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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
7
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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.
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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
After working through this section, you should be able to:
(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.
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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 . .
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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' .
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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
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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.
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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
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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
After working through this section, you should be able to:
(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.
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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
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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
After working through this section, you should be able to:
(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.
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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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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.
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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.
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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,