Chapter
COMPLEX NUMBERS AND
W. R. Hamilton
(1805-1865)
vMathematics is the Queen of Sciences and Arithmetic is the Queen of
Mathematics. – GAUSS v
5.1 Introduction
In earlier classes, we have studied linear equations in one
and two variables and quadratic equations in one variable.
We have seen that the equation x
2
+ 1 = 0 has no real
solution as x
2
+ 1 = 0 gives x
2
= – 1 and square of every
real number is non-negative. So, we need to extend the
real number system to a larger system so that we can
find the solution of the equation x
2
= – 1. In fact, the main
objective is to solve the equation ax
2
+ bx + c = 0, where
D = b
2
– 4ac < 0, which is not possible in the system of
real numbers.
5.2 Complex Numbers
Let us denote
1
by the symbol i. Then, we have
2
1
i
= −
. This means that i is a
solution of the equation x
2
+ 1 = 0.
A number of the form a + ib, where a and b are real numbers, is defined to be a
complex number. For example, 2 + i3, (– 1) +
3
i
,
1
4
i
+
are complex numbers.
For the complex number z = a + ib, a is called the real part, denoted by Re z and
b is called the imaginary part denoted by Im z of the complex number z. For example,
if z = 2 + i5, then Re z = 2 and Im z = 5.
Two complex numbers z
1
= a + ib and z
2
= c + id are equal if a = c and b = d.
5
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98 MATHEMATICS
Example 1 If 4x +
i(3xy) = 3 + i (– 6), where
x and y are real numbers, then find
the values of x and y.
Solution
We have
4x + i (3xy) = 3 +
i (–6) ... (1)
Equating the real and the imaginary parts of (1), we get
4x = 3, 3
xy = – 6,
which, on solving simultaneously, give
3
4
x
=
and
33
4
y =
.
5.3 Algebra of Complex Numbers
In this Section, we shall develop the algebra of complex numbers.
5.3.1 Addition of two complex numbers Let z
1
= a + ib and z
2
= c + id be any two
complex numbers. Then, the sum z
1
+ z
2
is defined as follows:
z
1
+ z
2
= (a + c) + i (b + d), which is again a complex number.
For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8
The addition of complex numbers satisfy the following properties:
(i) The closure law The sum of two complex numbers is a complex
number, i.e., z
1
+ z
2
is a complex number for all complex numbers
z
1
and z
2
.
(ii) The commutative law For any two complex numbers z
1
and z
2
,
z
1
+ z
2
= z
2
+ z
1
(iii) The associative law For any three complex numbers z
1
, z
2
, z
3
,
(z
1
+ z
2
) + z
3
= z
1
+ (z
2
+ z
3
).
(iv) The existence of additive identity There exists the complex number
0 + i 0 (denoted as 0), called the additive identity or the zero complex
number, such that, for every complex number z, z + 0 = z.
(v) The existence of additive inverse To every complex number
z = a + ib, we have the complex number – a + i(– b) (denoted as –
z),
called the additive inverse or negative of z. We observe that z + (–z) = 0
5.3.2 Difference of two complex numbers Given any two complex numbers z
1
and
z
2
, the difference z
1
z
2
is defined as follows:
z
1
z
2
= z
1
+ (– z
2
).
For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4
i
and (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99
5.3.3 Multiplication of two complex numbers Let z
1
= a +
ib and z
2
= c + id be any
two complex numbers. Then, the product z
1
z
2
is defined as follows:
z
1
z
2
= (ac
For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28
The multiplication of complex numbers possesses the following properties, which
we state without proofs.
(i) The closure law The product of two complex numbers is a complex number
,
the product z
1
z
2
is a complex number for all complex numbers z
1
and z
2
.
(ii) The commutative law For any two complex numbers z
1
and z
2
,
z
1
z
2
= z
2
z
1
.
(iii) The associative law For any three complex numbers z
1
, z
2
, z
3
,
(z
1
z
2
) z
3
= z
1
(z
2
z
3
).
(iv) The existence of multiplicative identity There exists the complex number
1 +
i 0 (denoted as 1), called the multiplicative identity such that z.1 =
z,
for every complex number z.
(v) The existence of multiplicative inverse For every non-zero complex
number z = a + ib or a + bi(a 0, b 0), we have the complex number
2 2 2 2
a b
i
a b a b
+
+ +
(denoted by
1
z
or z
–1
), called the multiplicative inverse
of z such that
1
1
z.
z
=
(the multiplicative identity).
(vi) The distributive law For any three complex numbers z
1
, z
2
, z
3
,
(a) z
1
(z
2
+ z
3
) = z
1
z
2
+ z
1
z
3
(b) (z
1
+ z
2
) z
3
= z
1
z
3
+ z
2
z
3
5.3.4 Division of two complex numbers Given any two complex numbers z
1
and z
2
,
where
2
0
z
, the quotient
1
2
z
z
is defined by
1
1
2 2
1
z
z
z z
=
For example, let z
1
= 6 + 3i and z
2
= 2 – i
Then
1
2
1
(6 3 )
2
z
i
z i
= + ×
=
(
)
6 3
i
+
( )
( )
( )
2 2
2 2
1
2
2 1 2 1
i
+
+ +
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100 MATHEMATICS
=
( )
2
6 3
5
i
i
+
=
( ) (
)
1 1
12 3 6 6 9 12
5 5
i i
+ + = +
5.3.5 Power of i we know that
(
)
3 2
1
i i i i i
= = = −
,
(
)
( )
2
2
4 2
1 1
i i
= = =
(
)
( )
2
2
5 2
1
i i i i i
= = =
,
(
)
( )
3
3
6 2
1 1
i i
= = = −
, etc.
Also, we have
1 2
2
1 1 1
, 1,
1 1
i i
i i i
i i
i
= × = = − = = = −
3 4
3 4
1 1 1 1
, 1
1 1
i i
i i i
i i
i i
= = × = = = = =
In general, for any integer k, i
4k
= 1, i
4k + 1
= i, i
4k + 2
= –1, i
4k + 3
= – i
5.3.6 The square roots of a negative real number
Note that i
2
= –1 and ( – i)
2
= i
2
= – 1
Therefore, the square roots of – 1 are i, – i. However, by the symbol
1
, we would
mean i only.
Now, we can see that i and –i both are the solutions of the equation x
2
+ 1 = 0 or
x
2
= –1.
Similarly
(
)
(
)
2 2
3 3
i =
i
2
= 3 (– 1) = – 3
(
)
2
3
i
=
(
)
2
3
i
2
= – 3
Therefore, the square roots of –3 are
3
i
and
3
i
.
Again, the symbol
3
is meant to represent
3
i
only, i.e.,
3
=
3
i
.
Generally, if a is a positive real number,
a
=
1
a
=
a i
,
a b
×
=
ab
for all positive real number a and b. This
result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0,
b < 0?
Let us examine.
Note that
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101
( ) ( )
2
1 1 1 1
i
= =
(by assuming
a b
×
=
ab
for all real numbers)
=
1
= 1, which is a contradiction to the fact that
=
2
1
i
.
Therefore,
a b ab
×
if both a and b are negative real numbers.
Further, if any of a and b is zero, then, clearly,
a b ab
× =
= 0.
5.3.7 Identities We prove the following identity
( )
2
2 2
1 2 1 2 1 2
2
z z z z z z
+ = + +
, for all complex numbers z
1
and z
2
.
Proof We have, (z
1
+ z
2
)
2
= (z
1
+ z
2
) (z
1
+
z
2
),
= (z
1
+ z
2
) z
1
+ (z
1
+ z
2
) z
2
(Distributive law)
=
2 2
1 2 1 1 2 2
z z z z z z
+ + +
(Distributive law)
=
2 2
1 1 2 1 2 2
z z z z z z
+ + +
(Commutative law of multiplication)
=
2 2
1 1 2 2
2
z z z z
+ +
Similarly, we can prove the following identities:
(i)
( )
2
2 2
1 2 1 1 2 2
2
z z z z z z
= +
(ii)
( )
3
3 2 2 3
1 2 1 1 2 1 2 2
3 3
z z z z z z z z
+ = + + +
(iii)
( )
3
3 2 2 3
1 2 1 1 2 1 2 2
3 3
z z z z z z z z
= +
(iv)
(
)
(
)
2 2
1 2 1 2 1 2
z z z z z z
= +
In fact, many other identities which are true for all real numbers, can be proved
to be true for all complex numbers.
Example 2 Express the following in the form of a + bi:
(i)
( )
1
5
8
i i
(ii)
(
)
(
)
2
i i
3
1
8
i
Solution (i)
( )
1
5
8
i i
=
2
5
8
i
=
(
)
5
1
8
=
5
8
=
5
0
8
i
+
(ii)
( ) ( )
3
1
2
8
i i i
=
5
1
2
8 8 8
i
× ×
× ×
=
( )
2
2
1
256
i
1
256
i i
=
.
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102 MATHEMATICS
Example 3 Express (5 – 3
i)
3
in the form a + ib.
Solution We have, (5 – 3i)
3
= 5
3
– 3 × 5
2
× (3i) + 3 × 5 (3i)
2
– (3i)
3
= 125 – 225
i – 135 + 27
i = – 10 – 198i.
Example 4
Express
(
)
(
)
3 2 2 3
i
+
in the form of a + ib
Solution We have,
(
)
(
)
3 2 2 3
i
+
=
(
)
(
)
3 2 2 3
i i
+
=
2
6 3 2 6 2
i i i
+ +
=
(
)
(
)
6 2 3 1 2 2
i
+ + +
5.4 The Modulus and the Conjugate of a Complex Number
Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined
to be the non-negative real number
2 2
a b
+
, i.e., | z | =
2 2
a b
+
and the conjugate
of z, denoted as
z
, is the complex number a ib, i.e.,
z
= aib.
For example,
2 2
3 3 1 10
i+ = + =
,
2 2
2 5 2 ( 5) 29
i = + =
,
and
3 3
i i
+ =
,
2 5 2 5
i i
= +
,
3 5
i
= 3i – 5
Observe that the multiplicative inverse of the non-zero complex number z is
given by
z
–1
=
1
a ib
+
=
2 2 2 2
a b
i
a b a b
+
+ +
=
2 2
a ib
a b
+
=
2
z
z
or z
2
z z
=
Furthermore, the following results can easily be derived.
For any two compex numbers z
1
and z
2
, we have
(i)
1 2 1 2
z z z z
=
(ii)
1
1
2 2
z
z
z z
=
provided
2
0
z
(iii)
1 2 1 2
z z z z
=
(iv)
1 2 1 2
z z z z
± = ±
(v)
1 1
2 2
z z
z z
=
provided z
2
0.
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 103
Example 5 Find the multiplicative inverse of 2 – 3i.
Solution Let z = 2 – 3
i
Then
z
= 2 + 3i and
2
2 2
2 ( 3) 13
z
= + =
Therefore, the multiplicative inverse of
2 3
i
is given by
z
–1
2
2 3 2 3
13 13 13
z i
i
z
+
= = = +
The above working can be reproduced in the following manner also,
z
–1
=
1 2 3
2 3 (2 3 )(2 3 )
i
i i i
+
=
+
=
2 2
2 3 2 3 2 3
13 13 13
2 (3 )
i i
i
i
+ +
= = +
Example 6 Express the following in the form a + ib
(i)
5 2
1 2
i
i
+
(ii) i
–35
Solution (i) We have,
5 2 5 2 1 2
1 2 1 2 1 2
i i i
i i i
+ + +
= ×
+
( )
2
5 5 2 2 2
1 2
i i
i
+ +
=
=
3 6 2 3(1 2 2 )
1 2 3
i i
+ +
=
+
=
1 2 2
i
+
.
(ii)
( )
35
35 17
2
1 1 1
i
i
i i
i
i i
= = = ×
=
2
i
i
i
=
EXERCISE 5.1
Express each of the complex number given in the Exercises 1 to 10 in the
form a + ib.
1.
( )
3
5
5
i i
2.
i
i
9
19
+
3.
i
39
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