M332_1

Complex numbers

About this free course

This free course provides a sample of Level 3 study in Mathematics: www.open.ac.uk/courses/find/mathematics

This version of the content may include video, images and interactive content that may not be optimised for your device.

You can experience this free course as it was originally designed on OpenLearn, the home of free learning from The Open University:
www.open.edu/openlearn/science-maths-technology/mathematics-and-statistics/mathematics/complex-numbers/content-section-0.

There you’ll also be able to track your progress via your activity record, which you can use to demonstrate your learning.

The Open University, Walton Hall, Milton Keynes, MK7 6AA

Copyright © 2016 The Open University

Intellectual property

Unless otherwise stated, this resource is released under the terms of the Creative Commons Licence v4.0 http://creativecommons.org/licenses/by-nc-sa/4.0/deed.en_GB. Within that The Open University interprets this licence in the following way: www.open.edu/openlearn/about-openlearn/frequently-asked-questions-on-openlearn. Copyright and rights falling outside the terms of the Creative Commons Licence are retained or controlled by The Open University.
Please read the full text before using any of the content.

We believe the primary barrier to accessing high-quality educational experiences is cost, which is why we aim to publish as
much free content as possible under an open licence. If it proves difficult to release content under our preferred Creative
Commons licence (e.g. because we can’t afford or gain the clearances or find suitable alternatives), we will still release
the materials for free under a personal end-user licence.

This is because the learning experience will always be the same high quality offering and that should always be seen as positive
– even if at times the licensing is different to Creative Commons.

When using the content you must attribute us (The Open University) (the OU) and any identified author in accordance with the
terms of the Creative Commons Licence.

The Acknowledgements section is used to list, amongst other things, third party (Proprietary), licensed content which is not
subject to Creative Commons licensing. Proprietary content must be used (retained) intact and in context to the content at
all times.

The Acknowledgements section is also used to bring to your attention any other Special Restrictions which may apply to the
content. For example there may be times when the Creative Commons Non-Commercial Sharealike licence does not apply to any
of the content even if owned by us (The Open University). In these instances, unless stated otherwise, the content may be
used for personal and non-commercial use.

We have also identified as Proprietary other material included in the content which is not subject to Creative Commons Licence. These
are OU logos, trading names and may extend to certain photographic and video images and sound recordings and any other material
as may be brought to your attention.

Unauthorised use of any of the content may constitute a breach of the terms and conditions and/or intellectual property laws.

We reserve the right to alter, amend or bring to an end any terms and conditions provided here without notice.

All rights falling outside the terms of the Creative Commons licence are retained or controlled by The Open University.

Head of Intellectual Property, The Open University

Designed and edited by The Open University

978-1-4730-1545-6 (.kdl)
978-1-4730-0777-2 (.epub)

Contents

- Introduction

- Learning outcomes

- 1 Introduction

- 2 The complex number system

- 2.1 Introduction

- 2.2 Defining the complex number system

- 2.3 Section summary

- 2.4 Self-assessment questions and problems

- 3 The complex plane

- 3.1 Introduction

- 3.2 Relationship between complex numbers and points in the plane

- 3.3 Section summary

- 3.4 Self-assessment questions and problems

- 4 Sets of points in the complex plane

- 4.1 Introduction

- 4.2 Defining useful subsets of the complex number system, and proving the Nested Rectangles Theorem

- 4.3 Section summary

- 4.4 Self-assessment questions and problems

- Conclusion

- Keep on learning

- Acknowledgements

Introduction

You may have met complex numbers before, but not had experience in manipulating them. This course gives an accessible introduction
to complex numbers, which are very important in science and technology, as well as mathematics. The course includes definitions,
concepts and techniques which will be very helpful and interesting to a wide variety of people with a reasonable background
in algebra and trigonometry.

This OpenLearn course provides a sample of Level 3 study in Mathematics

Learning outcomes

After studying this course, you should be able to:

- perform basic algebraic manipulation with complex numbers

- understand the geometric interpretation of complex numbers

- know methods of finding the nth roots of complex numbers and the solutions of simple polynomial equations.

1 Introduction

You have almost certainly met complex numbers before, but you may well not have had much experience in manipulating them.
In this course we provide you with an opportunity to gain confidence in working with complex numbers by working through a
number of suitable problems.

Perhaps the most striking difference between real numbers and complex numbers is the fact that complex numbers have a two-dimensional
character, arising from our definition of a complex number as an ordered pair of real numbers. This two-dimensional aspect of complex numbers leads to a most useful representation of them as points in the plane.

We know that the distance between points in a plane can be measured by the usual Euclidean measure of distance, and this leads
us to the important modulus function for complex numbers. This function will clearly play an important role in complex analysis
if the subject is to develop along lines resembling real analysis. Later we will see how the complex modulus can be used to
generalise many of the limiting processes of real analysis.

Many proofs in real analysis use the concept of least upper bound. This concept depends in turn on the relation of inequality
which is defined in terms of positive and negative numbers. We shall show in this course that it is not possible to define
such a thing as a “positive” complex number and so least upper bound arguments will not carry over to complex analysis.

In real analysis, the idea of a least upper bound is used to develop the method of proof by repeated bisection, whose validity
rests on the Nested Intervals Theorem. Because complex numbers are defined as pairs of real numbers, we are able to generalise
this theorem to a Nested Rectangles Theorem, which will play a similar role in complex analysis to that of the Nested Intervals
Theorem (or, equivalently, the least upper bound axiom) of real analysis.

There are three reading sections in this course, each of which includes a problems subsection. In Section 2 the complex number system is defined; in Section 3 the relationship between the complex numbers and points in the plane is developed; in Section 4 various useful subsets of the complex number system are defined and the Nested Rectangles Theorem is proved.

2 The complex number system

2.1 Introduction

In this section we shall define the complex number system as the set R × R (the Cartesian product of the set of reals, R, with itself) with suitable addition and multiplication operations. We shall define the real and imaginary parts of a complex
number and compare the properties of the complex number system with those of the real number system, particularly from the
point of view of analysis.

2.2 Defining the complex number system

In complex analysis we are concerned with functions whose domains and codomains are subsets of the set of complex numbers.
As you probably know, this structure is obtained from the set R × R by defining suitable operations of addition and multiplication. This reveals immediately one important difference between
real analysis and complex analysis: in real analysis we are concerned with sets of real numbers, in complex analysis we are
concerned with sets of ordered pairs of real numbers.

Whatever context is used to introduce complex numbers, one sooner or later meets the symbol i and the strange formula i^2 = −1.

Historically, the notion of a “number” i with this property arose from the desire to extend the real number system so that equations such as x^2 + 1 = 0 have solutions. There is no real number satisfying this equation so, as usual in mathematics, it was decided to invent a number system that did contain a solution. The remarkable fact is that having invented a solution of this one equation, we can use it to construct a system that contains the solutions of every polynomial equation!

For example, using the well-known formulas

for the solutions of the equation

we find that the solutions of

are apparently given by the expressions

These make no sense at all until we turn a blind eye to and just manipulate it formally, as though we knew what we were doing, to give

We then say that if i^2 = −1 then we might as well press on and replace by i and so the “solutions” of the equation are

We can check our formal manipulation of square root signs by substituting these “solutions” into the equation and see that
they work. But before we can do that, we need to know how to “add” and “multiply” “numbers” of the form x + iy, where x and y are real. If we assume that they add and multiply according to the ordinary laws of algebra, in other words that addition
and multiplication are associative and commutative and that multiplication is distributive over addition, then we have (a  + ib ) + ( c  +  id ) = ( a  + c)  + i( b + d ) and (a  + ib ) × (c  + id ) = ac  + i ( bc  + ad )  + i^2bd.

If we replace i^2 by −1, we obtain (a  + ib) × (c  + id) = ( ac − bd )  + i (bc  + ad ).

Now you can substitute into x^2  + x  +  1 and see that our formal juggling with square root signs appears to be justified.

In the same way, we can factorise formally the expression x^2 + bx + c into

and hence “solve” any quadratic equation by assuming also that the property αβ = 0 if and only if α = 0 or β = 0, which does
apply to R, carries over to our new system. (Note the use of “or” in the “or both” sense.)

It may disconcert you to see us play around with a symbol that behaves as though it were the square root of −1: it should
because, as yet, we have given no formal definition of this symbol, i. The problem that Pythagoras had in relating to the set of rationals is presumably similar to that which we have in relating i to the set of reals. Just as we feel more comfortable when we are able to define irrational numbers in terms of rational
numbers, so we must define numbers of the form a + ib in terms of the real numbers and justify our somewhat suspect manipulations. First we make the formal definition: then we
deduce that our manipulations are in order.

It is easy to do this once you realise that a + ib is completely specified by the ordered pair (a, b) of real numbers (putting aside for the moment the meaning of the symbol i) and so the set can be put into one-one correspondence with R × R. To turn this correspondence into an isomorphism, we need to define suitable addition and multiplication operations on R × R. Keeping an eye on the operations we have already found useful, we make the following definition.

Definition

The complex number system, denoted by C, is the set of all ordered pairs of real numbers (that is, R × R) with the operations of addition (denoted by +) and multiplication (denoted by ×) which satisfy

It is straightforward but tedious to check that + and × defined as above are both commutative and associative and that × is
distributive over +.

These definitions make perfect sense for any real numbers a, b, c and d and present no logical problems about square roots of −1 (no such square roots being mentioned in the definition). Thus we
have turned round the earlier process and have defined a complex number as an ordered pair of real numbers which obeys suitable
operations + and ×.

The notion of equality of complex numbers, comes from equality in R × R.

Two complex numbers, (a, b) and (c, d) are equal if and only if a = c and b = d.

Having made a proper definition of complex numbers, we now go back and restore our original notation, which, as we shall see, is far more convenient than the ordered pair notation. To do this, we note
first of all that the subset of C is isomorphic to R, with the complex number (a, 0) corresponding to the real number a. (“A is isomorphic to B” means that there is an isomorphism from A onto B.) It is obviously more convenient to write a instead of (a, 0) and furthermore, because

we can write

or, more concisely,

Then any complex number, (a, b), can be written as

We have already identified (1, 0) with the real number 1; all that remains is to replace (0, 1) by a symbol, i, of course. Thus the symbols

represent the same complex number, and we write

Definition

The symbol i represents (0, 1) C.

You may have noticed that we have just written a + bi instead of a + ib. It does not matter which you use.

As we remarked earlier, addition and multiplication of complex numbers (regarded as elements of R × R) are commutative and associative, and multiplication is distributive over addition. Numbers of the form a + ib can therefore be manipulated according to the ordinary laws of algebra to give

and

We now have to interpret the symbol i^2. It corresponds to (0, 1) × (0, 1) and direct application of the definition of multiplication shows that

which we are now abbreviating to −1. We are thus able to write

which corresponds to

Let us summarise these manoeuvres. Practical considerations suggest that a “number”, i, with the property i^2 = −1 might be very useful. But symbols such as obviously present logical difficulties. Nevertheless, brushing aside these difficulties, we can predict how numbers a + bi would behave. We use this information to define a perfectly logical structure on R × R that is isomorphic to the system we are trying to construct. By taking this structure on R × R as a starting point we are able to identify i with the ordered pair (0, 1) and a + bi with the ordered pair (a, b). The notation a + bi is thus nothing more than an alternative form for (a, b): its advantage is that expressions like (a  + bi ) × ( c  + di) can be manipulated according to the ordinary laws of algebra with i^2 replaced by −1 whenever it appears.

It is often necessary to refer to the first and second components of a complex number and so they are given names.

Definition

The real number a is called the real part of a+ ib and the real number b is called the imaginary part of a + ib.

Notation

We shall write

Conventions

A complex number which has imaginary part zero is often called “real”, because even though it is not a real number it is,
as we have seen, so closely identified with a real number that to insist on the distinction would be pedantic.

We shall find it useful to denote complex numbers by single letters. In particular, just as in real analysis x is often used to denote a variable real number, so, in complex analysis, z is a popular choice of symbol to denote a variable complex number. Furthermore, we shall nearly always implicitly assume
that x and y are real and that x, y and z are related by the equations:

In other words,

It is interesting to notice that since two complex numbers are equal if and only if their real parts are equal and their imaginary
parts are equal, a single equation connecting complex numbers leads to two equations connecting real numbers. Thus if w and z are complex numbers, we can deduce from the single equation

w = z

that

This process is usually referred to as equating real and imaginary parts.

As you must have realised, despite its name, the imaginary part of a complex number is a perfectly ordinary real number. The
name is an unfortunate one because it often leads to the mistaken impression that the imaginary part of a + ib is ib rather than b. The term “imaginary” arises from the original concern about complex numbers associated with the symbol . The way we have defined it, i is no more imaginary than the number .

Now that we all know what a complex number is, we need to agree on a list of properties of C that we can take as a starting point for complex analysis. We already know that C has some of the properties of R (such as associativity and commutativity of addition and of multiplication, and distributivity of multiplication over addition).
In fact, all the familiar arithmetic (field) properties of R carry over to C as theorems, in the sense that they can be proved from the properties of R. (The concept of a field is discussed fully in M. Spivak, Calculus. A field is a set with two closed binary operations (+ and ×) for which properties (i)–(ix) are satisfied.) In the list that
follows z,z_1,z_2,z_3 are any elements of C: in other words the properties hold for all .

Properties of C

- (i) Associative law for addition:

- (ii) Existence of an additive identity (denoted by 0):

- (iii) Existence of additive inverses:

- (iv) Commutative law for addition:

- (v) Associative law for multiplication:

- (vi) Existence of a multiplicative identity (denoted by 1):

- (vii) Existence of multiplicative inverses of non-zero numbers:

- (viii) Commutative law for multiplication:

- (ix) Distributive law:

(Note that in (ix) we have suppressed the multiplication symbol. This is common practice, as is the use of · for ×.)

We have already mentioned properties (i), (iv), (v), (viii) and (ix). Properties (ii), (iii), (vi) and (vii) are easily checked.

The additive identity, 0, is (0, 0) = 0 + 0i and the additive inverse of z = x + iy is −x + i(−y) which is written −x − iy or −z. As with the real numbers, we define the difference of two complex numbers by

or

The multiplicative identity, 1, is (1, 0) = 1 + 0i, and so if w = u + iv is the multiplicative inverse of z = x + iy then

that is,

and so

Thus, equating real and imaginary parts, we have

giving

since x and y cannot both be zero (because z ≠ 0). Thus the multiplicative inverse of x + iy is

All the other useful facts about C, such as the uniqueness of 0, wz = 0 if and only if w = 0 or z = 0, and so on, follow from the nine properties as for real numbers. The importance of this is that all the usual identities,
such as the difference of two squares, the sum of a (finite) geometric series, the binomial theorem for a positive integer
index go through as in real analysis. But now we come to the differences between real analysis and complex analysis. Most
real analysis courses rely heavily on other properties of the real numbers – properties that tell us that the real numbers
are ordered and enable us to define inequalities on the real numbers. (See M. Spivak, Calculus, pages 9 and 10.) This is a rather special use of the term “ordered”. We talk of a field being ordered if we can identify a subset of “positive elements’ such that one and only one of the following holds for each element, a, of the field: a is positive; −a is positive; a is zero (the trichotomy law) and, if a and b are positive then so are a + b and a × b. (The set of real numbers is an example of an ordered fieId.) In this sense of the word “ordered”, clearly there are problems
with trying to turn C into an ordered field. For how could we define “positive” and “negative” subsets in C which have the properties we would expect?

The concept of an ordered field is an attempt to characterise the special ordering of the real numbers and it is fairly clear
that the two-dimensional nature of the complex numbers is going to rule out the possibility of their being ordered in this
sense. But suspicion is not enough; we should prove that C is not an ordered field.

It is easy enough to show that in any structure satisfying (i)–(ix) and in which we can talk of “positive” and “negative” elements, the square of any non-zero element is positive. This is precisely the property
we set out not to have in constructing the complex numbers! (For, if a is positive then a × a = a^2 must be positive, by the trichotomy law, and if a is negative then −a is positive and so (−a) × (−a) = a^2 is positive, again by the trichotomy law.) If we wanted in some way to extend the concepts of positive and negative from
R to C we would presumably require (1, 0) to be positive, in which case, by the trichotomy law, (−1, 0) would be negative. Then
(0, 1) is non-zero, but (0, 1)2 = (−1, 0) is negative (and so we have the square of a non-zero element being negative). If we decided to get round this difficulty
by calling (−1, 0) “positive”, it follows that −(−1, 0) = (1, 0) must be non-zero: also, then (1, 0)2 = (1, 0) is negative (and again we have the square of a non-zero element being negative). We have shown that it is not possible
to find two classes which we could justifiably call “positive” and “negative”. It was precisely the existence of these two
classes in R that enabled us to define inequalities in R.

We therefore come to the conclusion that the relation < does not carry over from the real numbers to the complex numbers.

It is precisely the relation < that enables us to talk of least upper bounds (see Spivak) – and so in moving from real analysis
to complex analysis we have to throw overboard the least upper bound property that is so useful in real analysis.

Fortunately things are not as black as they might seem, because the proofs in which least upper bounds are used most frequently
in real analysis can also be done by the use of the technique by repeated bisection. This relies on a theorem (proved from
the least upper bound property) called the Nested Intervals Theorem (see Spivak). Happily, this theorem has an analogue in
complex analysis called the Nested Rectangles Theorem. We shall prove this theorem in Section 4.

2.3 Section summary

In this section we have seen that the complex number system is the set R × R together with the operations + and × defined by

From this, one can justify the performance of ordinary algebraic operations on expressions of the form a + ib where a, b and i is an undefined symbol with the property that i^2 is replaced by −1.

In the complex number system, any quadratic expression can be factorised into two linear factors.

The real part of a complex number z = x + iy is x and we write

The imaginary part of a complex number z = x + iy is y and we write

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

The complex number system obeys the usual algebraic laws of manipulation of numbers but the inequality relation, <, does not
carry over from the real numbers to the complex numbers.

Examples

(i) (3 + i) (4 −i)  = 12−i^2 + 4i−3i  = 13  + i.

(ii) (6 + 7i)−(3−i) = 3 + 8i.

(iii) (x + iy)(x−iy) = x^2 + y^2.

(iv) Re (6−4i)  = 6 , and Im (6−4i) = −4.

(v) xy−(x^2 + y^2)i = 1 = (x,yR).

then, by equating real and imaginary parts, we have xy =  1 and x^2 +  y^2 =  3

Thus the point (x, y) lies on the hyperbola xy = 1 and on the circle x^2 + y^2 = 3 and can take four possible values, corresponding to the points of intersection of the hyperbola and the circle.

2.4 Self-assessment questions and problems

Self-assessment questions are intended to test your immediate comprehension of a reading section. If you have difficulty with
them you should read again the appropriate parts of the material. Before checking the solution to any part of a question,
you should work through all the parts of that question.

SAQ 1

In each of the following cases, indicate whether the statement is True or False.

(i) Re(3 + 2i) = 3.

(ii) Im(3 + 2i) = 2i.

(iii) If z_1z_2 is a real number, then z_1 and z_2 must be real.

(iv) If z_1 + z_2 is a real number, then z_1 and z_2 must be real.

View answer - SAQ 1

SAQ 2

Express each of the following in the form a + bi where a and b are real numbers.

(i) (3  +  2i)  +  (2 −i ).

(ii) ( 4  + i ) −(2 - i).

(iii) (3 + 2i ) (2−i ).

(iv) Re (6−7i).

(v) Im (3−2i).

View answer - SAQ 2

SAQ 3

Find x and y if (x + 3i)(2 + yi) = 5 + 5i, and x, y R

View answer - SAQ 3

SAQ 4

Simplify each of the following expressions into the form a + ib, where a and b are real.

(i) (1 + 6i) (1−3i).

(ii) (1  + i) 3.

(iii) (1−i) (1 + i).

(iv) (3 + 2i) (3−2i).

(v) (3 + 7i) −(2 + i).

View answer - SAQ 4

SAQ 5

Find all possible values of z satisfying

(i) z^2 = −9

(ii) z^2 = −k^2, k real.

View answer - SAQ 5

SAQ 6

Solve the equation

( 1  + i) (x + iy)  = −2 + i.

where x, y R, by deducing simulationeous equations in x and y.

View answer - SAQ 6

As a consequence of the carry over from R to C of the usual arithmetic properties, many of the familiar definitions and results of the algebra of real numbers extend easily
to complex numbers. For example, an expression

where z and the ai's are complex is called a polynomial (in z). If an ≠ 0, n is called the degree of the polynomial.

SAQ 7

(i) Use mathematical induction to prove that if n is any positive integer, there are positive real numbers c_1……cn (each dependent upon n) such that

for all complex numbers z.

Identify the coefficients.

(ii) Generalise the results of (i) to (w + z)n.

View answer - SAQ 7

SAQ 8

(i) If n is a positive integer, show that

(Hint: Denote the right-hand side by p(z), and consider p(z) − zp(z).)

(ii) Deduce that

(Hint: See the solution to part (ii) of SAQ 7.)

View answer - SAQ 8

SAQ 9

(i) A polynomial p(z) is divisible by z − z_0 if a polynomial q(z) exists such that Show that for any positive integer n, zn − zn_0 is divisible by z − z_0.

(ii) If p(z) is a polynomial, show that the polynomial p(z) − p(z_0) is divisible by z − z_0.

(iii) Show that

(iv) The Fundamental Theorem of Algebra states that if p(z) is a non-constant polynomial, then the equation p(z) = 0 has a (complex) root. We shall prove this theorem later in the course. Assuming the theorem, prove that every polynomial
of degree n > 0 can be factorised into linear factors in the form

View answer - SAQ 9

When we wish to be explicit about the operations defined on C which are under consideration we sometimes write .

SAQ 10

Show that there is an isomorphism from (C, +, ×) to the set of matrices of the form a, b R, under the operations of matrix addition and multiplication.

View answer - SAQ 10

SAQ 11

(i) Show that there is an isomorphism from (C, +) to the set of two-dimensional geometric vectors with the operation of vector addition.

(ii) Show that the mapping from C to itself, where z_0 is a fixed complex number, is equivalent to a translation of R × R. (A mapping φ for which θ (x,y) = (x + α,y + β), where α and β are fixed, is a translation of R × R.)

View answer - SAQ 11

SAQ 12

Show that if z_0 = (x_0, y_0) is a fixed complex number, the mapping from C to itself defined by f:zz_0z is linear; that is, that

for all

View answer - SAQ 12

3 The complex plane

3.1 Introduction

In this section we shall develop the correspondence between C and R × R by obtaining a geometric representation of elements of C and operations on C. We shall define the polar form of a complex number and the modulus and argument of a complex number. We shall see that not
only does have some meaning in C but that every non-zero complex number has exactly n nth roots in C.

3.2 Relationship between complex numbers and points in the plane

We have seen in Section 2.2 that the complex number system is obtained by defining arithmetic operations on the set R × R. We also know that elements of R × R can be represented as points in a plane. It seems reasonable to ask what insight can be obtained by representing complex
numbers as points of the plane.

One obvious way of making the representation is to use Cartesian coordinates. Then the number x + iy is represented by the point with coordinates (x, y). Thus if a point P represents a complex number z then its x coordinate is the real part of z and its y coordinate is the imaginary part of z (hence the choice of notation z = x + iy made in Section 2.2).

The horizontal axis is called the real axis and the vertical axis is called the imaginary axis.

This way of representing complex numbers by points in the plane is used so often that we shall use phrases such as “the point
x + iy” and “the point z” as convenient shorthands for “the point representing z” and we shall speak of the complex plane. Figure 1 shows several points in the complex plane.

Figure 1

As we saw in SAQ 11 in Section 2.4 (and it is fairly obvious anyway), the geometric representation can be taken a stage further. Complex numbers also add in
the same way as number pairs regarded as vectors and so they add according to the parallelogram law (Figure 2).

Figure 2

Now points in the plane can also be represented by polar coordinates and this suggests another way of specifying a complex
number.

If you have forgotten about polar coordinates, a short revision will not come amiss. When we use polar coordinates – instead
of specifying a point in a plane by the distance from two lines, as in Cartesian coordinates – we specify a point P in the plane by its distance (r) from a fixed point O (the origin) and the angle (θ) that OP makes with a fixed half-line (the initial line) which has one end at O (Figure 3). (At present we consider only points P different from O.)

Figure 3

As is usual in mathematics, the angle θ is measured in an anticlockwise direction from the initial line (Figure 4.).

Figure 4

One snag that does arise with polar coordinates is that they do not give a one-one correspondence with the points of the plane. Thus a point with polar coordinates (r, θ) also has polar coordinates (r, θ + 2k) where k is any integer. But polar coordinates (r, θ) do specify a unique complex number, and from the relationship between polar coordinates and Cartesian coordinates we know that this number is
r cos θ + ir sin θ (Figure 5). So whether we use polar coordinates or Cartesian coordinates, a point in the plane specifies a unique complex number.

Figure 5

It is often useful to represent a complex number in terms of the polar coordinates of the corresponding point as in the following
definition.

Definition

A polar form of the non-zero complex number

where

The positive number r is called the modulus of z and θ is an argument of z. We denote the modulus of z by |z|, and so For the special case of z = 0, we define |z| = 0, but we do not need to define what is meant by an argument of 0.

As we know, there will be some difficulty in specifying arguments. This is manifested by the fact that the relations specify θ only to within an integer multiple of 2. (Notice that the equation is even less specific.)

Given any non-zero complex number, z, an argument can be any element from a whole set of numbers. For example, an argument of 1 + i can be any element from . Nevertheless, any one element from this set is sufficient to identify the set and whatever the number z(≠0) there is only one of its arguments in the interval (−, ].

Definition

If θ is an argument of z and θ(− ].θ is the principal value of the argument of z, and we write θ = Arg z. (Note the capital A.)

Thus, for example, (Figure 6).

Figure 6

It would have been just as reasonable to have chosen the interval [0.2) or (0.2] or [−, ) for Arg z, but we must settle for one option – we happen to have chosen (− ].

The polar form of a complex number can be very convenient and so we investigate now the effect of the arithmetic operations
on the polar forms of complex numbers.

Addition and subtraction of complex numbers given in polar form is not a pleasant affair if the answers are also required
to be in polar form. If

and

then

and that is about as far as you can get unless you are a wizard at trigonometry. (The polar form would be p cos ψ + ip sin ψ where p and ψ are expressions in r, ρ, θ, and φ.) Of course, in any particular case you can work out the polar form of the answer, but the general result is not worth working
out, let alone remembering.

On the other hand, multiplication is easy because

Thus

and an argument of zw can be obtained just by adding arguments of z and w. If we want to work with principal values, there is one minor difficulty: even if θ and φ are in (−, ], when we add θ and φ we might well get an answer outside (−, ]. So if we want to work with principal values we shall have to make sure to get the answer back into (−, ]. Thus although Arg z + Arg w is an argument of zw we may have to add or subtract 2 to obtain Arg(zw).

One possible snag that may have occurred to you is this: if θ and φ are arguments of z and w respectively, is θ + φ an argument of zw whatever θ and φ we choose? Well, the relation where θ_1 is related to θ_2 if θ_1 = θ_2 + 2k, k an integer, is an equivalence relation on R and all the arguments of a complex number will constitute an equivalence class. We are asking whether we can combine equivalence
classes (add arguments), in other words whether the equivalence relation is compatible with addition. Can we write

{arguments of z} + {arguments of w} = {arguments of zw}?

We can, of course, because if θ_1 and θ_2 are in the same equivalence class and φ_1 and φ_2 are in the same equivalence class, there are integers k and m such that

and so

that is, θ_1 + φ_1 is in the same equivalence class as θ_2 + φ_2.

We can also show that if α is a fixed argument of zw then we can choose an argument of z, θ say, and an argument of w, φ say, such that θ + φ = α.

Let α = Arg zw + 2n, where n is a fixed integer. Certainly,

and so

So choose

where p + q = k + n.

We define division of complex numbers (as for real numbers) as the inverse of multiplication, and so it is no surprise that polar forms are also very suitable
for division. If z−l is the number such that

then we define w ÷ z by

As for real numbers, we write

We now ask how to calculate l/z when z(≠0) is given. If θ is an argument of z and φ an argument of 1/z then θ + φ is an argument of . Since 0 is an argument of 1, we must be able to choose φ so that

and so −θ is an argument of 1/z. In fact, unless z is a negative real number, we can use principal values without trouble:

(Note that a statement like k > 0 implies that k is real.)

If z = −k, k > 0, then Arg z =  and Arg 1/z = .

What about |1/z|? We know that |1| = 1, and so

Thus

We can now interpret division quite easily in terms of polar forms:

and if θ and φ are arguments of z and w respectively, then θ − φ is an argument of w/z. More concisely,

To summarise: each of the arithmetic operations on C has a geometric interpretation. If z = x + iy is any complex number, corresponding to a point P, and if w = u + iv = p(cos φ +  sinφ) then:

1. the effect of adding w to z is to translate P to the point (x + u, y + v); the effect of subtracting w from z is to translate P to the point (x − u, y − v);

2. the effect of multiplying z by w is to rotate OP through the angle φ and “stretch” OP by a factor ρ;

the effect of dividing z by w is to rotate OP through the angle −φ and “stretch” OP by a factor 1/ρ.

The fact that when multiplying complex numbers their arguments are added suggests an interesting relationship between numbers
z and w for which

Multiplying these two numbers gives a number with a zero argument – a real number (Figure 7). What is more, the modulus of this number is |z| · |w| and this is just |z|2. In polar form we see that if

Figure 7

The relationship between z and w is important and leads us to the following definition.

Definition

If z = x + iy, the number x − iy is the conjugate of z, and we write

(Note that many authors write z* for .)

We have just seen that has the important property that .

This result is useful for dealing with quotients in Cartesian form, for an expression like

could be simplified without resorting to polar form if we could make the denominator real. We now know how to do that. Since
(c−id)/(c−id) = 1 , we can write

The details of this expression are not worth remembering, but the technique is: multiply numerator and denominator by the conjugate of the denominator. It is often
useful to remember that a denominator z becomes |z|2 after the process.

Perhaps the most impressive payoff from writing complex numbers in polar form arises in the problem of solving an equation
of the form

where a is a given non-zero complex number. Now it is easy to deduce from our knowledge of multiplication of polar forms that

and that if θ is an argument of z then nθ is an argument of zn. (See SAQ 21 in Section 3.4.) So if zn = a, z must be a number whose modulus is an nth root of a and which has an argument which is 1/n times an argument of a.

Example 1

Find a value of z such that z^4 = 16i.

View answer - Example 1

This example clearly suggests that we need a systematic method of finding all the solutions of the equation zn = a, a ≠ 0. We can see how to do this by considering the simpler problem of finding all the solutions of wn = 1. Since |1| = 1, we must have and if φ is an argument of w, nφ must be a multiple of 2 because 0 is an argument of 1. In other words, |w| = 1 and any of the values can be an argument of w – each leading to a different value of w. (If k takes integer values beyond n, we get repetitions of the values of

that have already been obtained.) Thus in the set of complex numbers, the number 1 has n nth roots:

The points representing the roots lie equally spaced around the unit circle. Figure 8 shows the roots for n = 8.

Figure 8

Now let us get back to the original problem: zn = a, a ≠ 0. We do not seem to have any difficulty in finding one solution of this equation: the problem is how to find the whole
set of solutions. Let us suppose we have one solution, z_0, so that z_0n = a. We now use the solution to the simplified problem, wn = 1. Suppose w is a solution of this problem, then (wz_0)n = wnz_0n = 1a = a, and so wz_0 is a solution to the original problem. But there are n possibilities for w and so there are n possibilities for wz_0 (one of which is z_0 because one of the values of w is 1). So knowing the solutions of wn = 1 (which is easy) and just one solution of zn = a (which is not too bad) we can generate n solutions of zn = a. The question is: “Have we now got them all?”. It would be most surprising if we had not, but we still have to check – or
rather you have to check, because that is SAQ 22 in Section 3.4.

In solving the equation zn = a in the set of complex numbers we are extending the notion of roots of a real number and so we make the following definition.

Definition

The solutions of zn = a are called nth roots of a.

We have proved the following theorem (taking SAQ 22 of Section 3.4 for granted).

Theorem 1

Every non-zero complex number has exactly n distinct nth roots.

We find the n nth roots of a ≠ 0 by using the following technique.

Technique

If then one solution of zn = a is

and all the others are obtained by multiplying z_0 in turn by each of the nth roots of wn = 1.

Now the effect of multiplying each of the n nth roots of 1 by z_0 will be to multiply each of their moduli (all equal to 1) by |a|1/n and add α/n to each of their arguments. In other words, the nth roots of the complex number a are obtained from the nth roots of 1 by expanding (or contracting) the unit circle by |a|1/n and rotating by (Arg a)/n. The nth roots of a will be equally spaced around a circle of radius |a|1/n. The roots are

Figure 9 illustrates this for n = 8 and |a| > 1.

Figure 9

Example 2

To find the solutions of z^4 = 1 + i, the fourth roots of 1 + i, we proceed as follows.

3. Thus one root is and the others can be found because the four of them are equally spaced around the circle of radius (Figure 10).

Figure 10

3.3 Section summary

In this section we have seen a correspondence between complex numbers and points in the plane using Cartesian coordinates;
the real part of the complex number is represented on the real axis (“horizontal”) and the imaginary part on the imaginary
axis (“vertical”). We can also use polar coordinates (r,θ) in which case r, the modulus of a non-zero complex number z is positive and θ is an argument of z, defined only to within an additive integer multiple of 2. The principal value of the argument is that value of θ in (− ], and is denoted by Arg z.

Multiplication and division are particularly easy operations to carry out when the numbers are in polar form. We discussed
the conjugate of a complex number and how to find the nth roots of a complex number.

Example 3

(i) If z = i, then |z| = 1 and

(ii) If z = −1, then |z| = 1 and Arg z = .

(iii) If z = −i then |z| =  = 2 and Arg = −

It is usually a good idea to scribble a little diagram like Figure 11 to make sure that you get the correct angle for Arg z.

Figure 11

(iv)

In Figure 12 we have taken α (0, /2).

Figure 12

Example 4

(i)

(Figure 13).

Figure 13

(ii)

(Figure 14).

Figure 14

Example 5

(i)

(ii)

Example 6

3.4 Self-assessment questions and problems

SAQ 13

Find |z| and Arg z in each of the following cases.

View answer - SAQ 13

SAQ 14

Express each of the following in the form a + ib where a and b are real.

View answer - SAQ 14

SAQ 15

Find the modulus (r) and an argument (θ) for each of the following numbers.

View answer - SAQ 15

SAQ 16

Express each of the following numbers in the form a + ib, where a and b are real numbers.

View answer - SAQ 16

SAQ 17

Show that for any complex numbers z and w

View answer - SAQ 17

SAQ 18

Illustrate the following sets of points of the complex plane where a, b, c, d α and β are fixed real numbers with b > a, d > c, − < α < β < .

View answer - SAQ 18

SAQ 19

If n is a multiple of 4, show that the coefficients, ci in SAQ 7 of Section 2.4 satisfy

and

Identify the cases when the positive sign should be taken on the right-hand side in the second case. (There is a hint after
the solutions.)

View answer - SAQ 19

SAQ 20

Let p(z) be a polynomial with real coefficients.

(i) Prove that .

(ii) If α is a root of p(z) show that is also a root.

View answer - SAQ 20

SAQ 21

(i) Prove by mathematical induction that if n is any positive integer, then |zn| = |z|n.

(ii) Show that if n is any positive integer, then

This result is known as de Moivre's Theorem. Abraham de Moivre (1667–1754) was born in France, but came to England when still a small boy. His interest in Mathematics started when he came
across a copy of Newton's Principia by chance. He is best known for the theorem stated above, and for calculating the various quadratic factors of x2n−2pxn + 1, and is also remembered as one of the founders of probability theory.

(iii) Show that if n is any positive integer, then

This extends de Moivre's Theorem to any integer n.

(iv) We have seen in the text that de Moivre's Theorem extends to the form: if m is a positive integer, one of the values of (cos θ + i sin θ)1/m is cos (θ/m) + i sin (θ/m). Extend de Moivre's Theorem to any rational index.

View answer - SAQ 21

SAQ 22

Show that if z_0 is a solution of zn = a, where a ≠ 0, then any other solution must be of the form z_0w where w is a solution of wn = 1.

View answer - SAQ 22

SAQ 23

(i) Find the fifth roots of −32.

(ii) Find the fourth roots of −16.

View answer - SAQ 23

SAQ 24

(i) Solve the equation z^4 + 4z^2 +  8 = 0.

(ii) Solve the equation z^4 + 4iz^2 +  8 = 0.

View answer - SAQ 24

4 Sets of points in the complex plane

4.1 Introduction

We have seen in SAQ 18 of Section 3.4 how some sets of points of the complex plane can be described algebraically in terms of operations on C. We now use the modulus function to take this a step further by defining discs in the complex plane. As we shall see, discs are extensively used in arguments involving limits. Another type of region that
is useful in analytic arguments is a rectangular region, and so we conclude this section with a discussion of the Nested Rectangles
Theorem.

4.2 Defining useful subsets of the complex number system, and proving the Nested Rectangles Theorem

You will no doubt recall that in real analysis extensive use is made of the modulus function . It gives us a way of measuring the “closeness” of two numbers, which we exploit in writing expressions such as

We are going to need a similar device in complex analysis and, of course, it is again provided by the modulus function, this
time with domain C. In Section 3 we defined |z|, the modulus of the complex number z. The modulus function is the function which maps a complex number to its modulus:

Note that the modulus function maps complex numbers to real numbers.

The modulus function with domain C is just an extension of the modulus function that we use in real analysis in the sense that if we restrict its domain to
R then the two functions are equal. The value at z of the modulus function, |z|, tells us the distance of the point z from the origin. Similarly, just as |a − b| is the distance between the two real numbers a and b, so |z − w| is the distance between the two complex numbers z and w.

More explicitly, if z = x + iy and w = u + iv, then

which is the usual (Euclidean) distance between the points (x, y) and (u, v).

The function is a distance function in the usual sense of the word in that it turns the set of complex numbers into a metric space (see SAQ 33 in Section 4.4). Among other things this means that we have the triangle inequality:

(Note that this is an inequality between real numbers (moduli): the modulus function enables us to write inequalities which
involve complex numbers.)

This inequality reduces to the usual triangle inequality of real analysis when z, w and v are real, but the two-dimensional nature of complex numbers explains why the triangle inequality is so called. The triangle
inequality is as vital to complex analysis as it is to real analysis: it also yields many more inequalities, some of which
are in the final problems section, Section 4.4.

In real analysis, the modulus function enables us to describe an interval around a point. For example, we talk of an . neighbourhood of a point x_0, as the interval The end points of this interval are simply the two elements of the set In the complex plane the analogous bounding set is which is the set of all points at a distance from z_0, that is to say the circle centre z_0, radius (Figure 25).

Figure 25

Thus in complex analysis, the analogue of an interval is the set of points whose boundary is a circle. Such a set is called
a disc. In particular, we make the following definitions.

Definition

The set is an open disc.

The set is a closed disc.

If we are not sure whether a disc is open or closed, or if we do not want to be specific, or if some but not all points of
the bounding circle are included we shall refer simply to the “disc”. We shall also sometimes drop the set notation and write,
for example, “the closed disc ” when there is no likelihood of misunderstanding. One other point of convention: we shall refer to the disc centre as the
origin and radius 1 as the (open or closed) unit disc.

Two concentric circles specify another type of set of complex numbers – the set of points between the two circles.

Definition

The set is an open annulus.

The set is a closed annulus.

If we do not wish to be specific, or if some but not all of the points of the bounding circles are included, we drop the adjective
“open” or “closed”. As for discs, we often write, for example, “the open annulus ”.

Another term which we shall find useful is “bounded set of points”.

Definition

A set of points is said to be bounded if all points of the set lie in a closed disc with centre the origin.

As an example of the way we can use the modulus function in complex analysis, consider how we can define the limit of a sequence
of complex numbers. In real analysis, an infinite sequence is simply a function from the set of natural numbers, N, to R; in complex analysis an infinite sequence is a function from N to C. Just as the terms of a real sequence can be plotted on the number line, so the terms of a complex sequence can be plotted
on the complex plane. Convergence is defined precisely as for real sequences.

A sequence {zn} converges to l if for every  > 0 there is a natural number N such that, for all natural numbers n,

The number l is called the limit of the sequence.

In other words, for all n > N every term in the sequence is in the (open) disc centre l radius

Notice that the analytic form of the definition looks just like that of real analysis converges to l if and only if — but the geometric interpretation is much more interesting. If the values of in this definition are made successively smaller, taking values then we get a succession of discs , and so on, each disc being a subset of its predecessor and containing its successor. Sequences of sets with this property
are often used in complex analysis; they are called nested sequences. They are used to give a method of proof similar to the repeated bisection arguments of real analysis.

Suppose S_1, and S_2 are two nonempty sets, with . Then there is at least one point common to both sets. Similarly, if are n nonempty sets with

– that is to say the sets form a nested sequence – then there is at least one point in common to all n sets. If we try to extend this idea to an infinite sequence of sets then a number of snags can arise which may result in ihe intersection of all the sets being empty.

(i) For example, taking sets in R, suppose Sn is the open interval . In this case the sets are nested, for for all n, but there is no point in common to every set – the obvious candidate, 0, has been excluded because the intervals are open.

(ii) In other circumstances the “obvious” candidate might be excluded not because the intervals are open but because the sets
are subsets of a set which is not complete – Q, the set of rationals, for example. The sets given by are nested but there is no number common to them all. The number is excluded because

(iii) The third snag that might arise with nested sets of real numbers concerns sets such as intervals . These sets are not open intervals and so are not subject to the first objection. They are also subsets of the complete set,
R, and so are not subject to the second objection. But there is no real number in every Sn.

In real analysis, we get over these difficulties by using nested sets that are closed intervals (and therefore subsets of
the complete set R) and that are such that the lengths of the intervals have limit 0. There are several ways of extending this idea to C and indeed to metric spaces in general, but a simple and useful way is to extend the idea of an interval to that of a rectangle.

Definition

The set

is a closed rectangle.

The set

is an open rectangle.

(Now look back at SAQ 18 part (v) inSection 3.4.)

If some but not all of the bounding points are included, we shall refer simply to a rectangle and we shall also use this term if we do not wish to be specific.

As a concept analogous to the length of an interval we can use the length of the largest side for a rectangle S. We denote this byℓ(S), so

Before we propose the Nested Rectangles Theorem we introduce the idea of the intersection of a collection of sets. Suppose that C is a collection of sets and that these sets may be indexed so that we can write C as

We extend the idea of the intersection of two sets to the intersection of the collection of sets in C by regarding the intersection of the Si as the set {x : x is in Si for every i N}; we denote this set by In other words, denotes the set of elements which are common to all the Si.

We can now propose the Nested Rectangles Theorem.

Theorem 2 (The Nested Rectangles Theorem)

Let be a sequence of nested closed rectangles, with , then

(i)

is nonempty and contains a unique complex number z_0, and

(ii) given  > 0, there is an N N such that

(This second part of the theorem tells us that we can always ensure that “eventually” all the rectangles (or, if you prefer,
“all but a finite number”) lie inside an arbitrarily small disc centred at z_0.)

Proof

(i) Probably the simplest proof of this part is to use the Nested Intervals Theorem from real analysis (see Spivak). If Sn is bounded by the lines Re z = an, Re z = bn, Im z = cn, Im 2  = dn, then the sets In = [ an,bn] form a sequence of nested intervals, and so there is a unique real number a belonging to every In. Similarly, the nested intervals Jn = [ cn,dn] have a unique real number b belonging to all of them. Since Sn = 1n×Jn, the unique number z_0 = a + ib belongs to all the Sn and so S contains z_0.

Figure 26

(ii) Let z Sn. By part (i) z_0Sn.

Now

But ; therefore, given  > 0, there is a number N such that

Thus, for z Sn and n > N,

In other words,

4.3 Section summary

The modulus function provides us with a measure of distance that turns the set of complex numbers into a metric space in much
the same way as does the modulus function defined on R. From the point of view of analysis the importance of this is that we can talk of the closeness of two complex numbers. We
can then define the limit of a sequence of complex numbers in a way which is almost identical to the definition of the limit
of a real sequence. Another analogue of real analysis arises with the Nested Rectangles Theorem. This will play a role in
complex analysis similar to the least upper bound property (or, equivalently, the Nested Intervals Theorem) of real analysis.

The modulus function also gives us a way of describing certain sets of points in the complex plane in a convenient algebraic
form. These descriptions are the subject of some of the problems in the next section.

4.4 Self-assessment questions and problems

SAQ 26

Find the distance between the numbers 2 − i and 1 + 3i.

View answer - SAQ 26

SAQ 26

In each of the following cases, classify the statement as True or False.

(i) The set , where a, b, c, d R, is a rectangle.

(ii) The set in (i) is a closed rectangle

(iii) The set in (i) is an open rectangle.

(iv) The setis a disc

(v) The set in (iv) is a closed disc.

(vi) The sets Sn, n = 1, 2, 3 …, where

form a nested sequence.

(vii) The sets Sn, n = 1, 2, 3…, where

form a nested sequence.

(viii) The sets Sn, n = 1, 2, 3…, where

form a nested sequence.

(ix) The sets Sn, n = 1, 2, 3…, where

form a nested sequence.

View answer - SAQ 26

SAQ 27

Sketch the curves corresponding to the following equations.

(i) |z + 1 | = 1.

(ii) |z + 3−i| = |z−2 |.

View answer - SAQ 27

SAQ 28

Find the Cartesian equation (that is an equation connecting x and y directly) of each of the curves in SAQ 27.

View answer - SAQ 28

SAQ 29

If you have ever tried to cut a flower bed by tying the ends of a piece of string to two canes so that the string is slack
and then fiddling about with another cane keeping the string taut, as in Figure 33, then you will know that the equation

is the equation of an ellipse. Find the Cartesian form of the equation

Figure 33

View answer - SAQ 29

SAQ 30

Sketch each of the following sets of points, where a, b,, c, d C:

View answer - SAQ 30

SAQ 31

(i) We sometimes want an algebraic description of a line segment in the complex plane, rather than a complete line. One way of doing this is to use a real parameter. Describe the set where t is real and z_1 and z_2 are fixed complex numbers. Identify the points corresponding to and 1 respectively.

(ii) Describe the set

View answer - SAQ 31

SAQ 32

(i) Prove the following inequalities.

(ii) Show that

(iii) Use the formula

together with part (ii) and a variation of (a) of part (i) to prove the triangle inequality:

(iv) By writing , prove that

Prove, similarly, that

and hence that

Interpret this result geometrically.

View answer - SAQ 32

SAQ 33

Show that the function where d(z,w) = |z−w| is a distance function: that is to say d satisfies:

View answer - SAQ 33

SAQ 34

Prove that

View answer - SAQ 34

SAQ 35

Find an upper bound for each of ihe following sets.

View answer - SAQ 35

The next problem deals with the Bolzano–Weierstrass Theorem. Bernard Bolzano (1781–1848) was professor of the Philosophy of Religion, in Prague. He wrote a treatise entitled Paradoxes of the Infinite, but this did not appear until after his death. Karl Weierstrass (1815–1897) was, together with Cauchy and Riemann, one of the founders of complex analysis. He led an uneventful life, much
of it spent in Berlin where he was the professor of Mathematics. He is remembered both for his work on elliptic functions,
and also for his attempts to put analysis on a firm logical foundation. You will come across his name in connection with the
Bolzano–Weierstrass Theorem, the Casorati–Weierstrass Theorem, and the Weierstrass M-test for uniform convergence.

To pose this problem we require the following definition.

Definition

Let S be any set, and z_0 any point in C. Then z_0 is called a cluster point of S if for all  > 0 there is z S such that . (Note that z_0 need not belong to S.)

SAQ 36

Prove the Bolzano–Weierstrass Theorem: if S is a bounded set and contains infinitely many complex numbers then S has at least one cluster point.

View answer - SAQ 36

Conclusion

This free course provided an introduction to studying Mathematics. It took you through a series of exercises designed to develop
your approach to study and learning at a distance and helped to improve your confidence as an independent learner.

Keep on learning

Study another free course

There are more than 800 courses on OpenLearn for you to choose from on a range of subjects.

Find out more about all our free courses.

Take your studies further

Find out more about studying with The Open University by visiting our online prospectus.

If you are new to university study, you may be interested in our Access Courses or Certificates.

What’s new from OpenLearn?

Sign up to our newsletter or view a sample.

For reference, full URLs to pages listed above:

OpenLearn – www.open.edu/openlearn/free-courses

Visiting our online prospectus – www.open.ac.uk/courses

Access Courses – www.open.ac.uk/courses/do-it/access

Certificates – www.open.ac.uk/courses/certificates-he

Newsletter – www.open.edu/openlearn/about-openlearn/subscribe-the-openlearn-newsletter

Acknowledgements

All written material contained within this course originated at the Open University

Course image: Håkan Dahlström in Flickr made available under Creative Commons Attribution 2.0 Licence.

Don't miss out:

If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover
our free learning resources and qualifications by visiting The Open University - www.open.edu/openlearn/free-courses

SAQ 1

Answer

(i) True

(ii) False. Im(3 + 2i) = 2.

(iii) False, consider z_1 = 1 + i, z_2 = 1 − i.

(iv) False, consider z_1 = x + iy and z_2 = x − iy.

Back

SAQ 2

Answer

(i) 5 + i.

(ii) 2 + 2i.

(iii) 8 + i.

(iv) 6 + 0i = 6.

(v) -2 + 0i = -2 (not 2i).

Back

SAQ 3

Answer

We have

Equating real and imaginary parts,

Back

SAQ 4

Answer

(i) 19 + 3i.

(ii)

(iii) 2.

(iv) 13.

(v) 1 + 6i.

Back

SAQ 5

Answer

(i) If z = x + iy, the equation z^2 = −9 becomes (x^2− y^2) + 2xyi = − 9.

Thus x^2 − y^2 = −9 and 2xy = 0. From xy = 0 we deduce that x or y is zero. From x^2 − y 2 = −9 we deduce that y cannot be zero. Thus x = 0, y = ±3 and so z = ± 3i.

(ii) z = ± ki.

Back

SAQ 6

Answer

The equation is equivalent to x−y + i(x + y) = − 2 + i. Thus x − y = −2 and x + y = 1, hence

Back

SAQ 7

Answer

(i) For n = 1, (1 + z)1 =1 + z and c_1 = 1.

If the result is true for n = k,

where c_1, c_2, … ,ck are positive real numbers (dependent on k). (In particular, note that the c_1 here is not the c_1 in the first line of this solution. The present c_1 is associated with n = k; the first one is associated with n = 1.) We thus deduce

and so the coefficients are real and positive.

Furthermore, they satisfy the familiar property of the binomial coefficients, often expressed as Pascal's Triangle, and so

(ii) Similarly,

(You can derive this from the previous result by replacing z by z/w,w≠0, and then multiplying by wn. The case w = 0 is dealt with independently, the result being trivial.)

Back

SAQ 8

Answer

(i) Since p(z) − zp(z) = 1 − zn, we have p(z) (1 −z) = 1− zn.

from which,

(ii) First, note that z ≠ w, that is z/w ≠ 1. Then replacing in (i) gives

provided z/w ≠ 1, that is z ≠ w. Thus

The result follows from multiplication by w n−1, and the fact that (wn− zn) / (w− z)  =  (zn− wn) / (z−w). The case w = 0 is dealt with independently: again the result is trivial.

Back

SAQ 9

Answer

(i) From SAQ 8, part (ii),

q(z) = zn−1 + zn−2z_0 + zn−3z_0^2 + ⋯ + z_0n−1,

so q(z) exists and zn − zn_0 is divisible by z − z_0.

(ii)

From (i), each of the terms on the right can be written in the form ai (zi−z_0i)  = qi where q_1(z) is a polynomial. Thus

where

(iii) If p(z) is divisible by z − z_0, there is a polynomial q(z) such that p(z) = (z − z_0)q(z) and so p(z_0) = 0.

If p(z_0) = 0, the equation p(z) −p(z_0)  = (z −z_0)q(z) becomes p(z) = (z − z_0)q(z) and so p(z) is divisible by z − z_0.

(iv) From the Fundamental Theorem of Algebra, there is a number zn for which p(zn) = 0 and so parts (iii) and (i) enable us to write

The polynomial q(z) is of degree n − 1 and the argument can now be applied to q(z). The result follows by induction.

Back

SAQ 10

Answer

The mapping

is clearly one-one and onto.

Also

and

Thus f is an isomorphism,

(As you would expect, the inverse matrix corresponds to the multiplicative inverse of a + ib.)

Back

SAQ 11

Answer

(i) The mapping f:a + ib→ ai + bjis clearly one-one and onto from C to the set of two-dimensional geometric vectors and

Note that we have used and for geometric vectors of unit length in the direction of the x-axis and y-axis respectively.

(This isomorphism gives a useful visualisation of addition and subtraction of complex numbers and also suggests how they may
usefully be represented geometrically. This is taken up in the next section.)

If z_0 = x_0 + iy_0 the effect of the mapping is to map (x, y) to (x_0 + x, y_0 + y), and it is therefore a translation.

Back

SAQ 12

Answer

We have

Thus f is linear.

Back

Example 1

Answer

Solution

Since |16i| = 16, we have

An argument of 16i is , so an argument of z is , because . So is a possible value of z. But and this is also an argument of a, so

is another possible value of z.

Back

SAQ 13

Answer

Back

SAQ 14

Answer

Back

SAQ 15

Answer

(i)

(We asked you only for an argument: you could have given any one of , where k is an integer. An alternative way of expressing this is to write θ = /4 (mod 2). Similar remarks apply to the other parts.)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Back

SAQ 16

Answer

Back

SAQ 17

Answer

(i)

(Note that we have taken z = x + iy; since this is the standard representation, we do not normally state it explicitly.)

(ii)

Back

SAQ 18

Answer

(i)

Figure 15

View description - Figure 15

(ii)

Figure 16

View description - Figure 16

(iii)

Figure 17

View description - Figure 17

(iv)

Figure 18

View description - Figure 18

(v)

Figure 19

View description - Figure 19

(vi)

and this is zero if and only if

Thus x and y satisfy

which is equivalent to

The locus is a circle with centre and radius (Figure 20). The point z = 2 on this circle must be excluded from the locus, because (z + 3)/(z − 2) is not defined for z = 2.

Figure 20

When you are used to manipulating complex numbers you will be able to work more elegantly as follows. gives since is real. Thus

and so

(vii)

Figure 21

View description - Figure 21

(viii)

Figure 22

View description - Figure 22

(ix) 3Re z + 2 Imz < 6 corresponds to 3x + 2y < 6 and so the set is as indicated in Figure 23.

Figure 23

View description - Figure 23

(x)

Figure 24

View description - Figure 24

(The regions in Figures 15, 16 and 23 are referred to as “half-planes”. Figures 17 and 18 illustrate “strips”; Figure 19 shows a “rectangle”, and Figure 24 a “triangle”. Note also that in Figures 15 to 19 we have illustrated only the case a, b, c and d positive. In Figure 21 we have chosen α < −/2 and β  > /2. In Figure 22 we have chosen 0 < α < /2.)

Style for figures: It is difficult to develop a completely consistent style for figures. However, you should find the following guidelines
useful. (i) Infinite (unbounded) sets are shown by squared-off shading as in Figures 15, 16, 21, 22 and 23. (ii) Sets which do not include their “boundary” lines have thin boundary lines (for example, Figures 15, 16, 17, 18, 21, 23 and 24). (iii) Thick boundary lines indicate that the set includes its boundary lines. (iv) Most other conventions are self-evident
or obvious from the context.

Back

SAQ 19

Answer

Substituting gives

Since is real if n is a multiple of 4, and so equating real and imaginary parts gives

Since , we have .

We have seen that if n is a multiple of 4 then (1 +i)n is real, that is Arg(1 + i)n is either 0 (in which case (1 + i)n is positive) or (in which case (1 +i)n is negative). But, since an argument of 1 + i is /4, an argument of (1 + i)n is n/4.

If n is an even multiple of 4, that is n = 2k × 4 for some integer k, an argument of (1 + i)n is 2k. In other words, Arg(1 + i)n = 0 and so (1 + i)n is positive.

If n is an odd multiple of 4, that is n = (2k + 1) × 4 for some integer k, an argument of (1 + i)n is (2k+ 1). In other words, Arg(1 + i)n =  and so (1 + i)n is negative.

Hint

Try substituting a suitable value for z in

Back

SAQ 20

Answer

(i)

By an extension of a result in SAQ 17(iii) Also, if ck is real, , and again from SAQ 17(ii), the conjugate of a sum is the sum of the conjugates and so

(ii) If α is a root of the polynomial then and so and thus from (i), ; in other words, is also a root.

Back

SAQ 21

Answer

(i) The result is trivially true for n = 1.

If the result is true for n = k, that is |zk| = |z|k, then

The result follows by mathematical induction.

(ii) The result is trivially true for n = 1. If the result is true for n = k, then

from the ordinary properties of multiplication of two complex numbers. The result follows by mathematical induction.

(iii) This follows from (ii) by noting that

Hence

(iv) Let r be any rational number. Then r = n/m where n is an integer and m is a positive integer. Since m is a positive integer, one of the values of (cosθ + isinθ )1/m is cos (θ/m) + i sin (θ/m) , and so, from (ii) and (iii), one of the values of (cos θ + isin θ)n/mis cos (n θ/m) + i sin(nθ/m).

Back

SAQ 22

Answer

Since a ≠ 0, we have z_0 ≠ 0 and so any complex number z can be written in the form z_0w, just by choosing . If z is a solution of zn = a, then

(z_0w)n = a

and so

z_0nwn = a.

But z_0n = a and so we must have wn = 1.

Back

SAQ 23

Answer

(i) Since −32  = 32 [cos (-)  + isin (-)] , one fifth root of −32 is

Since the five fifth roots of −32 are equally spaced around the circle with centre at the origin and radius 2, the other roots
are

(Note that k = 3 gives the real fifth root −2. The above method is the systematic one; we could, however, have made use of −2 as the one
obvious fifth root of −32 from the start.)

(ii) Since −16 = 16[cos (−)  + i sin (−)]. , one fourth root of −16 is

Since the four fourth of −16 are equally spaced around the circle with centre at the origin and radius 2, the other roots
are

Back

SAQ 24

Answer

(i) Putting w = z^2 gives w^2 + 4w + 8  = 0,

and so

. So z can take the values of the two square roots of −2 + 2i and the two square roots of −2 −2i. Since

the two square roots of −2 + 2i are

and

Similarly, we can show that the two square roots of −2 −2i are

(Alternatively, we could use the result of SAQ 20 (ii) to find the last two roots, since is a polynomial with real coefficients.)

(ii) Putting w = z^2 gives

and so

that is

Since , the two square roots of are

and

Since

and

Note that the four roots of do not occur in conjugate pairs: the coefficient of z^2 is not real.

Back

SAQ 26

Answer

Back

SAQ 26

Answer

(i) True.

(ii) False.

(iii) False.

(iv) True.

(v) True.

(vi) False (Figure 27).

Figure 27

(vii) False (Figure 28).

Figure 28

(viii) True (Figure 29).

Figure 29

(ix) True (Figure 30).

Figure 30

Back

SAQ 27

Answer

(i)

Figure 31

View description - Figure 31

(ii)

Figure 32

View description - Figure 32

Back

SAQ 28

Answer

(i) (x + 1)2 + y^2 = 1.

(ii) 5x−y + 3 = 0.

Back

SAQ 29

Answer

You have to tread with caution to avoid as much arithmetic as possible. One way is to write

| z−1 | = 3−| z−2 |

and so

| z−1 |2 = 9−6 | z−2 | + | z + 2 |2

giving

36 | z−2 |2 =  ( 9  + | z−2 |2−| z−1 |2)2.

The right-hand side simplifies considerably and (after a few days’ work) you should arrive at

(It is possible to go straight to x and y in the first equation, but in initial manipulations it is often simpler to work in terms of z.)

Back

SAQ 30

Answer

(i)

Figure 34

View description - Figure 34

(ii)

Figure 35

View description - Figure 35

(iii)

Figure 36

View description - Figure 36

(Note that is the set of all points whose distance from a is less than or equal to their distance from b.)

Back

SAQ 31

Answer

(i) The set is the set of points on the line segment joining z_1 and z_2, which is denoted by [z_1, z_2] by analogy with an interval on the real line. The points corresponding to correspond to the points: z_2, the point of trisection of [z_1, z_2] nearer to z_2, the midpoint, and z_1, respectively.

(ii) Since

is real and . Hence A is the line which includes the line segment of part (i).

Back

SAQ 32

Answer

(i) (a)

(b)

(c) Since, for any real numbers x and y, we have

we can deduce that

and so the result follows.

(d) We have

Also

So

The result follows because moduli are nonnegative.

(ii)For any complex number α,

( see Problem 3(i) of Section 1.4)

Put and note that ; then

(iii)We have

Also

Thus

since both sides are nonnegative.

(iv)From (iii), we have . Thus . Interchanging z and w, gives and combining the two results gives

The result corresponds to the fact that the difference between the lengths of any two sides of a triangle does not exceed
the length of the third side.

Back

SAQ 33

Answer

(i)

(ii) If z = x + iy and w = u + iv then

Thus d(z, w) = 0 if and only if x= u and y = v; that is if z = w, d(z, w) > 0 otherwise, because |z − w| is nonnegative.

(iii) We have

using the result of SAQ 32 (iii), with z replaced by z − a and w replaced by a − w. Thus

(Note that the set C and the function d with domain C × C constitutes a metric space.)

Back

SAQ 34

Answer

(i) by part (a) of SAQ 32 (i).

(ii) Since

we know that

and so

from which

Back

SAQ 35

Answer

(i) We know that |z| = 2 and we want to replace by something larger. This means replacing by something smaller. We can write and use the result of SAQ 32 (iv) to give So

(ii) A similar argument to that in (i) gives

Back

SAQ 36

Answer

Since S is bounded there is a closed disc with centre the origin and radius K, say, which contains S. Therefore there is a closed square, S_0, with vertices at ±K ±iK which encloses S. The imaginary axis divides this square into two closed rectangles. At least one of these must contain infinitely many points.
Choose this rectangle. If both rectangles contain infinitely many points, choose the right-hand one. The real axis divides
this rectangle into two closed squares. At least one of these contains infinitely many points of S. Choose this one, or the upper one if they both do. Call this square S_1. If we continue in this way, dividing each square by a vertical line and then by a horizontal line, we get a nested sequence
of closed squares {Sn} where Sn has side of length .

Also . By the Nested Rectangles Theorem, there is one and only one point z_0 in every square. This is a cluster point, because given  > 0, the disc |z − z_0| < contains the squares Sn where , that is, (Figure 37). (Notice that z_0 could be at the corner of Sn: this position giving the maximum size of disc required.) Thus the disc contains an infinite number of points of the set.

Notice that there may be more than one cluster point. The cluster point will be unique if at each stage of the process only
one of the two rectangles contains an infinite number of points of S.

Figure 37

Back

Figure 15

Description
Figure 15

Back

Figure 16

Description
Figure 16

Back

Figure 17

Description
Figure 17

Back

Figure 18

Description
Figure 18

Back

Figure 19

Description
Figure 19

Back

Figure 21

Description
Figure 21

Back

Figure 22

Description
Figure 22

Back

Figure 23

Description
Figure 23

Back

Figure 24

Description
Figure 24

Back

Figure 31

Description
Figure 31

Back

Figure 32

Description
Figure 32

Back

Figure 34

Description
Figure 34

Back

Figure 35

Description
Figure 35

Back

Figure 36

Description
Figure 36

Back