thus

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!+ . . .

= 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!+ . . .

= 1− θ2

2!+θ4

4!+ . . .+ i

(θ − θ3

3!+θ5

5!+ . . .

)

= cos θ + i sin θ

which gives Euler’s relation,

eiθ = cos θ + i sin θ

Note

(a) e−iθ = cos θ − i sin θ,

(b) eiπ/2 = i,

(c) eiπ = −1,

(d) ei3π/2 = −i,

(e) ei2pπ = 1 for any integer p.

From the results

eiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ

it follows that

cos θ =1

2

(eiθ + e−iθ

)and

sin θ =1

2i

(eiθ − e−iθ

)Also, de Moivre’s theorem can now be written as(

eiθ)q

= eiqθ

for any rational number q.

Example: Find the real and imaginary parts of

z =(√

3 + i)6

(i− 1)3.

Solution

Let z1 = (√

3 + i) and z2 = (i− 1), and write z1 and z2 in polar form:

|z1| = |√

3 + i| =√

(√

3)2 + 12 =√

4 = 2

11

and if Argz1 = θ then

cos θ =

√3

2, sin θ =

1

2, − π < θ ≤ π ⇒ θ =

π

6

Thus z1 = (√

3 + i) = 2eiπ/6.

Also

|z2| = |i− 1| =√

(−1)2 + 12 =√

2

and if Argz2 = φ then

cosφ = − 1√2, sinφ =

1√2, − π < φ ≤ π ⇒ φ =

3π

4

Thus z2 = (i− 1) =√

2ei3π/4.

The numerator of z is therefore

(√

3 + i)6 =(2eiπ/6

)6= 26e6iπ/6 = 26e−iπ = −26

Also, the denominator of z is

(i− 1)−3 =(√

2ei3π/4)−3

= 2−3/2e−9iπ/4 = 2−3/2e−iπ/4

= 2−3/2[cos(π

4

)− i sin

(π4

)]= 2−3/2

[1√2− i 1√

2

]= 2−2(1− i)

Therefore

z =(√

3 + i)6

(i− 1)3= −26 × 2−2(1− i) = 16(i− 1) =⇒ Re(z) = −16, Im(z) = 16.

1.7 Applications of De Moivre’s Theorem

Example: Show that each of

z0 = 〈1, 0〉, z1 = 〈1, 2π/3〉, z2 = 〈1, 4π/3〉

satisfies the equation z3 = 1. Plot all 3 numbers in the Argand diagram.

Solution

Firstly z0 = 1⇒ z30 = 1.

Also, since z = 〈r, θ〉 = r(cos θ + i sin θ)

⇒ z3 = 〈r3, 3θ〉 = r3(cos 3θ + i sin 3θ)

it follows that

z31 = 〈13, 3× 2π/3〉 = 〈1, 2π〉 = 1

and z32 = 〈13, 3× 4π/3〉 = 〈1, 4π〉 = 1

Since all of the z’s have modulus 1, they lie on the unit circle. They are called the

cube roots of unity.

12

2π/3

2π/3

..

.z

z

z

0

1

2

In the last example we verified that three given numbers were the cube roots of unity. We

now show how to find them directly. The method used will allow us to find all the roots of

any complex number.

Example: Find the cube roots of unity, i.e. the three values of 11/3.

Solution

First write 1 in the most general polar form. Since |1| = 1 and Arg(1) = 0 it follows that

arg(1) = 2pπ for any integer p and that

1 = 1[cos(2pπ) + i sin(2pπ)] is the most general polar form of the (complex) number 1. Now

we require z = 11/3 and so, by de Moivre’s theorem,

z = {1[cos(2pπ) + i sin(2pπ)]}1/3

= cos

(2pπ

3

)+ i sin

(2pπ

3

)for any integer p. Setting p equal to 0, 1 and 2 in turn gives the required 3 values. It is usual

to denote the corresponding z values by the subscript p.

p = 0 ⇒ z0 = cos 0 + i sin 0 = 1

p = 1 ⇒ z1 = cos2π

3+ i sin

2π

3= −1

2+ i

√3

2

p = 2 ⇒ z2 = cos4π

3+ i sin

4π

3= −1

2− i√

3

2

These are the 3 numbers in the previous example.

Note

(a) Any other value of p will give one of z0, z1, z2. Thus there are exactly 3 distinct

values of z for which z3 = 1.

(b) z1 and z2 form a complex conjugate pair. This illustrates a general result, viz. that if

z is a root of a polynomial with real coefficients, the z is also a root. (Of course, if z is

real then this says nothing.)

13

1.7.1 The nth Roots of Unity

If n is any positive integer, we say that z is an nth root of unity if zn = 1. To find z, proceed

as before:

1 = 1[cos(2pπ) + i sin(2pπ)] (p any integer)

z = 11/n = [cos(2pπ) + i sin(2pπ)]1/n

= cos

(2pπ

n

)+ i sin

(2pπ

n

)Taking p = 0, 1, 2, . . . , (n− 1) will give the n roots. (Other values of n will just repeat the

values.) They all have modulus 1 and so lie on the unit circle.

..z

z

0

1

.

.

z2

zn -1

θ

θ

θ

The angle θ is 2π/n. The roots are equally spaced around the unit circle.

Example: Find all values z for which

z3 = 1− i√

3

Solution: First write 1− i√

3 in the most general polar form. Since

|1− i√

3| =√

12 + (−√

3)2 =√

4 = 2

and, for the principal argument θ,

cos θ =1

2, sin θ = −

√3

2, − π < θ ≤ π ⇒ θ = −π

3

Thus

z3 = 2[cos(−π

3+ 2pπ

)+ i sin

(−π

3+ 2pπ

)]where p is any integer. De Moivre’s theorem now gives

z = 21/3

[cos

(−π

9+

2pπ

3

)+ i sin

(−π

9+

2pπ

3

)]

14

Now set p equal to 0,1,2 in turn to give the three values for z:

z0 = 21/3[cos(−π

9

)+ i sin

(−π

9

)]z1 = 21/3

[cos

(5π

9

)+ i sin

(5π

9

)]

z2 = 21/3

[cos

(11π

9

)+ i sin

(11π

9

)]z0, z1, z2 are equally spaced around the circle |z| = 21/3.

1.7.2 Deriving multi-angle trigonometric formulae using de Moivre’s theorem

De Moivre’s theorem (and the Binomial expansion) can be used to easily find expressions for

multi-angle trigonometric functions. For any positive integer n, de Moivre’s theorem states

that

(cos θ + i sin θ)n = cosnθ + i sinnθ.

Equating real and imaginary parts, we see that

cosnθ = Re [(cos θ + i sin θ)n] , sinnθ = Im [(cos θ + i sin θ)n] .

Example: Expand cos 2θ and sin 2θ as powers of cos θ and sin θ.

Solution: Since cos 2θ + i sin 2θ = (cos θ + i sin θ)2, we see that cos 2θ and sin 2θ are the

real and imaginary parts of (cos θ + i sin θ)2. Thus we expand (cos θ + i sin θ)2 by the

binomial theorem and equate real and imaginary parts:

(cos θ + i sin θ)2 = cos2 θ + 2i cos θ sin θ + (i sin θ)2

= cos2 θ − sin2 θ + 2i sin θ cos θ.

Hencecos 2θ = cos2 θ − sin2 θ

sin 2θ = 2 sin θ cos θ.

Using cos2 θ + sin2 θ = 1, we can also write cos 2θ as

cos 2θ = 2 cos2 θ − 1

or cos 2θ = 1− 2 sin2 θ.

Example: Expand cos 3θ as a sum of powers of cos θ.

Solution: From de Moivre’s theorem with n = 3 we have

cos 3θ + i sin 3θ = (cos θ + i sin θ)3

15

and so cos 3θ is the real part of (cos θ + i sin θ)3. Again we use the Binomial Theorem to

expand this expression:

(cos θ + i sin θ)3 = (cos θ)3 + 3(cos θ)2(i sin θ) + 3 cos θ(i sin θ)2 + (i sin θ)3

= cos3 θ + 3i sin θ cos2 θ − 3 sin2 θ cos θ − i sin3 θ.

Hence cos 3θ = cos3 θ − 3 sin2 θ cos θ. Since cos2 θ + sin2 θ = 1, we can write cos 3θ as

cos 3θ = cos3 θ − 3 cos θ(1− cos2 θ)

= 4 cos3 θ − 3 cos θ.

Exercise: Show thatsin 3θ = 3 sin θ − 4 sin3 θ

cos 4θ = 8 cos4 θ − 8 cos2 θ + 1

sin 4θ = 4 cos θ(sin θ − 2 sin3 θ).

1.7.3 Finding the mth roots of a complex number

De Moivre’s theorem holds when n is rational (and not just when n is an integer). However, in

this case, cosnθ + i sinnθ is just one of several values that (cos θ + i sin θ)n may have.

Suppose that n = 1/m (where m is an integer), then

(cos θ + i sin θ)1/m = cosθ

m+ i sin

θ

m.

This gives one of the mth roots of the complex number z = cos θ + i sin θ. However, there are

a total of m distinct mth roots of z. We can obtain them all by writing z in the form

z = cos θ + i sin θ = cos(θ + 2πp) + i sin(θ + 2πp)

where p is any integer. De Moivre’s theorem now gives

z1/m = [cos(θ + 2πp) + i sin(θ + 2πp)]1/m

= cos

(θ + 2πp

m

)+ i sin

(θ + 2πp

m

).

Taking p = 0, 1, 2, . . . ,m− 1 gives all the different roots. (Other values of p could be used,

but they will just repeat previous roots.) We can extend this method to any complex number

by writing it in the form

z = r(cos θ + i sin θ), r > 0

(modulus-argument form) or

z = r{cos(θ + 2πp) + i sin(θ + 2πp)}, r > 0

where again p is any integer. Applying de Moivre’s theorem now gives

z1/m = r1/m [cos(θ + 2πp) + i sin(θ + 2πp)]1/m

= r1/m[cos

(θ + 2πp

m

)+ i sin

(θ + 2πp

m

)].

16

Alternatively we may use the exponential notation and then

z = reiθ = rei(θ+2πp),

so that

z1/m = r1/m exp

[i

(θ + 2πp

m

)]= r1/m

[cos

(θ + 2πp

m

)+ i sin

(θ + 2πp

m

)],

where p = 0, 1, 2, . . . ,m− 1.

Example: Find the two square roots of z = 4i.

Solution: In modulus-argument (polar) form,

z = 4[cos(π

2+ 2πp

)+ i sin

(π2

+ 2πp)]

for any integer p. Therefore, using de Moivre’s theorem,

z1/2 = 41/2[cos(π

4+ πp

)+ i sin

(π4

+ πp)], p = 0, 1.

p = 0 ⇒ z1/2 = 2(

cosπ

4+ i sin

π

4

)= 2

(1√2

+i√2

)=√

2 + i√

2

p = 1 ⇒ z1/2 = 2

(cos

5π

4+ i sin

5π

4

)= 2

(− 1√

2− i√

2

)= −√

2− i√

2

Alternatively, we may use the exponential notation:

z = 4ei(π/2+2πp) ⇒ z1/2 = 41/2ei(π/4+πp) = 2ei(π/4+πp), p = 0, 1

and now substitute in the values for p to get

p = 0 ⇒ z1/2 = 2eiπ/4 = 2(

cosπ

4+ i sin

π

4

)=√

2 + i√

2

p = 1 ⇒ z1/2 = 2ei5π/4 = 2

(cos

5π

4+ i sin

5π

4

)= −√

2− i√

2

Example: Find the cube roots of z =√

3 + i and show them on an Argand diagram.

Solution: In polar form

z = 2[cos(π

6+ 2πp

)+ i sin

(π6

+ 2πp)]

⇒ z1/3 = 21/3

[cos

(π

18+

2πp

3

)+ i sin

(π

18+

2πp

3

)], p = 0, 1, 2

p = 0 ⇒ z0 = z1/3 = 21/3(

cosπ

18+ i sin

π

18

)p = 1 ⇒ z1 = z1/3 = 21/3

(cos

13π

18+ i sin

13π

18

)p = 2 ⇒ z2 = z1/3 = 21/3

(cos

25π

18+ i sin

25π

18

)17

Argand Diagram to show the roots

on the circle |z| = 21/3.

z

z

z

0

1

2

Example: Find the four distinct fourth roots of z = −1.

Solution: In exponential form,

z = ei(π+2πp)

⇒ z1/4 = ei(2p+1)π/4

for any integer p. Therefore

p = 0 ⇒ z0 = z1/4 = eiπ/4 = cosπ

4+ i sin

π

4=

1 + i√2

p = 1 ⇒ z1 = z1/4 = ei3π/4 = cos3π

4+ i sin

3π

4=−1 + i√

2

p = 2 ⇒ z2 = z1/4 = ei5π/4 = cos5π

4+ i sin

5π

4=−1− i√

2

p = 3 ⇒ z3 = z1/4 = ei7π/4 = cos7π

4+ i sin

7π

4=

1− i√2

To summarise, there are three steps to carry out to find the mth roots of a complex number:

1. Write the complex number z in general polar form

z = r[cos(θ + 2πp) + i sin(θ + 2πp)], r > 0

or z = rei(θ+2πp)

2. Use de Moivre’s theorem

z1/m = r1/m[cos

(θ + 2πp

m

)+ i sin

(θ + 2πp

m

)]or z1/m = r1/mei(θ+2πp)/m

3. Let p = 0, 1, 2, . . . ,m− 1 to get the m roots.

18

1.7.4 Expressing Powers of Trigonometric Functions as Multiple Angle

Functions

It is sometimes useful to be able to express cosn θ and sinn θ, where n is a positive integer, in

terms of cosines and sines of multiples of θ. This is necessary when such expressions have to

be integrated. To achieve this we use the Binomial expansion and the identities

cos θ =1

2(eiθ + e−iθ), sin θ =

1

2i(eiθ − e−iθ)

.

Example: Express cos2 θ and sin2 θ in terms of cos 2θ.

Solution:

cos2 θ = (cos θ)2 =

[1

2(eiθ + e−iθ)

]2=

1

4

[(eiθ)2 + 2eiθ.e−iθ + (e−iθ)2

]=

1

4(e2iθ + 2 + e−2iθ).

But e2iθ + e−2iθ = 2 cos 2θ, and so

cos2 θ =1

2(1 + cos 2θ).

We can then use the relation sin2 θ = 1− cos2 θ to deduce that

sin2 θ =1

2(1− cos 2θ).

Example: Express cos3 θ in multiple angles.

Solution:

cos3 θ = (cos θ)3 =

[1

2(eiθ + e−iθ)

]3

=1

8(e3iθ + 3e2iθ.e−iθ + 3eiθ.e−2iθ + e−3iθ)

=1

8(e3iθ + 3eiθ + 3e−iθ + e−3iθ)

=1

8

[(e3iθ + e−3iθ) + 3(eiθ + e−iθ)

]=

1

8[2 cos 3θ + 6 cos θ]

=1

4(cos 3θ + 3 cos θ).

19

Exercise: Find

1. the fifth roots of −6√

3 + 6i

2. the sixth roots of −1

3. the fifth roots of i

4. the cube roots of 1 + i.

Except for (2), leave the answer in polar form.

Answers

1. 1·64(cosφ+ i sinφ), where φ = π6, 17π

30, 29π

30, 41π

30, 53π

30.

2. ±i,√32± i

2,−√32± i

2.

3. cosφ+ i sinφ, where φ = π10, π2, 9π10, 13π

10, 17π

10.

4. 1·12(cosφ+ i sinφ), where φ = π12, 3π

4, 17π

12.

Exercise: Show that

1. sin3 θ = 14(3 sin θ − sin 3θ)

2. cos4 θ = 18(cos 4θ + 4 cos 2θ + 3)

3. sin4 θ = 18(cos 4θ − 4 cos 2θ + 3)

In all of these examples, the expansion of a power of cosine gives a series involving only

cosines of multiples of the angle. This is also true for an even power of sine, but an odd

power of sine gives a series with sines of multiple angles. This reflects the fact that both

cosn θ and sin2n θ are even functions (i.e. their value is unchanged if θ is replaced by −θ) but

sin2n+1 θ is an odd function (its value changes sign if θ is replaced by −θ).

1.8 Circular and Hyperbolic Functions of Complex Numbers

We have already shown that

cos θ =1

2

(eiθ + e−iθ

)sin θ =

1

2i

(eiθ − e−iθ

)

20

Also the definitions of the basic hyperbolic functions are

coshx =1

2

(ex + e−x

)sinhx =

1

2

(ex − e−x

)If these last two definitions are also taken to apply to complex numbers, then we have

cosh iθ = cos θ

sinh iθ = i sin θ

and also

cos iθ = cosh θ

sin iθ = i sinh θ

The results are true for any complex number θ (not just real and pure-imaginary).

All of the standard trigonometric identities are valid for all complex numbers. For

example, from

sin(θ + φ) = sin θ cosφ+ cos θ sinφ

it follows that

sin(x+ iy) = sinx cos(iy) + cos x sin(iy)

= sin x cosh y + i cosx sinh y

So, if x and y are real, then

Re(sin(x+ iy)) = sinx cosh y

Im(sin(x+ iy)) = cosx sinh y

Example: Find a z for which sin z = 2.

Solution: Let z = x+ iy, then we require

Re(sin(x+ iy)) = sinx cosh y = 2

Im(sin(x+ iy)) = cosx sinh y = 0

and these are satisfied by x = π/2 = 1.57079 and y = cosh−1 2 = 1.31695.

21