Chapter
COMPLEX NUMBERS AND
QUADRATIC EQUATIONS
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
11
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(3x – y) = 3 + i (– 6), where
x and y are real numbers, then find
the values of x and y.
Solution
We have
4x + i (3x – y) = 3 +
i (–6) ... (1)
Equating the real and the imaginary parts of (1), we get
4x = 3, 3
x – y = – 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
(the additive identity).
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
– bd) + i(ad + bc)
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
,
We already know that
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
= a – ib.
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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104 MATHEMATICS
Fig 5.1
4. 3(7 + i7) + i (7 + i7) 5. (1 – i) – ( –1 +
i6)
6.
1 2 5
4
5 5 2
i i
+ − +
7.
1 7 1 4
4
3 3 3 3
i i i
+ + + − − +
8. (1 – i)
4
9.
3
1
3
3
i
+
10.
3
1
2
3
i
− −
Find the multiplicative inverse of each of the complex numbers given in the
Exercises 11 to 13.
11. 4 – 3i 12.
5 3
i
+
13. – i
14. Express the following expression in the form of a + ib :
(
)
(
)
( ) ( )
3 5 3 5
3 2 3 2
i i
i i
+ −
+ − −
5.5 Argand Plane and Polar Representation
We already know that corresponding to
each ordered pair of real numbers
(x, y), we get a unique point in the XY-
plane and vice-versa with reference to a
set of mutually perpendicular lines known
as the x-axis and the y-axis. The complex
number x + iy which corresponds to the
ordered pair (x, y) can be represented
geometrically as the unique point P(x, y)
in the XY-plane and vice-versa.
Some complex numbers such as
2 + 4i, – 2 + 3i, 0 + 1i, 2 + 0i, – 5 –2i and
1 – 2
i which correspond to the ordered
pairs (2, 4), ( – 2, 3), (0, 1), (2, 0), ( –5, –2), and (1, – 2), respectively, have been
represented geometrically by the points A, B, C, D, E, and F, respectively in
the Fig 5.1.
The plane having a complex number assigned to each of its point is called the
complex plane or the Argand plane.
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 105
Obviously, in the Argand plane, the modulus of the complex number
x + iy =
2 2
x y
+
is the distance between the point P(x, y) and the origin O (0, 0)
(Fig 5.2). The points on the x-axis corresponds to the complex numbers of the form
a + i 0 and the points on the y-axis corresponds to the complex numbers of the form
Fig 5.2
Fig 5.3
0 + i b. The x-axis and y-axis in the Argand plane are called, respectively, the real axis
and the imaginary axis.
The representation of a complex number z = x + iy and its conjugate
z = x – iy in the Argand plane are, respectively, the points P (x, y) and Q (x, – y).
Geometrically, the point (x, – y) is the mirror image of the point (x, y) on the real
axis (Fig 5.3).
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106 MATHEMATICS
5.5.1 Polar representation of a complex
number Let the point P represent the non-
zero complex number z = x + iy. Let the
directed line segment OP be of length r and
θ be the angle which OP makes with the
positive direction of x-axis (Fig 5.4).
We may note that the point P is
uniquely determined by the ordered pair of
real numbers (r, θ), called the polar
coordinates of the point P. We consider
the origin as the pole and the positive
direction of the x axis as the initial line.
We have,
x = r cos θ, y = r sin θ and therefore,
z = r (cos θ + i sin θ). The latter
is said to be the polar form of the complex number. Here
2 2
r x y z
= + =
is the
modulus of z and θ is called the argument (or amplitude) of z which is denoted by arg z.
For any complex number z ≠ 0, there corresponds only one value of θ in
0 ≤ θ < 2π. However, any other interval of length 2π, for example – π < θ ≤ π, can be
such an interval.We shall take the value of θ such that – π < θ ≤ π, called principal
argument of z and is denoted by arg z, unless specified otherwise. (Figs. 5.5 and 5.6)
Fig 5.4
Fig 5.5
(
)
0 2
≤ θ < π
Fig 5.6 (– π < θ ≤ π )
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 107
Example 7 Represent the complex number
1 3
z i
= +
in the polar form.
Solution
Let 1 = r cos θ,
3
= r sin θ
By squaring and adding, we get
(
)
2 2 2
cos
θ sin θ 4
r
+ =
i.e.,
r
=
=
4
2
(conventionally, r >0)
Therefore,
1
cos θ
2
=
,
3
sin θ
2
=
, which gives
θ
π
=
3
Therefore, required polar form is
π π
2 cos sin
3 3
z i
= +
The complex number
z
i
=
+
1
3
is represented as shown in Fig 5.7.
Example 8 Convert the complex number
16
1 3
i
−
+
into polar form.
Solution The given complex number
16
1 3
i
−
+
=
16 1 3
1 3 1 3
i
i i
− −
×
+ −
=
(
)
( )
(
)
2
–16 1 – 3 –16 1– 3
=
1+ 3
1 – 3
i i
i
=
(
)
– 4 1 – 3 = – 4 + 4 3
i i
(Fig 5.8).
Let – 4 = r cos θ,
4
3
= r sin θ
By squaring and adding, we get
16 + 48 =
(
)
2 2 2
cos
θ + sin θ
r
which gives r
2
= 64, i.e., r = 8
Hence cos θ =
−
1
2
, sin θ =
3
2
π 2π
θ = π – =
3 3
Thus, the required polar form is
2
π 2π
8 cos sin
3 3
i
+
Fig 5.7
Fig 5.8
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108 MATHEMATICS
EXERCISE 5.2
Find the modulus and the arguments of each of the complex numbers in
Exercises 1 to 2.
1. z = – 1 – i
3
2. z = –
3
+ i
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 – i 4. – 1 + i 5. – 1 – i
6. – 3 7.
3
+ i 8. i
5.6 Quadratic Equations
We are already familiar with the quadratic equations and have solved them in the set
of real numbers in the cases where discriminant is non-negative, i.e., ≥ 0,
Let us consider the following quadratic equation:
ax
2
+ bx + c = 0 with real coefficients a, b, c and a ≠ 0.
Also, let us assume that the b
2
– 4ac < 0.
Now, we know that we can find the square root of negative real numbers in the
set of complex numbers. Therefore, the solutions to the above equation are available in
the set of complex numbers which are given by
x =
2 2
4 4
2 2
b b ac b ac b i
a a
− ± − − ± −
=
A
Note At this point of time, some would be interested to know as to how many
roots does an equation have? In this regard, the following theorem known as the
Fundamental theorem of Algebra is stated below (without proof).
“A polynomial equation has at least one root.”
As a consequence of this theorem, the following result, which is of immense
importance, is arrived at:
“A polynomial equation of degree n has n roots.”
Example 9 Solve x
2
+ 2 = 0
Solution We have, x
2
+ 2 = 0
or x
2
= – 2 i.e., x =
2
± −
=
2
±
i
Example 10 Solve x
2
+ x + 1= 0
Solution Here, b
2
– 4ac = 1
2
– 4 × 1 × 1 = 1 – 4 = – 3
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 109
Therefore, the solutions are given by x =
1 3 1 3
2 1 2
i
− ± − − ±
=
×
Example 11 Solve
2
5 5 0
x x
+ + =
Solution Here, the discriminant of the equation is
2
1 4 5 5
− × ×
= 1 – 20 = – 19
Therefore, the solutions are
1 19 1 19
2 5 2 5
i
− ± − − ±
=
.
EXERCISE 5.3
Solve each of the following equations:
1. x
2
+ 3 = 0 2. 2x
2
+ x + 1 = 0 3. x
2
+ 3x + 9 = 0
4. – x
2
+ x – 2 = 0 5. x
2
+ 3x + 5 = 0 6. x
2
– x + 2 = 0
7.
2
2 2 0
x x
+ + =
8.
2
3 2 3 3 0
x x
− + =
9.
2
1
0
2
x x
+ + =
10.
2
1 0
2
x
x
+ + =
Miscellaneous Examples
Example 12 Find the conjugate of
(3 2 )(2 3 )
(1 2 )(2 )
i i
i i
− +
+ −
.
Solution We have ,
(3 2 )(2 3 )
(1 2 ) (2 )
i i
i i
− +
+ −
=
6 9 4 6
2 4 2
i i
i i
+ − +
− + +
=
12 5 4 3
4 3 4 3
i i
i i
+ −
×
+ −
=
48 36 20 15 63 16
16 9 25
i i i
− + + −
=
+
=
63 16
25 25
i
−
Therefore, conjugate of
(3 2 )(2 3 ) 63 16
is
(1 2 ) (2 ) 25 25
i i
i
i i
− +
+
+ −
.
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110 MATHEMATICS
Example 13 Find the modulus and argument of the complex numbers:
(i)
1
1
i
i
+
−
, (ii)
1
1
i
+
Solution (i) We have,
1
1
i
i
+
−
=
1 1 1 1 2
1 1 1 1
i i i
i
i i
+ + − +
× = =
− + +
= 0 + i
Now, let us put 0 = r cos θ, 1 = r sin θ
Squaring and adding, r
2
= 1 i.e., r = 1 so that
cos θ = 0, sin θ = 1
Therefore,
π
θ
2
=
Hence, the modulus of
1
1
i
i
+
−
is 1 and the argument is
π
2
.
(ii) We have
1 1 1 1
1 (1 )(1 ) 1 1 2 2
i i i
i i i
− −
= = = −
+ + − +
Let
1
2
= r cos θ, –
1
2
= r sin θ
Proceeding as in part (i) above, we get
1 1 1
; cosθ , sin θ
2 2 2
r
−
= = =
Therefore
π
θ
4
−
=
Hence, the modulus of
1
1
i
+
is
1
2
, argument is
π
4
−
.
Example 14 If x + iy =
a ib
a ib
+
−
, prove that x
2
+ y
2
= 1.
Solution We have,
x + iy =
( ) ( )
( ) ( )
a ib a ib
a ib a ib
+ +
− +
=
2 2
2 2
2
a b abi
a b
− +
+
=
2 2
2 2 2 2
2
a b ab
i
a b a b
−
+
+ +
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 111
So that, x – iy =
2 2
2 2 2 2
2
a b ab
i
a b a b
−
−
+ +
Therefore,
x
2
+ y
2
= (x + iy) (x – iy) =
2 2 2 2 2
2 2 2 2 2 2
( ) 4
( ) ( )
a b a b
a b a b
−
+
+ +
=
2 2 2
2 2 2
( )
( )
a b
a b
+
+
= 1
Example 15 Find real θ such that
3 2 sin
θ
1 2 sin
θ
i
i
+
−
is purely real.
Solution We have,
3 2 sin
θ
1 2 sin
θ
i
i
+
−
=
(3 2 sin
θ)(1 2 sinθ)
(1 2 sin
θ) (1 2 sinθ)
i i
i i
+ +
− +
=
2
2
3+ 6 sin
θ + 2 sinθ – 4sin θ
1+ 4sin θ
i i
=
2
2 2
3 4sin
θ 8 sinθ
1 4sin
θ 1 4sin θ
i−
+
+ +
We are given the complex number to be real. Therefore
2
8sin
θ
1 4sin
θ
+
= 0, i.e., sin θ = 0
Thus θ = nπ, n ∈ Z.
Example 16 Convert the complex number
1
π π
cos sin
3 3
i
z
i
−
=
+
in the polar form.
Solution We have, z =
1
1 3
2 2
i
i
−
+
=
(
)
2 3 1 3
2( 1) 1 3
1 3
1 3 1 3
i i
i i
i i
+ − +
− −
× =
+
+ −
=
3 1 3 1
2 2
i
− +
+
Now, put
3 1 3 1
cos , sin
2 2
r r
θ θ
− +
= =
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112 MATHEMATICS
Squaring and adding, we obtain
2 2
2
3 1 3 1
2 2
r
− +
= +
=
( )
2
2 3 1
2 4
2
4 4
+
×
= =
Hence,
2
r
=
which gives
3 1 3 1
cosθ , sinθ
2 2 2 2
− +
= =
Therefore,
π π 5π
θ
4 6 12
= + =
(Why?)
Hence, the polar form is
5
π 5π
2 cos sin
12 12
i
+
Miscellaneous Exercise on Chapter 5
1. Evaluate:
3
25
18
1
i
i
+
.
2. For any two complex numbers z
1
and z
2
, prove that
Re (z
1
z
2
) = Re z
1
Re z
2
– Imz
1
Imz
2
.
3. Reduce
1 2 3 4
1 4 1 5
i
i i i
−
−
− + +
to the standard form .
4. If
a ib
x iy
c id
−
− =
−
prove that
( )
2 2
2
2 2
2 2
a b
x y
c d
+
+ =
+
.
5. Convert the following in the polar form:
(i)
( )
2
1 7
2
i
i
+
−
, (ii)
1 3
1 – 2
i
i
+
Solve each of the equation in Exercises 6 to 9.
6.
2
20
3 4 0
3
x x
− + =
7.
2
3
2 0
2
x x
− + =
8.
2
27 10 1 0
x x
− + =
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 113
9.
2
21 28 10 0
x x
− + =
10. If z
1
= 2 – i, z
2
= 1 + i, find
1 2
1 2
1
– 1
z z
z z
+ +
+
.
11. If a + ib =
2
2
( )
2 1
x i
x
+
+
, prove that a
2
+ b
2
=
( )
2 2
2
2
( 1)
2 1
x
x
+
+
.
12. Let z
1
= 2 – i, z
2
= –2 + i. Find
(i)
1 2
1
Re
z z
z
, (ii)
1 1
1
Im
z z
.
13. Find the modulus and argument of the complex number
1 2
1 3
i
i
+
−
.
14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
15. Find the modulus of
1 1
1 1
i i
i i
+ −
−
− +
.
16. If (x + iy)
3
= u + iv, then show that
2 2
4( – )
u v
x y
x y
+ =
.
17. If α and β are different complex numbers with
β 1
=
, then find
β α
1
αβ
–
–
.
18. Find the number of non-zero integral solutions of the equation
1 2
x
x
– i
=
.
19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a
2
+ b
2
) (c
2
+ d
2
) (e
2
+ f
2
) (g
2
+ h
2
) = A
2
+ B
2
20. If
1
1
1
m
i
– i
+
=
, then find the least positive integral value of m.
2020-21
114 MATHEMATICS
Summary
®
A number of the form a + ib, where a and b are real numbers, is called a
complex number, a is called the real part and b is called the imaginar
y part
of the complex number.
®
Let z
1
= a + ib and
z
2
= c + id. Then
(i) z
1
+ z
2
= (a + c) + i (b + d)
(ii) z
1
z
2
= (ac – bd) + i (ad +
bc)
®
For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists 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 (a + ib)
2
2 2 2 2
a b
i
a b a b
−
+
+ +
= 1 + i0 =1
®
For any integer k, i
4k
= 1, i
4k + 1
= i, i
4k + 2
= – 1, i
4k + 3
= – i
®
The conjugate of the complex number z = a + ib, denoted by
z
, is given by
z
= a – ib.
®
The polar form of the complex number z = x + iy is r (cosθ + i sin
θ), where
r =
2 2
x y
+
(the modulus of z) and cosθ =
x
r
, sinθ =
y
r
. (θ is known as the
argument of z. The value of θ, such that – π < θ ≤ π, is called the principal
argument of z.
®
A polynomial equation of n degree has n roots.
®
The solutions of the quadratic equation ax
2
+ bx + c = 0, where a, b, c ∈ R,
a ≠ 0, b
2
– 4ac < 0, are given by x =
2
4
2
b ac b i
a
− ± −
.
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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 115
Historical Note
The fact that square root of a negative number does not exist in the real number
system was recognised by the Greeks. But the credit goes to the Indian
mathematician Mahavira (850) who first stated this difficulty clearly. “He mentions
in his work ‘Ganitasara Sangraha’ as in the nature of things a negative (quantity)
is not a square (quantity)’, it has, therefore, no square root”. Bhaskara, another
Indian mathematician, also writes in his work Bijaganita, written in 1150. “There
is no square root of a negative quantity, for it is not a square.” Cardan (1545)
considered the problem of solving
x + y = 10,
xy = 40.
He obtained x = 5 +
15
−
and y = 5 –
15
−
as the solution of it, which
was discarded by him by saying that these numbers are ‘useless’. Albert Girard
(about 1625) accepted square root of negative numbers and said that this will
enable us to get as many roots as the degree of the polynomial equation. Euler
was the first to introduce the symbol i for
1
−
and W.R. Hamilton (about
1830) regarded the complex number a + ib as an ordered pair of real numbers
(a, b) thus giving it a purely mathematical definition and avoiding use of the so
called ‘imaginary numbers’.
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