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Various conventions are commonly used with complex numbers, as you can see

in the table:

(a) any real number x can be thought of as a complex number whose

imaginary part is zero, and the complex number x is then said to be purely 1+ O i = 1

real (thus IRis a subset of C);

(b) if the real part of a complex number is 0, but the imaginary part is

non-zero, then we write the complex number in purely imaginary form;

1 IN TRODUC ING COMPLEX NUMBERS

After working through this section, you should be able to:

(a) determine tile real pad, the imaginary part and the complex conjugate of a

given complex number;

(b) perform addition, subtraction, multiplication and division of complex

numbers;

(c) use the Binomial Theorem and the Geometric Series Identity to simplify

complex expressions.

1 .1 W ha t is a com p lex num ber?We assume that you are already familiar with various different types of

numbers, such as the natural nurnbers N = {I, 2, 3, ... }, the integers

7 l . . = {... , -2, -1,0,1,2, ... }, the rational numbers (or fractions)

Q = {p ig: p E 7 l . . , g EN}, and the real numbers IR, which can be represented

by decimals (terminating or non-terminating). We assume also that you are

familiar with the usual arithmetic operations of addition, subtraction,multiplication and division of real numbers.

We are now going to introduce the idea of a complex number and we begin

with some definitions.

Definit ions A complex number z is an expression of the form J; + iy,

where x and yare real numbers and iis a symbol with the property that

i = -1. We write

z = x + iy or, equivalently, z = x + yi,

and say that z is expressed in Cartesian form. The real number x is the

real part of z (written x = Re z) and the real number y is theimaginary part of z (written y = Imz).

Two complex numbers are equal if their real parts are equal and their

imaginary parts are equal.

The set of complex numbers is denoted by I C .

Here are some examples of complex numbers z which correspond to given real

numbers x and y.

z = x + iy 1+ 2i .j2+ i7r 3i 1 l+i ° 1- 2i

Rez=x 1 .j2 o 1 1 ° 1

Imz = y 2 7r 3

°o -2

(c) 0 + Oi is written 0, the zero complex Humber;

(d) we usually abbreviate Iito i;

(e) if y is negative, then we usually write z as x -IYli.

For example:

~ = 0.5,

tt =3.1415 ...

0+ 3i =3i

1+ (-2)i = 1 - 2i

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1 .2 A rithm etic in CThe definition of a complex number contains the symbol '+' and refers to the

'square' of i. This suggests that arithmetic operations can be performed with

complex numbers; the following definitions are made.

Def ini t ions The binary operations of addition, subtraction and

multiplication of complex numbers are denoted by the same symbols as

for real numbers and are performed by the usual procedure - that is,

treating the complex numbers as real expressions involving an algebraic

symbol i with the property that i =-1.

Some examples will help to make this definition clear.

Exam ple 1 .1

Express each of the following numbers in Cartesian form.

(a) (1+2i)+(~+ni)

(b) (1+2i)(i+ni)

(c ) 2(1+ 2 'i) - 2i (~+ ni)

(d) ( J + 2i)(1 - 2i)

Solut ion

(a) By the usual procedure:

(l+ 2'i) + (~ + ni) =1+ 2i + i + ni

=~ 1 - (2 + n)i.

(b) By the usual procedure:

(1 + 2i) (~ + ni) =1+ ni + i+ 2ni2;

applying the extra property that 'i2 = -1, we obtain

(1+ 2i) (~+ ni) = 0- 2n) + (n + l)i.

(c) By the usual procedure and the property that i = -1:

2(1 + 2 'i) - 2 i(i + n'i) = 2 + 4i - i - 2ni2

= (2 + 2n) + 3i.

(d) By the usual procedure and the property that i =-1:

(1+ 2 'i)(1 - 2i) =1 - 2i + 2i - 4i2

= 1 + 4 =5. •

The following problems provide practice at such manipulation of complex

numbers.

P rob lem 1.1 _

(a) Express each of the following in Cartesian form.

(i) (2+ i) + 3i(-1 + 3i)

(ii) (2+i)(-1+3i)

(iii) (-1+3-i)(-1-3'i)

(b) Write down the real and imaginary parts of z = (2 + i) + 3i( -1 + 3i).

8

In some contexts, e.g. electrical

engineering (where iis used for

current), it is common practice

to write j for i.

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P rob lem 1 .2 _

Express each of the following ill Cartesian form.

(a) (X l + iYl) + (X 2 + iY2) (b) (X l + iyr) - (X 2 + iY2)

(c) ( X l + iYl)(X2 + iY2) (d) (x + iy)(x - iy)

As with real numbers, the negative, -z, of a complex number z is defined insuch a way that z + (-z) =O.

Def ini t ion The negative, -z, of a complex Humber z =x + iy is

-z= (-x)+i(-y),

usually written -z =-x - iy.

Next, we discuss division of complex numbers. As with real numbers, the

reciprocal, it», of a non-zero complex number z is defined in such a way that

z(l/z) = 1.

Def ini t ions The reciprocal, 1/z, of a non-zero complex number

z =x + iy is

1 x - zy=

x2 + y2 '

The alternative notation Z-l for the reciprocal is also used.

The quotient, Zr/Z2, of a complex number Z1 by a non-zero complex

number Z2 is

This definition of 1 /z works because

(x + iy)(x - iy) = x2 + y2 ,

so that

= (x+iy) ( X2

-iY2

)x +y

(x + iy)(x - iy)

x2 + y2

x2 +y2--:c-~~ = 1x2 + y2 .

The above definition of quotient suggests that in order to evaluate zd Z2 one

must first evaluate 1/ Z2 and then multiply by Zl' In practice, it is easier to do

both operations at once using the following strategy.

S tra te gy fo r o bta in in g a q uo tie nt

To obtain the quotient

·1:1 + iY 1

X 2 + iY2'

in Cartesian form, multiply both numerator and denominator by X 2 - iY2,

so that the denominator becomes real.

where X 2 + iY 2 # 0,

If x and y, with or without

subscripts, appear in the

specification of a complex

number, then it is presumed

that x and yare both real.

For example,

-(l+)=-l-i.

Notice that x2 + y2 is strictly

positive since Z is non-zero.

Problem 1.2(d)

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E xam ple 1 .2

Express the following numbers in Cartesian form.

]

(a.) 1 + 2i

3+ 4i(b) 1+ 2i

Solut ion

(a) By the stra tegy,

1 1- 2i

(1+ 2i)(1 - 2i)

1- 2i

1 + '1

1 2= - - -z.

5 5

(b) By the strategy,

1+ 2i

3+ 4i (3 + 4i)(1 - 2i)

] + 2i (1+ 2i)(1 - 2'i)

3 - 2i - 8i2

1+4

11 - 2i 11 2= -- - = ---i .•

5 5 5

P rob lem 1 .3 _

(a.) Express the following numbers in Cartesian form.

(i) ~2

1(ii) -.

1+.~(iii) 1+ 2i111 --

2 + 3i

(b) Express the quotient

:£1+ iY l

X2 + iY2 '

in Cartesian form.

where X2 + iY 2 = 1 = 0,

The above process of changing the sign of the imaginary part of a complex

number is often used, and so we introduce the following terminology and

notation.

Defin i t ion The complex conjugate, Z, of a complex number

z =x + iy is

Z = x - iy.

The complex conjugate of z satisfies the simple identities

Re Z =Re z and Im z =- Im z.

J O

Itwould be acceptable to leave

this solution in the form

(11 - 2i)/5, since this can

readily be red uced to Cartesian

form.

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Several more complicated identities involving complex conjugates are given in

the following result.

T he ore m 1 .1 P ro perties o f th e c om p le x c on ju ga te

(a) If z is a complex number, then

(i ) z + z = 2 Re z;

(ii) z - z = 2ilmz;

(iii) (z ) = z.

(b) If z] and Z2 are complex numbers, then

(i ) Zl + Z2 =Zl + Z2;

(ii) z] -Z2=Zl-Z2;

(iii) ZlZ2 =Zl Z2;

(iv) ZdZ2 = ZdZ2' where Z2 : : J o .

Proof

(a) If z = x + iy, then z = x - w , so that

(i ) z+z=(x+iy)+(x-iy)=2x=2Rez;

(ii) z - z = (x + iy) - (x - iy) =2iy =2ilmz;

(iii) (z ) =(x - iy ) =x + iy =z.

(b) The proofs of these identities all follow from the results of Problems 1.2

and 1.3(b). To illustrate the method we prove (iii).

Let Zl =Xl + iY1 and Z2 =X2 + iY2, so that

ZlZ2 = (X1X2 - Y1Y2) + i(XlY2 + :J:2Y1),

by Problem 1.2(c). Also, Zl =Xl - iY1 and Z2 =X2 - iY2, so that

Zl Z2 = (X1X2 - (-yd( -Y2)) + i(X1( -Y2) + X2( -yd)= (X1X2 - Y1Y2) - i(XIY2 + x2yd

(1.1 )

= Z1Z2,

as required. •

P rob lem 1 .4 _

Prove the identities stated in Theorem 1.1(b), parts (i) and (iv).

Now that we have explained how to perform the usual arithmetic operations

with complex numbers, it is natural to ask the following question. Do these

operations have the usual properties which are known to hold for real numbers?

It is a straightforward matter to check that, for example, addition of complex

numbers is associative; that is, for all Zl, Z2, Z3 in 1[,

It is also straightforward, but more tedious, to show that multiplication of

complex numbers is associative; that is, for all Zl, Z2, Z3 in 1[,

Part (b) says: 'the conjugate of

a sum is the sum of the

conj ugates', etc.

Note the use of the long

conjugate bar over expressions

involving several symbols.

Replace Yl, Y2 by -Yl, -Y2 inEquation (1.1).

1 1

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Tn fad, it t nrns out that all Ihe IISl131 aritluuetic properties do hold for complex

numbers. These are summarized in the following table.

A rith me tic in C

Addition Multiplication

A L For all ZI, Z2 in I C ,

ZI + Z2 E C.

A2 For all z in C,

Z + 0 = 0 + z = z.

MJ For all ZI, Z2 in C,ZI Z2 E C.

M2 For all z ill C,

z1=lz=z.

1 1 3 Por all Z in C,

z+(-z)=(-z)+z=O.

M3 For all non-zero z in C,

ZZ-l =z-lz =1.

A4 For all zI, Z2, Z3 in C,

(ZI + Z2) + Za = ZI + (Z2 + Z3).

A5 F'or all ZI, Z2 in C,

ZJ + Z2 = Z2 + Zl·

M4 For all ZI, Z2, Z3 in C,

(ZIZ2)Z3 = ZI(Z2Z3).

M5 For all ZI, Z2 in C,

ZI Z2 =Z2Z1·

D For all ZI, Z2, Za in C,ZI (Z2 + Z3) = ZI Z2 + Zl Z3·

Once all these properties have been proved (and we shall not, give the details),

then the contents of the table can be described in algebraic terms as follows:

C is an Abelian grou p under the operation of addition, with identity 0;

the set of non-zero complex numbers is an Abelian group under the operation

of multiplication, with identity 1;

these two structures are linked by the distributive properly.

Because C has all these properties, it is called a field; Q and IR are also fields.

Notice that in property M3 we have used Z-1 to denote the reciprocal L'z. It is

also standard practice to use the notation z ", where n E 2, for integral powers

of a non-zero z; in particular, zO = 1for all non-zero z. The zero complex

number has powers 0·:=0 for k =1,2,3, .... Vve shall discuss the meaning of

fractional powers, such as zl/2 =Jz, in Section 3, and a.lso in Unit A2.

1 .3 Id en tities w ith com plex num bersBecause complex numbers satisfy the usual arithmetic properties, we can prove

and then use all the usual algebraic identities. For example, if ZI and Z2 areany complex numbers, then

and

2 2ZI - Z2 = (ZI - Z2)(ZI + Z2).

Thus, for example, if z2 + 9 = 0, then

z2 + 0= (z - 3i)(z + 3i) = 0,

so that Z =3'i or Z = -3i.

12

Closure

Identity

Inverse

Associa.tive

Commutative

Distributive

AI-A5

MI-M5

D

For example,

'i2 = -I, 'i

3 = -i, i4 = 1;

i- = -i, i-2

= -1, 'i-3 = i,i-4 = 1; iO =1.

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Prob lem 1.5 _

The identities in Problem 1.5 are, ill fact, special cases of two important

general identities which will often be used ill the course. The first of these is

the Binomial Theorem, which we slate ill two forms. The proof is the same as

in the real case, so we omit it.

Theorem 1.2 B inom ia l Theorem

(a) If z E ICand n E N , then

(1+ z) " =~ (~) zk

n(n-1) 2 TI.

=1+ nz + I Z + ... + Z •2.

(b ) If zl,Z2 E ICand n EN , then

( ) " ~ ( n ) ",-k_kZl + Z2 =~ k Zl "'2

k=O

. l~(n- 1 )-z"'+nzn-lz + zn-2z2+···+zn- 1 I 2 2! 1 2 2 .

Remark It is worth recalling that the coefficients which appear in the

Binomial Theorem can be arranged in the form of Pascal's Triangle, as follows.

(l+z)O 1

(1+ z )l 1 1

('I + Z ) 2 1 2 1

(1+z)3 1 3 a 1

(1+ Z ) 4 1 4 6 4 1

Prob lem 1.6 _

Use the Binomial Theorem to simplify the following expressions.

(a) (l

+i)4 (b) (3

+2i)3

Next we state the identity which is used to sum a f nite geometric series. Once

again we give two forms. The proof is the sauie a .s ill the real case, so again we

omit it.

Theorem 1.3 G eom etric S eries Identity

(a) If z E ICand n E N, then

1- z " =(1- z)(l + z + z2 + ... + zn-l).

(b) If Zl, Z2 E ICami n E N, then

11 . n ( )( 7l-j + ,,-2 + 7t-3 2 I- + 11-1)Zl - Z2 = Zl - Z2 Z 1 Zl Z2 Z I Z2 - ... Z2 ..

The binomial coefficient

n!

k!(n-k)!

n (7 1 , - 1 ) ... ( n - k + 1 )

k!

is sometimes written as "'Ck.

Note that, by convention, O!= 1

and 0° = 1 in these formulas.

Problem 1.5(a) is the special

case with n = 3.

Problem Ui(b) is the special

case with n = 3.

13

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Remark The first of these two ident.ities call he written as

.) 1-z71+ 1

I + z + z: +.. .+ z" = -- - -1 - zfor z o f . 1.

This is t .he familiar formula for summing a finite geometric series.

P rob lem 1 .7 _

(a) Use the Geometric Series Identity to simplify the expression

1+ (1+ i) + (1+ i) 2 + ( J + ' i)3.

(b) Use the Geometric Series Identity to find one linear factor of

5 .Z - l ,

(H'int: ,£ 5 = i.)

2 THE COM PLEX PLANE


(a) determine the modulus of a given complex number;

(b) determine the principal argument and other arqumetiis of a given non-zero

complex number;

(c) convert a complex number in Cartesian form to polar form, and vice versa;

(d) interpret geometrically the sum, product and quotient of two complex

numbers;

(e) state de Moivre's Theorem, and use it to evaluate powers of complexnumbers.

2.1 C artes ian co ord in atesIn this section we describe a. geometric interpreta.tion of complex numbers, and

we see how this interpretation leads to useful insights concerning the properties

of complex numbers.

Cartesian coordinates can be used to represent the complex number z = x + iy

by the ordered pair (x , y ) in ~2 For example, the number 4 + 3i is representedby (4,3) and in Figure 2.1 this point is labelled 4 + 3i.

3 ---- ...4 +3iII

I

4

Figure 2.1

Thus we often speak of 'the point z =x + iy' and, with this interpretation,

refer to the plane as the complex plane or the z-plane. The horizontal axis is

called the Teal axis and the vertical axis is called the imaginary axis, and they

are sometimes labelled x and y, respectively, in the usual way.

7 4

Hamilton's definition of complex

numbers as ordered pairs (see

the Introduction) is based on

this Cartesian representation.

The complex plane is often

called the Argand diagmm, after

Jean-Robert Argand

(1768-1822), a French-Swissmathematician, although both

Gauss and Wessel (1745-1818),

a Norwegian surveyor and

cartographer, used the idea

before Argand.

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The various operations 011 complex numbers described in Section 1 can all be

giveu a geometric interpretation in the complex plane. For example, if z is a

complex number then, as shown in Figures 2.2 and 2.3,

-z is obtained by rotating z through the angle 7f about the origin;

z is obtained by reflecting z in the real axis.

FI

I

I

II

I

h"

Figure 2.2 Figure 2.3

Since a complex number can be thought of as a based vector, the sum of two

complex numbers, and also their difference, satisfy the parallelogram law [or

vectors, as shown in Figure 2.4.

z~ __ --

_---Zl - Z 2

=ZI+(-Z2)

Figure 2.4

P ro b lem 2 .1

With Z1 =3 + iand Z2 = -1 + 2i, plot the following numbers,

(a) Z1, Z2, -Z1, -Z2, Z1 + Z2, Z1 - Z2·

(b) Z1, Z2, Z1, Z2, Z1 + Z2, Z1 + Z2 (on a separate diagram).

Multiplication and division of complex numbers also have useful geometric

interpretations. Before describing these, however, we need to introduce some

other geometric concepts.

2.2 Po la r formThe modulus, or absolute value, of a real number x is defined as follows:

{

X

I x l = - x : x 2 : u ,

x < o .

Equivalently, I x l is the distance along the real line [rom 0 to x. The modulus of

a complex number z is similarly defined to be the distance from 0 to z.

Definit ion The modulus, or absolute value, of a complex number

Z

=x

+iy is the distance from 0 to z; it is denoted by

14Thus

Iz l = Ix + iyl = J X 2 + u'.

The complex number x + iy

corresponds to the vector from

the point (0,0) to the point

(x , y).

For a 2 : 0, .; a means lhe

non-negative square root of a.

7 5

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For example:

1 3+ ..J il= J 3 2 +..J 2 =5 , 1 -:31= J ( -3)2 =3 , 1 -2il = J ( -2)2 =2.

These moduli are shown as dis Lances in Figure 2.5.

/5

/

I

3+ -Ii

o .)

-2i "j

o -3 n

Pigure 2.5

P rob lem 2.2 _

(a) Evaluate the following moduli.

(i ) 11+il (ii) 12 - 4il (iii) I i I (iv) 1 -5 + 12il

(b) Prove that Izl = Izi and I-zi = [z].

If Z l, Z2 a.re any two complex numbers, then, by definition, IZ l - z21 is the

distance from 0 to Z] - Z2. Using the parallelogram law to add Z2 and Z l - Z2

(see Figure 2.6), we deduce that

IZ l - z21 is the distance from Z l to Z2·

Beca.use Z l + Z2 = Z l - (-Z2)' there is a similar geometric interpretation for

IZ I + z21:

IZ J + z21 is the distance from Z l to -Z2·

P rob lem 2.3 _

With Z l =3 +iand Z2 =-] + 2i, determine:

(a ) IZ I - z21i

(b ) IZ J + z21i

(c) the distance from Z2 to -ZI.

We now collect together various basic properties of the modulus,

T heo rem 2.1 P ro perties o f th e m o du lu s

(a) Izi ~ 0 , with equality if and only if Z = o .

(b ) Izl = Izi and I-zi = Izi.

(c ) IzI2 =zZ.

(d) IZ I - z21= IZ 2 - z .].

(e) IhZ21 = IZIIIz21 and Izdz21 = Izd/hl, where Z2 = f . o .

7 6

Eiqure 2.6

Property (d) says, algebraically,

that the distance from Zl to Z2

is the same as the distance from

Z2 to Zl.

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If the modulus I z i of a complex number z is equal to 0 , then z itself must

equal 0 (and vice versa). However, the modulus of a non-zem complex number

does not determine the number completely; all the points which lie on the

circle of radius r centred at the origin have the same modulus, namely T. We

can determine the non-zero complex number z completely by giving its

modulus I z l =T together with:

the angle e that the line from the origin to z makes with the

positive real axis.

Angles used to determine position in this way are conventionally taken to be

positive when measured in an anticlockwise direction from the positive real

axis, and negative when measured in a clockwise direction.

For example, 1 + ihas modulus v ' 2 " and the (positive) angle that the line from Eiqure 2.7

the origin to 1 + imakes with the positive real axis is K /4 (see Figure 2.7). Of

course, K/4 is not the only angle which, along with the modulus v ' 2 , specifies1+ ij anyone of the angles

KKK K

. . . , 4 " - 2K, 4' 4 " + 2K, 4 " + 4K,

would do just as well ~ ill particular the (negative) angle K /4 - 27 r = -h/4

(see Figure 2.8). This feature is reflected in the following definition by the use

of the sine and cosine functions. Figure 2.9 illustrates the definition, showing

P r o o f Property (a) follows from the fact that I z l = J x 2 + y2 , if z = x + iy ,

and property (b) was proved in Problem 2.2(b). To prove property (c), note

that if z = x + iy then

zz = (x + iy)(x - iy) = x2 + y2 = I z 1 2 .

Property (d) follows from property (L), since Z2 - Zl = -(Z l - Z 2)'

Each of the identities in property (e) can be proved by writing Z 1 = :);1+ iY1,

Z2 = X2 + iY 2 and then calculating both sides. However, it is neater Louse

property (c) and Theorem 1.1, as follows:

2 --I Z 1 z 2 1 = ( Z 1 Z 2 ) ( Z 1 Z 2 )

= ( Z 1Z 2 ) ( Z 1 Z 2 )

= ( z 1 z d ( Z 2 Z 2 )

1 1

2 2= Z 1 hi

(property (c))

(Theorem l.l(b), part (iii))

(associativity and commutati vity)

(property (c)),

so that I Z 1 z 2 1 = I Z 1 1 1 z 2 1 . Similarly, if z21 0 (so that Z 2 10 and I Z 2 1 > 0 ) , then

h / z 2 1 2 = ( Z 1 / Z 2 ) ( Z 1 / Z 2 ) = ( z 1 / z 2 ) ( z d z 2 ) = ( z 1 z d / ( Z 2 Z 2 ) = I Z 1 1 2 / l z 2 1 2 ,

so that I z d z 2 1 = I Z 1 1 / l z 2 1 . •

one argument of z =x + iy.

sine = 1 j _ .T

Definit ion Anargument of a non-zero complex number z = x + iy

with I z l =T is all angle e (measured in radians) such that

xcos e =- ami

7'

Remarks

1 No argument is assigned to O .

1+i

1T /4

-71T/4

l+i

Figure 2.8

/

/I

II

\ I\ I\ /\ /

" /<c, _- _--/

2 Each non-zero complex Humber has inflnitely many arguments, all differing

by integer multiples of 2K. For example, the arguments of 1 + i(see above) are Figure 2.9 An argument of z

-77r/4, K/4, 971"/4, Ih/4, ... ,.. ,

which may be written as

where k : E 7 1 . .

7 7

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3 For some complex numbers, arguments are easily obtained by plotting the

point. For example, Figure 2.]0 shows that 37f/4 is an argument of -1 + i,

7f/2 is an argument. of i and -7f/4 is an argument of 1- i. The calculation of

arguments is dealt with later in the sect.ion.

-1-1.

1- i

rr/2rr/ -1___ -"!--1_

-rr/-1

Figure 2.10

Since any non-zero complex number is completely determined by its modulus

and any Olle of its arguments, these two quantities can be used to define an

alternative coordinate system for non-zero complex numbers.

Definit ion The ordered pair ( 1 ' , 0 ) , where r is the modulus of a non-zero

complex number z and 0 is an argument of z, are called polar

coordinates of z , The expression

z = 1 '( cos 0 + i sin 0 )

is called a representation of z in polar form.

Rema rks

1 Some alternative notations for polar coordinates are ( 1 ' , e ) and [ r , 0 ]. We

shall rarely use polar coordinates, preferring almost always to use polar form.

2 It follows from the definition of polar form that if z =x + iy, then

.7 ; = 'r cos 0 and y = r sin e.

E x amp le 2 .1

Represent -1-i in polar form.

Solut ion

Here

r = 1-1- il= )(-1)2 + (-1)2= V2

and, from Figure 2.11, one choice for e is 57f/4. Thus

-1 - i = V2(cos 57f/4 + i sin 57f/4)

is in polar form. •

P rob lem 2.4 _

(a) Represent the complex number i in polar form.

(b) Represent each of the following complex numbers in Cartesian form.

(i) 2(cos7f/3+isin7f/3)

(ii) 3(cos( -7f/4) + i sin( -7f/4))

The terminology 'arg z ' is often used to denote an argument of a non-zero

complex number z. Without further information, however, the expression arg z

is ambiguous, since z has infinitely many arguments, and so we shall use it

rarely. Instead, we select one argument for special attention and call this the

principal argument (a shortened version of the more conventional 'principal

value of the argument').

18

Note that r > 0 and f) E R

-l-i

Eiqure 2.11

Another polar form for -1 - iis

v'2(cos( -37f/4) + isin( -37f/4)).

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Def ini t ion The principal argument of a non-zero complex number z

is the unique argument e of z satisfying - 7 1 " < e : : : : 1 " ; it is d e n o t e d by

e =Argz.

(Note the capital A in Arg ..)

For example, as you have seen, the arguments of 1+ iare

... , - 7 7 r / 4 , 7 1 " / 4 , 9 7 1 " / 4 , 1 7 7 r / 4 , ... ,

and hence Arg(l + i) =7 1 " / 4 because - 7 1 " < 7 1 "/ 4 : : : : 1 r.

For complex numbers z such as 1+iit is easy to determine Arg z by

inspection. In general, the following strategy may be applied.

Eiqure 2.14

S tra te gy fo r d ete rm in in g p rin cip al a rg um e nts

Siuce the arguments of a

non-zero complex number differ

by multiples of 2 7 1 " , exactly oneof them satisfies - 7 1 " < () : :::7 1 " .

There are other equally valid

strategies.

To determine the principal argument e of a non-zero complex number

z = x + iy.

Case (i) If z lies on one of the axes, then e is evident (see Figure 2 . 1 2 ) . _{}"--=_:7r;__-()- ..

Case (ii) If z does not lie on one of the axes, then

(a) decide in which quadrant z lies (by plotting z if necessary), and then

calculate the angle

¢=tan-1( i y i / i : . c 1 )

{}=o

(see Figure 2 . 1 3 ) ;

(b) obtain e in terms of ¢ by using th e appropriate formula in Figure 2 . 1 4 .

Remarks

1 Having Iound Arg z , other arguments may be obtained by adding integer

multiples of 2 7 1 " to it.

2 The requirement that the principal argument Arg z should satisfy

- 7 1 " < Arg z :::: 7 1 "may seem rather arbitrary. It is, b u t if some choice h a s to be

made, then this one is marginally better than others. One reason for tbe choice

- 7 1 " < Arg z :::: 1 "will become clear in Units A3 a n d A4.

E xam ple 2.2

Find the principal argument of each of the Iollowiug complex numbers.

(a) 1+2i (b) -1 - J3 i (c) -1 + J3 i

{}= 7r2

Eiqure 2.12

Figure 2.13

Some texts prefer

0::; A rgz < 271' .

19

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Solut ion

Following t he st rategy (case (ii) each Lime), we have the following.

(a) l -I 2i lies ill Ihe first quadrant (Figure 2.15), and

< t> = tan-I(2/l).= tall 12;

thus the principal argument () is

(Figure 2.1 -1 )

(about 1.11 radians).

(b) -I-V 3 - i lies ill the third quadrant (Figure 2.16), and

c p = (a11-1(I-hl/l-ll) = (.an-I J3=7r/3;

thus the principal argument 8 is

() = -(7r - 7r/3)

= -27r/:1.

(c) -1 + V3i lies in the second quadrant (Figure 2.17), and

(Figure 2.14)

thus the principal argument 8 is

(Figure 2.14)= 7 r - C P

= 27r/3.

(Of course, knowing that Arg(-l - V3i) = -27r/3 from part (b) and that

- ] - I - V3i is the conjugate of -1-V3i, it follows immediately that

Arg( -1+ h·i) = -( -27r/3) =27r/3). •

P rob lem 2.5 _

For each of the following complex numbers z, write down Arg z and express Z

in polar form.

(a) -4 (b) 3V3-1-3i (c) .J3-i (d) -l-i

2 .3 T he g eom etric in terpre tation o f m ultip lica tion

a nd d iv is io nA geometric interpretation of the multiplication of complex numbers can be

given using the polar form of complex numbers. Indeed, if ZJ and Z2 are

non-zero complex numbers with polar forms

Z] = '/'1(cos 81 + i sin ()1) and Z2 = 1'2 (cos 82 + i sin 82),

then

ZJZ2 = 1'lr2(cos8] -I - isin8d(cos82 -I - isin82)

=1']1'2(( cos 8j cos 82 - sin e j sin 82) + i(sin 81 cos 82 -I - cos 81 sin 82))

= 'I'11'2(cos(8l +82) +'isin(8, +82)),

by usiug the formulas for the sine and cosine of the sum of two angles. The

above formula shows that I z , z 2 1 =1']1'2 = I Z 1 1 1 z 2 1 , which we knew already, and

also that the number e 'l -I - () 2 is an argument of ZIZ2. Thus we can describe the

effect of multiplying Z1 by Z2 (both non-zero) as follows.

The modulus of ZI Z2 is the modulus of Z] multiplied by the modulus of Z2;

all argument of ZI Z2 is an argument of Z1 plus an argument of Z2.

20

1+ 2i

Figure 2,15

-1- V 3i

Figure 2.16

-1+ V 3i

Figure 2.17

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Thus the geometric effect 011 ZI of multiplying it by Z2 is to scale it by the

factor IZ21and rotate it about ° through the angle Arg Z2. (This rotation is

anti clockwise if Arg Z2 > 0, clockwise if Arg Z2 < 0.) This is illustrated il l

Figure 2.18 for the case where Zl, Z2 and Z1Z2 are in the first quadrant, el, ()2

and e 1 + ()2 being their principal arguments. (Note that in Figure 2.18 we have

omitted the arrowheads from the arguments. Henceforth, this will be our usual

practice. )

Unfortunately, it is not always true that the principal argument ofZ1Z2

is thesum of the principal arguments of Z1 and Z2. Itmay differ from this SUlll by

± 2 7 T . For example, if Arg Z1 =7T /2 aud Arg Z2 =3 7 T / 4, then

Arg Z1+ Arg Z2= 5 7 T / 4 .

Thus 5 7 T / 4 is an argument of Z1Z2 but (because 5 7 T / 4 > 7 T ) it is not the principal

argument of ZIZ2. In fact, since -7 T < Arg(zlz2) :S 7 T ,

Arg(z1z2) = 5 7 T / 4 - 2 7 T = - 3 7 T / 4 .

On the other hand, if Arg Z1 = - 7 T / 4 and Arg Z2= - 7 7 T /8, then

Argz1 + Arg z-, = - 9 7 T / 8 ,

which is an argument of ZIZ2, but

In general, since - 2 7 T < Arg Z1 + Arg Z2 :S 2 7 T , we have the following property of

Argz.

If Z1 and Z2 are (non-zero) complex numbers, then

Arg(z1z2) = Arg ZI + Arg Z2 + 2n7T,

where n is -1, 0 or 1 according as Arg zj + Arg z-, is greater than 7 T , lies

in the interval j-or, 7T] , or is less than or equal to -7T.

P rob lem 2 .6 _

Use polar forms of the complex numbers

Z1 = -1- V3i, Z2 = 3V3 + 3i,

to evaluate Z1Z2 and z r , (You will find Example 2.2(b) and Problem 2.5(0)

useful. )

P rob lem 2 .7 _

Describe the geometric effect on a complex number z of multiplying Z by 2i.

As you might expect, the polar form of complex numbers is useful also for

division. Indeed, if Z1 and Z2 are non-zero complex numbers with polar forms

Z1 = 1'1 (cos ()l + isin ()1) and Z2= 7'2(cos e 2 + isin ()2),

then

Z1 1'1(COS()l+isin()r)

Z2 7'2(cos ()2 + i sin ()2)

~ (cosel + isin ()d(cos ()2 - isin()2)

1'2(cos (h + i Sill ()2) (cos ()2 - i sin ()2)

' rl (( ·os ()1cos ()2 + Sill( )[ Sill( )2)+ i(sin ()1cos ()2 - cos ()1 sin ()2))

7'2 cos? ()2 + Sil12 ()2T

= ____!:_(cos(()[(h) + iSill(()l - ( ) 2 ) ) , (2.1)1'2

using the formulas for the sine and cosine of the difference of two angles. This

formula shows that Izdz21 = 1'dT2 = IZ11/lz21and also that the number e [ - e 2

( ; 1 1 + ( ; 1 2

F i g ' I J . 1 · e 2.18

21

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is an arguurent of z I /22, Thus we call describe the effect of dividing non-zero

complex numbors as follows.

The modulus of ZI/Z2 is t.he modulus of ZI divided by the modulus of Z2;

an argurueut of ZI / Z2 is an argument of Zt minus an argument of Z2.

Thus t he geometric effect on ZI of dividing it hy Z2 is to scale it by the fa.ctor1/lz21 and rotate it about. 0 through the angle - Arg Z2. (This rotation is

clockwise if Arg z , > 0, aul.iclockwise if Argz2 < 0.)

Prob lem 2.8 _

Use pola.r forms of the complex numbers

ZI =1+ y'3'i, Z2 =h-,

to eva.luate ZJ/Z2.

P rob lem 2.9 _

Descrihe the geometric effect on a complex number z of dividing z by 2i.

An important special case of Formula (2.1) for the quotient zd Z2 is obtained

when

zl=1 and z2=r(cosB+isinB),

so that

In this case we find that

1(cos 0 + i sin 0)

T(COS 0 + i sin 0) r(cos 0 + isinO)

1= -(cos(O - 0) + isin(O - 0) )

r'

1= -(cos(-O)+isin(-O)).

T

Thus the reciproca.l of a. non-zero complex number z can be described as

follows.

The modulus of Z-1 is the reciprocal of the modulus of z ;

an argument of Z-1 is the negative of an argument of z.

Notice tha.t if z lies outside the circle of radius 1 centred at 0, so that Izl > I,then Z-1 lies inside this circle (because Iz-J I < I), as shown in Figure 2.19, and

vice versa. If Z lies on this circle, then z is of the form z = cos 0 + i sin 0 and

Z-1 = (cos 0 + i sin 0)-1

= cos( -8) + i sine -8),

so Z-l also lies on the circle (see Figure 2.20); moreover, in this case

z-1 = cos( -0) + i sine -8)

= cos 0 - i sin 0

= z.

In general, for all non-zero z,

-1 Z 1 _Z = zz = r ; p z ,

22

z

-1

-i

Figure 2.19- - - - - - - - - - - - - - - - - - -

z=ose + isine

-J

....__._--z-t

-, = cost-e) -\-'isin(-O)

Figure 2.20

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-

and so (since 1/lz12 is real)

Argz-1 = Arg z.

Also, if -7r < Arg Z < 7r, then

Arg z' =- Arg z,

since z is the reflection of Z in the real axis. Thus we have the following

properties of Arg z.

If Z is non-zero and -7r < Arg Z < rr , then

Argz =Arg z-1 =- Arg z.

P rob lem 2.10 _

Use a polar form of 1+ i to evaluate (1+ i)-I.

The product of several complex numbers Z1, Z2,'" ,Zn has a similar

interpretation.

The modulus of ZlZ2 Z" is the product of the moduli of Zl, Z2,··., Zn;

an argument of Z1 Z2 Zn is the sum of arguments of ZI, Z2, ... , Zn·

In other words, the product of the n complex numbers

Z k = T k ( C O S e k + isinek),

is given by

ZlZ2··· Z" =1'11'2··· rn(cos(e1

+ e2

+ ... + e n )+ isin(e1 + e2 + ... + en))·

k =1 ,2, . .. , n ,

P ro b lem 2 .1 1

Use polar forms of the complex numbers

Z1 = 1+ i, Z2 = 1+ V 3 i , Z3 = V 3 + i,

In the next subsection, polar form is used to calculate powers.

2 .4 de M oiv re 's Theo rem

An important special case of Formula (2.2) for the product Z1Z2 ... Zn is

obtained when

Z1 =Z2 =... =Zn =cos e + i sin e ,

so that

1'1=r2 = .. . =Tn =1 and e 1 =e 2 =... =e n = e .

In this case, Formula (2.2) becomes

(cos e + i sin e ) n =cos n e + i sin n e ,

this identity is due to Je Moivre.

n = 1 ,2, . . . ;

(2 .2)

Abraham de Moivre, who

worked mostly in England, was

an 18th century probabilist. Heknew this identity as early as

1707.

23

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T heo re m 2 .2 d e M o iv re 's T heo rem

Figure 2.2] shows the geometric interpretation of de Moivre's identity. The

powers of cos 0 + isin 0 are equally spaced around the circle with centre 0 and

radius 1 , t li e angle between adjacent powers being O . Each multiplication by

cos f) + i sin f) gives rise to a rotation through f) (radians) about O.

(ms{l I i si nO )"

_._-,-..._= cos 2 {1 + iin 2 0

cos {I + iin () cos {I + i sin ()

(cos 1 1 + is inO) I

= cos( -0) +i i n( - () )

Figure 2.21 Figure 2.22

In Figure 2.22, the position of (cos f) + i sin 0)-1 suggests that de Moivre's

identity holds also for negative integer powers; we now show that this is true.

If 11 , is an integer and 0 is a real number, then

(cos 0 + isin 0) " =cos nO + isin nO .

Proof We have already proved de Moivre's Theorem for a positive integer n,

and it is also true for:

11 , =0, since (cos 0 + i sin 0) ° = 1 =cos 0 + i sin 0,

andn = -I, since (cos8+isinO)-1 = cos(-8) +isin(-O).

To complete the proof, note that if m is a positive integer, then

(cos8 + i sin o)-m = ((cosO + i sine)-l)m

=cos( - e ) + isin( _ e ) ) 1 1 1

=cos( -me) + isin( -me).

Hence de Moivre's Theorem holds also if n=-m, where m is a positive

integer. •

P rob lem 2.12 _

Use de Moivre's Theorem to evaluate the following powers.

(a ) (V3+i)4 (b) (1-V3i)3 (c ) (l+i)lO

(d) (-1+i)-8 (e ) (V3+i)-6

24

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au argument of z2 is double an argument of z.

Since ihas modulus 1 and argument tt/2, one solution of z2 =iis obtained by

taking z to have modulus V I = 1 and argument ~ ( 1 f / 2 ) = 1 f / 4 (see Figure 3.1). Fiqure 3.1

This gives

3 SOLV ING EQUATION

NUMBER

W ITH COM PLEX


(a) calculate the nth roots of a complex number;

(b) solve certain polynomial eq uatious with complex coefficients.

As we explained in the Introduction, the use of complex numbers allows both

quadratic and cubic equations with real coefficients to be solved. You will see

in this course that complex numbers enable us to solve many equations which

may not have real solutions. In this section we describe various polynomial

equations whose complex solutions can be found explicitly.

3 .1 C a lcu la tin g nth roots

If a is a non-negative real uuiuber and n is a positive integer, then 0 / 0 . or al/ndenotes the non-negative nth root of a; that is, the unique non-negative

number x such that z" =a. In this subsection, we discuss the nth roots of a

complex number, beginning with square roots.

The simplest quadratic equation which has a complex solution but 110 real

solutions is

Z2 + 1 =0, that is, z2 =-l.

One solution to this equation is z =i, since i = -1; another sol ution is

z =-i, since (_i)2 = (_J)2i2 =-1.

A more general quadratic equation is

(3,1)

where w is a given complex number. Any solution z of Equation (3.1) is called

a square root of w; for example, both iand -i are square roots of -1.

In fact, we shall show shortly that each non-zero complex number w has

exactly two square roots. The following example shows [lOW to find square

roots geometrically.

E xam ple 3 .1

Find (the) two solutions of the equation

Z2 =i.

Solut ion

By the geometric properties of complex multiplication, described in

Subsection 2.3,

the modulus of z2 is the square of the modulus of z

and

z = 1 (ros ~ + iSill ~)4 /1

I 1.

= /2 + /2t.

Later we shall introduce v w or

W1/2 to denote a particular

square root of 11).

z1

IT/4

Check:

( ~ + ~ ir1 1 1. 1 .

=-+2--1--=12 J2 J2 2 25

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S· 1 J . . I· f·) . I (·l)? 1mce In+ In! IS a so utron 0 z: = Z ant - - = ,v2 v2

Z =_ ( _ 1 + _ 1 i ),j2 ,j2

is another solution of z2 =i. Therefore the required solutions are

( 1 1 )=± ,j2 + l2i ,

illustrated ill Figure 3.2. •

Remarks

1 No t i ce that. the second solution could also have been found geometrically. If

we had begun by taking i to have modulus 1and argument 7f/2 + 27f=57f/2,

then the corresponding solution z would have modulus Jl= ] and argument

t(57T-j2) = 57f/4 (see Figure 3.3). This gives

z = 1 (cos 5 4 7f'isin 5 : ) =- ~ - ~i.

2 Note that if 11) is any complex number and z is a square root of ui, then -z

is also a square root of w.

3 An alternative method of solving z2 =i is to write z =x + ' iV, equate the

rea.l pa.rts and imaginary parts of

( : r . : + i y ) 2 = x 2 - y2 + 2xyi = 'i,

and then solve the resulting equations for x and y (see Exercise 3.2). This

method is, however, not suitable for finding the nth roots of complex numbers

if n > 2.

P roblem 3.1 _

Find (the) two solutions of the equation

z2 =-1+ hi.

We now turn to the more general equation

where w is a given complex number and n is any positive integer with n 2' 2.

Each solution of z" =W is called an nth root of ' 1 1 1 . We shall show shortly that

each non-zero complex number 11 J has exactly n nth roots.

As a simple example, consider the equation

z3 = -8,

which has just one real solution, z = -2. To discover other solutions, we put-8 in pola.r form. Since

-8 = 8(cos7f+isin7f),

a solution of z3 = -8 is obtained by taking z to have modulus W =2 and

argument 7f/3. This gives

z =2(cos tt/3 + isin 7f/3) = 1 + hi.

Another polar form of -8 is

-8 = 8(cos(-7r) +isin(-7f)),

and so

z =2(cos( -7f /3) + i sin( -7f/3)) = 1 - hi

also sa.tisfies z3 = -8. Thus (the) three solutions of z3 =-8 are

z =-2, 1+hi, 1- hi.

26

»</'

/

I

/I

1 J.

.j2 + .j2!\\\

I/I/

-/"/

Eiqure 3.2

5rr/4

Figure 3.3

Recall that Zl = Z2 means that

Re a, =Rez2

and

Im zj = 1mZ2.

If 'W =0, then Z =0 is the only

solution.

For a 2 0, ija means the

non-negative cube root of a.

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-2(cos rr/3 +isin rr/3)= 1 -1 . J 3 i

Figure 3.4

Notice that these solutions all lie 011 the circle with centre 0 and radius 2 (see

Figure 3.4), and that the angle between adjacent solutions is 2 7 r /3. Thus these

three cube roots form the vertices of an equilateral triangle. This is a special

case of the following general result for nth roots.

T he orem 3 .1 Let

w =p ( cos if; +isin if;)

be a non-zero complex number in polar form. Then w has exactly n nth

roots, given by

Zk = pl/n ( cos (~ + k ~ ) +isin (~ + k : 2 :) ) ,k = 0, 1 , . . . , 7 1 , - 1.

These n nth roots form the vertices of an n-sided regular polygon

inscribed in the circle of radius pl/n centred at O .

Proof We seek the solutions of z" ='W in polar form, z =r(cos 0 +isin 0 ) .

Since w = p ( cos if; + isin i f ; ) , the equation z" =w takes the form

rn(cos B - \ - isine)" =p(cos i f; - \- isin i f ; ) ;

that is,

r" (cos n O + isin n O ) =p ( cos if; + isin i f ; ) ,

by de Moivre's Theorem.

Equating the moduli of both sides, and using the fact that the arguments of

the two sides differ by an integer multiple of 2 7 r , we obtain

TTl. = P and n O = if; + 2 k 7 r , where k E 7 L .

The only possible value of r is pi/n (since r must be lion-negative), and the

only possible values of 0 are

if; 2 7 r0=- + k-,

n nwhere k E 7 L .

Hence the solutions of z" = ware all of the form

( (if; 2 7 r ) (if; 2 7 r ) )

k =//n cos ;;;,+ k - - ; ; : +isin ;, + A :- - ; ; : , where A : E 7 L .

At first sight, it might appear that we have Iouud infinitely many solutions, one

for each value of k, However, not all these solutions are distinct. Indeed, if kl

and k2 differ by an integer multiple of n, say

where m E 7 L ,

then

(( P 2 7 r )- + kl - -\-2mn,

11 , n

We are using the Greek letters p

and t jJ here because 7' and e areneeded in the proof.

Sometimes the loose phrase

'equating moduli and

arguments' is used.

For a ::::0, al/n means the

non-negative nth root of a.

27

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( P ' ) 7 T ¢ ? 7 Tso that - -I J .: 2::_ and - + /':I::_ differ by an integer multiple of 2 7 T . Hence the

71 Il n nsolutions arising from 1 . : [ and 1 . : ' 2 are identical and so all possible solutions of

zi t = w arise from lite integers k = 0,1, ... , n - 1. These It solutions are clearly

distinct, since they lie on Ihe circle of radius (il/n centred at 0, with the angle

2 7 T I n between adjacent solutions, Thus they do form the vertices of a regular

l1-sicled polygon (Figure 3.5 illustrates this in the case n = 6). • z~

If 1lJ = p(cos ¢+ iin ¢), where ¢ is the pr incipal arqumetit of 'W, then

I/Il ( ¢ . . ¢ )o = p cos - 1 - 2 Sill -

11 nis called the principal nth root of 'W, denoted by l jUi or W

lln. Note that if w

is a positive real number, with principal argument ¢=0, then the principal

nth root of' 'W lias argument 0 and so is positive. Hence this LIse of the notation

tyW is consistent with the familiar real case. This consistency is taken further

because for 0 E IC , W or 0 lin is defined to be O .

A particularly important. case of Theorem 3.1 occurs when w =1, so that p =1

and ¢=o .

Corol lary The number 1has exactly n nth roots, given by

(2 7 T k ) ( 2 7 T k )

Zk = cos --;-;- + isin ---;;- ,

These are called the nth roots of unity.

k=0,1, ... ,n-1.

The nth roots of unity lie on the circle of radius 1 centred at 0, with the angle

2n-jn between adjacent roots. The cases n =2,3,4 are illustrated in Figure 3.6.

Zj

=-J

Z o = Io =1

71 = 2 71 =3 n=4

Figure 3.6 Roots of unity

E xample 3 .2

Determine the fourth roots of -8 + 8V3i in polar and Cartesian forms, plot

them in the complex plane, and indicate the principal fourth root.

Solution

Since 1-8 + 8V3il = J ( -8)2 + (8V3)2= ]6 and -8 + 8V3i has principal

8V3 7 r 27 rargument 7 r - tan-

1-8- = 7 r - '3 = 3' we deduce that

-8 + 8V3i =]6 (cos 237 r+ i sin 23

7 r) .

Hence, by Theorem 3.1 with p = 16 and ¢=2 7 r / 3 , the four fourth roots of

-8 + 8V3i are

/ ( ( 2 7 r /3 27r) ( 2 7 r /3 27r))ZI . : =16

4cos -4- + 1 .:4 + isin -4- + 1 .:4 .

= 2 (cos (~ + k ~ ) +'i sin (~ + k ~ ) ) , k =0,1,2,3.

28

Figure 3.5

In particular, the principal

square root of tv is denoted by

.jW or W1/2.

Note tha.t Zo = 1is the principa.l

nth root of unity for each n.

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Thus the arguments of the four fourth roots of -8 + 8 V 3 i are

i, i+ ~ = 2;, i+ 2 (~) = 767r, i+ 3 (~) = 5

37r,

and so the polar and Cartesian [arms of the fourth roots are as given below.

Zk Polar form Cartesian form

Zo 2(cos 7r/6 + i Sill 7r/6) V 3 +iZl 2(cos 27r/3 + i sin 27r/3) -1+ V 3 iZ2 2(cos 7 7 r /6 + i sin 7 7 r /6) - V 3 - iZ; J 2(cos 57r/3 + isiu Iitr /3) 1- V 3 i

The fourth roots are plotted in Figure 3.7.

Since the principal argument of - 8 + 8 V 3 i is 2n/3, its principal fourth root is

Zo = V 3 + i. •

The above solution illustrates the following strategy.

S tra te gy fo r fin d in g nth roots

To find the 71, nth roots, zo , Z1, ... , Zn-l, of a non-zero complex number w:

(a) express w in polar form, with modulus p and argument ¢;

(b) substitute the values of p and ¢ in the formula

Zk = pi/n (cos (~ + k 2:) + sin (~ + k ~~) ) ,

k=O,l, ... ,n-l;

(c) if required, convert the roots to Cartesian form.

Remarks

1 In step (a) you should normally choose if ) to be Argw (as in Example 3.2);

this has the advantage that the root Zo obtained in step (b) is the priucipal nthroot of w.

One disadvantage of this choice is the appearance of minus signs when Arg w is

nega.tive. This can be avoided by choosing ¢ to be (Arg w) + 2 7 r (> 0 ), but,

with this choice, Zo in step (b) will not be the principal nth root of w, which

has to be identified separately (see the solution to Problem 3.2(b)).

2 Where possible you should try to use the fact that the nth roots of w Iorui a

regular n-slded polygon to check your calculation of nth roots. For example, in

Example 3.2 note that

Zl = iz o , ·2Z2 = t Zo = -Zo and

corresponding to the fact that multiplying Z by i rotates Z about 0 through n/ 2

anti clockwise.

Prob lem 3.2 _

(a) Determine the cube roots of 8i in Cartesian form, plot them in the complex

plane, and indicate the principal cube root.

(b) Determine the sixth roots of -iill polar form, plot them ill the complex

plane, and indicate the principal sixth root.

Prob lem 3 .3 _

(a) Use the Geometric Series Identity to prove that if Z is an nth root of unity

( 7 1 , ~ 2) and z I- I, then

1 + Z + Z2 + ... + z,,-l =O .

(b) Deduce [Torn part (a) that the n ntb roots of unity have sum O.

Figure 3.7

29

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3 .2 S olutions of po lynom ia l equationsThe quadratic equation

az2 + bz + c =0,

where a, b , care complex numbers and a = I- 0, can be solved by the methods

which are available ill the real case. For example, we may be able to factorize

the quadratic expression, as in the following cases:

Z2 + 9 = (z - 3'i)(z + 3i) = 0,

Z2 + (1 - i)z - i = (z + 1 )(z - i) =0 ,

so that. z =±3i;

so that z = -l,i.

If there is lIO easy factorization, then the formula

-b ± Jb2 - <lac(3.2)=

2a

can be used. The justification of this formula (hy completing the square and

rearranging) is identical to the real case.

P rob lem 3.4 _

Solve the following equations.

(a) z2-Tiz+8=0 (b) z2+2z+1-i=0

In the previous subsection we saw bow to find the n solutions of the equation

However, it is only in exceptional cases that we can find an explicit algebraic

solution of the polynomial equation (of degree n)

anz'" + an_lz.".-l + ... + a"lz + ao= 0,

where ao, al, ... , an are complex numbers and an = I- O . For example, it may be

possible to reduce a given polynomial equation to a quadratic equation by

making a . substitution (as in the next example).

E xample 3 .3

Solve the equation

z4 + 4z2 + 8 = o .

Solut ion

Substituting 1V = z2 gives

w2 + < lw + 8 = 0,

which has solutions

-4± J16 - 32w = =-2 ± 2i

2 .

Thus z = ±J-2 + 2i or z = ±J-2 - 2i. Since

-2+2i =v'8(cos311"/4+isin311"/4),

we have

J -2 + 2·i= 81/4( cos 311"/8+ i sin 311"/8);

thus two solutions of Z4 + 4z2 + 8 =0 are ±81/4( cos 3 7 r /8 + isin 311"/8).

Similarly, since

-2 - 2i =v'8 (cos (-311"/4) + isin (-311"/4)) ,

we have

J -2 - 2i = 81/4 (cos( -311"/8) + i sin( -311"/8)) ;

thus two further solutions are ±81/4 ( C O S ( -311"/8) + isin( -311"/8)).

30

Since 371-j 4 is the principal

argument of -2 + 2i, this is the

principal square root of -2 + 2i.

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-So the four solutions are

Remark Since cos(-371"/8)=cos 371"/8and sin( -371"/8) = - sin 371"/8,the foursolutions in Example 3.3 form two complex conjugate pairs. Itcall be shown

that non-real roots of a polynomial equation with real coefficients must occur

in complex conjugate pairs; see Exercise 3.4.

P rob lem 3 .5 _

(a) Solve the equation

z6 - 7 iz3 + 8 =O .

(Hint: Use Problems 3.2(a) and 3.4(a). Also, you may find the following

fact useful: if z is a cube root of 8i, then - 4 z is a cube root of -i.)(b) Solve the equation

z4 + 4 iz2 + 8 =O .

4 SETS OF COMPLEX NUMBERS


(a) understand the meanings of inequalities between real expressions involving

complex numbers;

(b) understand the specification of subsets of the complex plane in terms of

such inequalities;

(c) recognize certain basic open and closed sets.

4 .1 In eq ua litie sThroughout the course we shall use many inequalities involving complex

numbers, and you will need to become adept at interpreting them. Here are

some simple inequalities involving a complex number z and SOlUeexamples of

values of z for which they are true (.() or fa.lse (x).

l+i 2-i _1 + 1i -1- 3i2 2

Rez> 1 x .( x x

Izl S; 1 x x .( X

I Il ll z l > 2 x x x .(

Arg z < 7f/2 .( .( x .(

Notice that these inequalities are all between expressions which are real-valued.

We never write inequalities between complex-valued expressions such as 2+ i

or z2 + l.

The inequalities

Zl < Z2 and ZJ S; Z2

have no meaning unless both Z1 and Z2 are real. IR is all ordered field , bu Leis

not.

31

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P rob lem 4 .1 _

Complete the following true/false table.

1+ 2i -1-2i I -2

Rez < 0

Iz l > ~

Irn z S -1Arg z ;: ::: ()

4.2 S ketch ing subsets o f the com plex p lane

{audio- tape}

In real analysis we often lise intervals, and these are defined by using

inequalities. For example, the open interval with endpoints 1, 3 is

] 1, 3 [ ={ :r; : 1 < x < :l }

and the closed interval with endpoints -2, 2 is

[-2,2] = {x : -2 ~.'1; S 2}.

An interval such as

] - 7 r , 7 r ] = {x : +: < X S 7 r }

is called half-open (or half-closed), and it is often convenient to use unbounded

open and closed intervals, such a.s

]O,oo[ = {:r: x> O} (open)

a.nd

[l,oo[ = { : r ; : x ; : : : : 1 } (closed) .

In the audio-tape section, which follows, a similar method is used to define

va.rious subsets of the complex plane.

NOW START TIlE TAPE.

32

Some texts use 'round brackets'

for open intervals.

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35

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36

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37

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In the audio-tape frames we used various conventions for sketching subsets of C:

• the interior of a set is shown by 'tone';

• boundary curves which belong to the set are drawn unbroken;

• boundary curves which do not belong to t he set are drawn broken;

• isolated boundary points which belong to the set are drawn as filled-in

circles;

• isolated boundary points which do not belong to the set are drawn as

empty circles.

These conventions will remain in force throughout the course, although we

shall not always include the tone. When sketching sets by hand you should

replace tone by hatching.

P rob lem 4.7 _

Sketch the following sets, using the conventions listed above.

(a ) {z: lm z > O}

(b) {z:lz+ll~l}

(c ) {z: 0 < 1 z + 1 + 2·i 1 < l}

(d ) {z: 1 Arg ( z + . L - ' i ) 1 < 7r/3}

(e) {z:lz-ll~lz-21}

(Hint: Interpret the inequality in terms of distances.)

(f ) C - {z : Re z ; : : > I}

(g ) {z:IHlz>Ol-{z:lz+ll~l}

(h ) {z: Arg z =7 r/6 } U {z: A rg (z - J 3 - i) =O }

( i) { z : Arg z = 7r/6} n { z : A rg (z - J 3 - 'i ) = O }

5 PRO VING INEQUAL IT IES

After working through this section, yon should be able to:

(a) use the rules for rearranging inequalities and the rules for obtaining new

inequalities from old ones;

(b) prove inequalities involving the moduli of complex numbers by using

various forms of the Triangle Inequality.

5 .1 R ules fo r rearrang ing ineq ua litiesIn Section 4 we used equalities and inequalities to define subsets of the complex

plane. In this section we show you how to prove new inequalities by deducing

them from simpler known inequalities (such as Iz l ~ 0, which holds for all z )

using various rules. We begin by reminding you of the rules for rearranging a

given inequality into an equivalent form; such equivalent inequa.lities are linked

by the symbol '~', which may be read as 'is equivalent to' or 'if and only if'.

38

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-

Ru le s f or r ea rr ang in g in e q uali tie s

For all a, b , e ill ~, the following rules apply.

Rule 1 a<b ¢=} b - a > 0.

Rule 2 a.<b ¢=} 0.+ e < b + c.

Rule 3 If e > 0, then a<b ¢=} ae < be.If e < 0, then a<b ¢=} ae> be.

Rule 4 If a, b > 0, then a<b1 1

¢=} - >-.a b

Rule 5 If a, b ? ': ° and ]J > 0, then

Rule 6 1 0 . 1 < b ¢=} -b < a < b.

a < b ¢=} {L P < bP.

There are corresponding versions of Rules 1-6 in which the strict inequality '<'is replaced by the weak inequality t = ; ' .

The next two rules can be used to deduce new inequalities from given ones.

Here, however, the new inequalities are not equivalent to the old ones, since theold inequalities cannot be deduced from the new ones. Such deductions are

written using the symbol '=>', which may be read as 'implies'.

T r an s itiv e Ru le

For all a, b , e in ~,

a < band b < e => a < e.

Comb in a tio n R u le sFor all a, b , e, d in ~, if a < band e < d , then

Sum Rule a + e < b + d,

ae < bd (provided that a, e ~ 0 ).roduct Rule

There are also weak ami weak/strict versions of the Transitive Rule and the

Combination Rules, which you should be able to work out as they arise.

The following example illustrates how the various rules are used in practice.

E xample 5 .1Prove that

27,2> (1'+1)2,

Solut ion

for r ? ': 3.

We rearrange the given inequality ill order to find an equivalent, but simpler

one:

(1'+1)2

2> --T

1h> 1+-

T

1

h-1> -r (Rule 2)

(Rule 3)

(Rule 5 )

1r > -- = v '2 + 1

J 2 -1(R\lle 4).

Examples

x<3¢=>3-x>O

x<3¢=>~;-1<3-1

2< x ¢=> 1< x/2

2< 3¢=> -2 > -3

2<3¢=>~>1

2<3¢=>V2<V3

1-21< 3¢=> -3 < -2 < 3

For example, if x < 2, then

x < 3 (because 2 < 3).

For example, if n < 5, then as

2 < 3,

n+2<5+:3=8

and

2n < 3 x 5 = 15.

III future we shall not usually

indicate which rule for

rearranging a given inequality is

being used.

39

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Since V 2 +1 = 2.L1 I< J .. . , t he filial inequality is true for T : : : : 3 (by the

Transitive Rule: ,. ::: :: l a nd 3 > V 2 + 1 =* r > J2 + 1 ). Hence l .he first

inequality must be 'me fotr :::;)also. •

Remar k Example 5.1 could he solved, a l terna t ive ly , by using Rule 1 to obtain

r2 - 2r - I > 0 and then completing the square. There is often more tha n one

way to deal with a given inequality.

P rob lem 5 .1 _

Prove that

3 1 '-~-- < I,r: + 2

for T > 2.

5 .2 T he T rian gle In eq ua lityMany inequalities have a geometric interpretation. For example, the two

inequali ties

1:1:1< J x 2 + y 2 and Iyl < J x 2 + y 2

state that, in a right-angled triangle, the hypotenuse is the longest side.

can be written in complex form as

Theyz

I

IIz l

I lm z l )

(5.1)IIRezi ~ Izi and [Im z] ~ Izl

(see Figure 5.1) or, equivalently, as

-Izl ~ Rc z ~ Izl and -izi ~ Im z ~ 14 (5.2)

Another elementary fact from plane geometry is that the length of any side of a

triangle is less than or equal to the sum of the lengths of the other two sides. If

Zl, Z 2 are complex numbers, then 0, Zl and Z l + Z 2 form the vertices of a

triangle (see Figure 5 .2) with sides IZII, IZ21and IZI + z 2 1 , and so

IZI+ z 2 1 ~ IZII+ I Z 2 1 ·

This is one form of an inequality called the Triangle Inequality, which will be

used frequently throughout the course.

T he orem 5 .1 T ria ng le In eq ua lity

If Zl, Z 2 E C, then

(a) IZJ+ z 2 1 ~ IZll + hi

(b) IZI - z21 ::: Ilzll- IZ211

(usual form);

(backwards form).

40

Eiqure 5.1

Figure 5.2

Part (b) gives

I Z l - z 2 1 2 I Z l l - l z 2 1and

I Z l - z 2 1 2 I Z 2 1 - I Z ll ·

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The backwards form of the Triangle Inequality also has a useful geometricinterpretation, concerning the two circles centred at 0 through Z1 and Z2. It

says that the distance from Zl to Z2 is at least as large as the difference

between the radii of these circles, as shown in Figure 5.3 for the case IZ11> IZ21. Figure 5.3

Several other versions of the Triangle Inequality are given in the following

corollary. Each is a variant of one of the forms of the Triangle Inequality.

Proof Although part (a) follows from plane geometry, we give a proof using

complex numbers which illustrates the use of several results from this unit. An

alternative proof using polar form is given in Exercise 5.4.

We have

IZ1 + z212

= (Z1 + Z2)(Zl + Z2)

= (Z1 + Z2)(Z1 + Z2)

(Theorem 2.1(c))

(Theorem l.1(b)(i))

= Z1Z1 + Z1Z2 + Z2Z1 + Z2Z2

= IZ112+ Z1Z2 + Z1Z2 + IZ212

(Theorem 2.1(c); Theorem 1.1, parts (b) (iii) and (a)(iii))

= hl2 + 2Re(zlz2) + IZ212 (Theorem l.1(a)(i))

~ IZ112+ 21z1z21+ IZ212 (by Equation (5.1))

= IZ112+ 2hllz21 + IZ212 (since IZ1z21= IZ111z21= IZ111z21)

= (lz11 + IZ21)2,

and so part (a) follows.

Part (b) can be proved by a similar method; alternatively, note that

IZ11= IZ1- Z2 + z21

:::::: IZ1 - z21 + IZ21 (by part (a)),

so that

IZ1 - z21 2 : IZ11- IZ21·

Similarly,

IZ2 - zll 2 : IZ21-lz11,

but Iz z - z11 = IZ1 - z21, so that

IZ1 - z21 2 : IZ21-lz11·

Part (b) follows [rom inequalities (5.3) and (5.4). •

(5.3)

(5.4)

Corol lary If Z, Z1, Z2, ... , Zn E ! C , then

(a) I~I::::::IRe zl + [Im z];

(b) IZ1 - z21 ::::::Z11+ IZ21;

(c) IZ1 + z212: Ilz11-lz211;

(d) IZ1 ± Z2 ± ± znl ::::::IZ11+ IZ21+ + Iz"l;

(e) IZ1 ± Z2 ± ± z,,1 2 : IZ11- IZ21- -Iznl·

Proof Part (a) is obtained by taking Z1 = Re z and Z2 = ilm z in the usual

form of the Triangle Inequality.

Parts (b) and (c) are obtained by substituting -Z2 for Z2 in Theorem 5.1.

Parts (d) and (e) are obtained from Theorem 5.1 and parts (b) and (c) of this

corollary by applying the Principle of Mathematical Induction - we omit the

details. •

Parts (a), (b) and (d) are

variants of the usual form of the

Triangle Inequality, whereas

parts (c) and (e) are variants of

the backwards form.

41

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The Triangle Inequality can be used to obtain estimates for the modulus of a

complex expression involving z when we know that z lies in a certain set (such

as a circle). Here are some typical applicat.ions.

E xam ple 5 .2

(a) Prove that

(i) Iz2 - 4z - 31 ::; 15, for Izl =2;(ii) Iz2 - 712:: 3, for Izl =2;

(iii) Iz2 + 21 2::2, for Izl =2.

(b) Find a number 1 1 1 such that

I ( z : ~ ~ ; ( : ; : 2) I < M, for [z ] =2.

Solut ion

(a) (i) By the Triangle Inequality,

Iz2 - 4z - 31 ::; Iz21+ 1-4zl + 1-31

=zl2 + 41z1+ 3;

so, for Izl = 2,

Iz2 - 4z - 31::; 4 + 8+ 3 =15.

(ii) By the backwards form of the Triangle Inequality,

Iz2 - 71 2:: IIzl2 - 71 ;

so, for Izi =2,

Iz2 - 71 2: 14- 71=3.

(iii) By the backwards form of the Triangle Inequality,

Iz2 + 212:: IIzl2 - 21;

so, for Izl = 2,

Iz2 + 21 2:: 14 - 21=2.

(b) From part (a) we have, for Izl =2,

Iz2 - 4z - 31 ::; 15, Iz2 - 71 2::3, Iz2 + 21 2: 2. (5.5)

Now

Iz2 - 4z - 3 I

( Z 2 - 7) (z2 + 2)

Iz2 - 4z - 31

Iz2 - 71 X Iz2 + 21

1

21= z - 4z - 31 X Iz2 _ 71

So, for Izi =2, using Inequalities (5.5), we have

Iz2 - 4z - 3 I 1 1 5< 15 x - x - =-

( z 2 - 7) (z2 + 2) - 3 2 2'

because

Iz2 - 71 2: 3 ==?

and Iz2 + 21 2::2 ==?

1/lz2 - 71 ::; 1/3

1/lz2 + 21::; 1/2.

Thus we can take M =5/2. •

Remarks

1 Example 5.2(b) illustrates the fact that to obtain an upper estimate for a

quotient, we need an upper estimate (15) for the numerator and a lower

estimate (3 x 2) for the denominator.

42

As indicated in these solutions

it is not usual to refer to any of

the variants (in the corollary) of

the Triangle Inequality.

However, use of the backwardsform should be distinguished.

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2 The inequality

IZ2 + 2 1 ~ 2 , for Izl = 2,

is said to be 'best possible' because it holds with equality if z = 2i or -2i.

However, the inequality

IZ2 - 4z - 3 1 : : : ; 15, for Izl = 2,

is not 'best possible'. With more work it is possible to prove the best possibleinequality

IZ2 - 4z - 3 1 : : : ; 7.j7f3 (= 10.69 ... ), for Izl = 2.

P ro b lem 5 .2

Prove that

(a)~<ll 1<1 forlzl=l',7 - 3 + 4Z2 - ,

1

z3 + 2z + 1 1 17(b) 2 :::; z2 + 1 :::;4' for Izl = 3.

EXERC ISES

Section 1

Exe rc is e 1 .1 Complete the following table.

z -zm zez

2 + 3i

-3 - i

4i

5

o

Exe rc is e 1 .2 Express each of the following complex numbers in Cartesianform.

(c) (l+i)2 (d) (1- i)21

(e) - .1- t

(i ) 3 + 5i2 - 3i

l+i(f) 1_ i

(j) 3+ 2i1+ 4i

(m)l-i+i2_···+ilO

(g ) (1 + i)4 (h) (2 + i)2 - (2 - i) 2

(k) (3 + 4i)4 - (3 - 4i)4 (I) 1+ i + i2 + ... + i10

Exe rc is e 1 .3 Write down the real part, imaginary part and complex conjugate

of the complex numbers in parts (a), (e), (g) of Exercise 1.2.

Exe rc is e 1 .4 Prove that Irn Z = - Im z.

[(2i)2 + 2[ = [- 4 + 2[ = 2

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(a ) 5 (b ) ~

(f) -J3 - i

(c ) -3i

(g ) 3 + 4i

(d) 2 + 2i

(h) 3-4i

(e) -2 + 2i

S ection 2

Ex er cis e 2 .1 Plot each of the following complex numbers, and express each

one in polar form, using the principal argument in each case.

Ex er cis e 2 .2 Plot each of the following complex numbers, and express each

one in Cartesian form.

(a) cos11"+isin11" (b) 4(cos(-11"/2)+isin(-11"/2))

(c) 3(cos 311"/4+ i sin 311"/4) (d) 3( cos 11"/6+ i sin 11"/6)

(e) cos( -211"/3) + 'I sin( -211"/3)

Ex er cis e 2 .3 Find the distance from Z2 to Zl in each of the following cases.

(a ) Zl =1+ i, Z2 =2+ 3i

(b) Zl = -2 + 3i, Z2 =1- 7i

(c ) Zl = i, Z2 = -i

Ex er cis e 2 .4 Use polar form and de Moivre's Theorem to evaluate the

following expressions, giving your answers in Cartesian form.

(1+ i)6(a ) (1+ J3i)5 (b) (1+ i)-4 (c ) (V3 _ i)3

Ex er cis e 2 .5 Use de Moivre's Theorem and the Binomial Theorem to prove

that

sin 38 =3 sin 8 - 4 sin'' 8,

where 8 is a real number.

Ex er cis e 2 .6 Prove that if (x + iy)4 =a + ib, where x + iy and a + ib are in

Cartesian form, then

(x 2 + y2)4 = a2 + b2 .

Ex er cis e 2 .7 Prove that if z =Z-l, then I z i =1.

S ec tion 3

Ex er cis e 3 .1 For each of the following complex numbers determine, in

Cartesian form where convenient, the nth roots indicated, and plot them. In

each case, identify the principal nth root.

(a) The square roots of: (i) -i; (ii) 4i.

(b) The cube roots of: (i) -1; (ii) -2 + 2i.

(c) The fourth roots of: (i) V ; (-1 - i); (ii) -1 + i.

(d) The fifth roots of: (i) -1; (ii) -16 + 16V3i.

Ex er cis e 3 .2 Use the method of equating real parts and imaginary parts to

solve each of the following equations.

(a) (x + iy)2 = 3 + 4i (b) (x + iy)2 = -5 + 12'1

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Exe rc is e 3 .3 Solve each of the following equations, and plot their solutions.

(a) z4-z2+1+i=0 (b) z3-4z2+6z-4=0

Exe rc is e 3 .4 Let p(z) =an zn + an_1Z n -1 + ... + at Z + ao , where

ao , al,· .. ,a n are real numbers. Prove that if z satisfies p(z) =0, then p(z) =O.

(This shows that non-real roots of a polynomial equation with real coefficients

must occur in complex conjugate pairs.)

Sect ion 4

Exe rc is e 4 .1 Draw a diagram of each of the following sets of points in the

complex plane. In each case indicate which points of the boundary belong to

the set, and which points do not.

(a) {z: IRezl < 1, [Im z] < I}

(b) { z : [z - il ::::;2, Izl 2 : I}

( c) {z: Re z + 2 Im z + 3 > O }

(d) {z: Rez 2 : o} U {z: Imz > o}

(e) { z : Izl > 1,1 Arg z] ::::;7r/4}

(f ) { z : -7r < Arg(z + 2)} n { z : Arg(z + 2) < 7r/2}

(g) {z : 1 z + 1+ 2i I < I}

(h) {z: Rez > 1, [z - il < 2}

(i) {z : Iz + il < Iz + 2il}

(j ) {z:Re(:~~) : : : : ; o }

(k) {z: Izl < 3} - {z : Izi ::::;2}

(1 ) C - {z: z2 + z - 2 =o}

S ection 5

Exe rc is e 5 .1 For Izi =2, find an upper estimate for each of the following

moduli.

(a ) [z + 31

(d) 13 z2

- 5 1

(b) [z - 4il (c) 13z+ 21

(e) Iz2 + z + 11

Exe rc is e 5 .2 For Izi =5, find a positive lower estimate for each of the

following moduli.

(a) Iz - 21 (b) Iz + 3il (c) Iz - 71 (d) 12z - 71

Exe rc is e 5 .3 Find (positive) numbers m and M such that

for Izl = 4.

Exe rc is e 5 .4 Prove the usual form of the Triangle Inequality:

if Z l ,Z 2 E C, then IZI + z21 ::::;IZII + IZ21,

m337 1 Workbook

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