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5

Chapter

COMPLEX NUMBERS AND
QUADRATIC EQUATIONS
5
...
g
...
Therefore, square root of 16 is ± 4
...
So
we need to extend the system of real numbers to a system in which we can find out the
square roots of negative numbers
...
e
...


5
...
1 Imaginary numbers
Square root of a negative number is called an imaginary number
...
1
...


To compute in for n > 4, we divide n by 4 and write it in the form n = 4m + r, where m is
quotient and r is remainder (0 ≤ r ≤ 4)
Hence
in = i4m+r = (i4)m
...
(i)3 = i3 = – i
and
(i)–435 = i – (4 × 108 + 3) = (i)– (4 × 108)
...
3 = 4 = i
(i )
(i) (i )
(i) If a and b are positive real numbers, then
− a × −b =

(ii)

a
...
However,

a b ≠

ab if a and b, both are negative
...
1
...


(b) If z = a + ib is the complex number, then a and b are called real and imaginary
parts, respectively, of the complex number and written as Re (z) = a, Im (z) = b
...

(d) If the imaginary part of a complex number is zero, then the complex number is
known as purely real number and if real part is zero, then it is called
purely imaginary number, for example, 2 is a purely real number because its
imaginary part is zero and 3i is a purely imaginary number because its real part
is zero
...
1
...

(b) Let z1 = a + ib and z2 = c + id be two complex numbers then
z1 + z2 = (a + c) + i (b + d)
...
1
...
As the sum of two complex numbers is again a complex number, the set of
complex numbers is closed with respect to addition
...
Addition of complex numbers is commutative, i
...
, z1 + z2 = z2 + z1
3
...
e
...
For any complex number z = x + i y, there exist 0, i
...
, (0 + 0i) complex number
such that z + 0 = 0 + z = z, known as identity element for addition
...
For any complex number z = x + iy, there always exists a number – z = – a – ib
such that z + (– z) = (– z) + z = 0 and is known as the additive inverse of z
...
1
...
Then
z1
...
As the product of two complex numbers is a complex number, the set of complex
numbers is closed with respect to multiplication
...
Multiplication of complex numbers is commutative, i
...
, z1
...
z1
3
...
e
...
z2)
...
(z2
...
For any complex number z = x + iy, there exists a complex number 1, i
...
, (1 + 0i)
such that
z
...
z = z, known as identity element for multiplication
...
For any non zero complex number z = x + i y, there exists a complex number
z
1 1
1
a − ib
=
such that z ⋅ = ⋅ z = 1 , i
...
, multiplicative inverse of a + ib =

...
For any three complex numbers z1, z2 and z3 ,
z1
...
z2 + z1
...
z3 = z1
...
z3
i
...
, for complex numbers multiplication is distributive over addition
...
1
...
Then

z1 a + ib (ac + bd )
(bc − ad )
+i 2
=
= 2
2
z2 c + id
c +d
c +d2
5
...
8 Conjugate of a complex number
Let z = a + ib be a complex number
...
e
...

Note that additive inverse of z is – a – ib but conjugate of z is a – ib
...
( z ) = z
2
...

4
...


z + z = 2 Re (z) , z – z = 2 i Im(z)
z = z , if z is purely real
...
z = {Re (z)}2 + {Im (z)}2
...
( z1 + z2 ) = z1 + z2 , ( z1 − z2 ) = z1 – z2

z1
(z )
= 1 ( z2 ≠ 0)
z2
( z2 )
5
...
9 Modulus of a complex number
Let z = a + ib be a complex number
...
( z1
...
e
...

5
...
10 Properties of modulus of a complex number
1
...
e
...


z = z = −z

3
...
z z = z , z = z

2

z
z1
= 1 ( z2 ≠ 0)
z2
z2

5
...
z2 ,

6
...


z1 − z2 = z1 + z2 − 2 Re ( z1 z2 )

8
...


z1 − z2 ≥ z1 − z2

10
...
As stated earlier multiplicative inverse (reciprocal) of a complex number
z = a + ib (≠ 0) is

a − ib
z
1
= 2 2 = 2
z
a +b
z

5
...
The complex number 0 + 0i
represent the origin 0 ( 0, 0)
...
e
...
Therefore, x-axis is called real axis
...
e
...
Therefore, y-axis is called
imaginary axis
...
The plane representing complex numbers as points is called complex
plane or Argand plane or Gaussian plane
...
2
...
If OP makes an angle θ with the positive direction of x-axis,
then z = r (cosθ + isinθ) is called the polar form of the complex number, where
r= z =

a 2 + b2 and tanθ =

b

...

The unique value of θ such that – π ≤ θ ≤ π is called the principal argument
...
z2) = arg (z1) + arg (z2)

 z1 
arg  z  = arg (z1) – arg (z2)
 2
5
...
2 Solution of a quadratic equation
The equations ax2 + bx + c = 0, where a, b and c are numbers (real or complex, a ≠ 0)
is called the general quadratic equation in variable x
...

The quadratic equation ax2 + bx + c = 0 with real coefficients has two roots given
by

–b + D
–b – D
and
, where D = b2 – 4ac, called the discriminant of the equation
...
When D = 0, roots of the quadratic equation are real and equal
...

Further, if a, b, c ∈ Q and D is a perfect square, then the roots of the equation
are rational and unequal, and if a, b, c ∈Q and D is not a perfect square, then
the roots are irrational and occur in pair
...

2
...

a
a

3
...
Then the quadratic equation is given by x2 – Sx + P = 0
...
2 Solved Exmaples
Short Answer Type
Example 1 Evaluate : (1 + i)6 + (1 – i)3
Solution (1 + i)6 = {(1 + i)2}3 = (1 + i2 + 2i)3 = (1 – 1 + 2i)3 = 8 i3 = – 8i
and
(1 – i)3 = 1 – i3 – 3i + 3i2 = 1 + i – 3i – 3 = – 2 – 2i
Therefore,
(1 + i)6 + (1 – i)3 = – 8i – 2 – 2i = – 2 – 10i
1

Example 2 If ( x + iy ) 3 = a + ib, where x, y, a, b ∈ R, show that

x y
− = – 2 (a2 + b2)
a b

1

Solution ( x + iy ) 3 = a + ib
⇒
i
...
,

⇒

x + iy = (a + ib)3
x + iy = a3 + i3 b3 + 3iab (a + ib)
= a3 – ib3 + i3a2b – 3ab2
= a3 – 3ab2 + i (3a2b – b3)
x = a3 – 3ab2 and y = 3a2b – b3

Thus

x
y
2
2
2
2
a = a – 3b and b = 3a – b

So,

x y
− = a2 – 3b2 – 3a2 + b2 = – 2 a2 – 2b2 = – 2 (a2 + b2)
...
(1)
and
2xy = – y


...
e
...

When x = −

1
1 1
, from (1), we get y2 = +
2
4 2

or y2 =

3
3
, i
...
, y = ±

...

2
2
2
2

2 z +1
is – 2, then show that the locus of the point
iz +1
representing z in the argand plane is a straight line
...
Then

2 z +1
2( x + iy ) + 1 (2 x + 1) + i 2 y
=
=
iz + 1
i ( x + iy ) + 1
(1 − y ) + ix
=

{(2 x +1) + i 2 y} {(1− y) − ix}
×
{(1− y) + ix}
{(1− y) − ix}

(2 x + 1− y ) + i (2 y − 2 y 2 − 2 x 2 − x )
=
1+ y 2 − 2 y + x 2

Thus

But
So
⇒
i
...
,

 2 z +1  2 y − 2 y 2 − 2 x2 − x
Im 
=
1+ y 2 − 2 y + x2
 iz +1 
 2 z +1 
Im 
(Given)
= –2
 iz +1 
2 y − 2 y 2 − 2x2 − x
=−2
1+ y 2 − 2 y + x 2
2y – 2y2 – 2x2 – x = – 2 – 2y2 + 4y – 2x2
x + 2y – 2 = 0, which is the equation of a line
...


Solution Let z = x + iy
...
e
...

Example 6 Let z1 and z2 be two complex numbers such that z1 + i z 2 = 0 and
arg (z1 z2) = π
...

Solution Given that z1 + i z 2 = 0
⇒
Thus

z1 = i z2 , i
...
, z2 = – i z1
arg (z1 z2) = arg z1 + arg (– i z1) = π

⇒

arg (– i z12 ) = π

arg (– i ) + arg ( z12 ) = π
arg (– i ) + 2 arg (z1) = π
−π
+ 2 arg (z1) = π
⇒
2
3π
⇒
arg (z1) =
4
Example 7 Let z1 and z2 be two complex numbers such that z1 + z2 = z1 + z2
...

Solution Let z1 = r1 (cosθ1 + i sin θ1) and z2 = r2 (cosθ2 + i sin θ2)
r1 = z1 , arg ( z1 ) = θ1, r2 = z2 , arg (z2) = θ2
...
e
...

1
2
3
z1 z2 z3

Solution z1 = z2 = z3 =1

18/04/18

COMPLEX NUMBERS AND QUADRATIC EQUATIONS

2

2

81

2

⇒

z1 = z2 = z3 =1

⇒

z1 z1 = z2 z2 = z3 z3 = 1

⇒

z1 =

Given that

1
1
1
+
+
=1
z1 z2 z3

⇒

z1 + z2 + z3 = 1 , i
...
, z1 + z 2 + z 3 = 1

⇒

z1 + z2 + z3 = 1

1
1
1
, z 2 = , z3 =
z1
z2
z3

Example 9 If a complex number z lies in the interior or on the boundary of a circle of
radius 3 units and centre (– 4, 0), find the greatest and least values of z +1
...

According to given condition z + 4 ≤ 3
...

Since least value of the modulus of a complex number is zero, the least value of

z +1 = 0
...
Hence 3 < z < 4 is the portion between two circles x2 + y2 = 9 and
x2 + y2 = 16
...


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82

EXEMPLAR PROBLEMS – MATHEMATICS

Example 12 Find the value of P such that the difference of the roots of the equation
x2 – Px + 8 = 0 is 2
...
β = 8
...
e
...

Example 13 Find the value of a such that the sum of the squares of the roots of the
equation x2 – (a – 2) x – (a + 1) = 0 is least
...
e
...

Long Answer Type
Example 14 Find the value of k if for the complex numbers z1 and z2,
2

2

2

2

1− z1 z2 − z1 − z2 = k (1− z1 )(1− z2 )
Solution
2

L
...
S
...
H
...
= k (1 – z1 ) (1 − z2 )
⇒

k=1

18/04/18

COMPLEX NUMBERS AND QUADRATIC EQUATIONS

83

Hence, equating LHS and RHS, we get k = 1
...

Solution Let z = x + iy, z1 = x1 + iy1 and z2 = x2 + iy2
...
and 2x2 = 1 + y22
⇒
2 (x1 – x2) = (y1 + y2) (y1 – y2)

Again

 y −y 
2 = (y1 + y2)  1 2 
 x1 − x2 
z1 – z2 = (x1 – x2) + i (y1 – y2)

Therefore,

tan θ =

⇒

π y −y
tan = 1 2
4 x1 − x2

i
...
,

1=

⇒


...
(2)

y1 − y2
, where θ = arg (z1 – z2)
x1 − x2
π

 since θ= 
4


y1 − y2
x1 − x2

From (2), we get 2 = y1 + y2, i
...
, Im (z1 + z2) = 2
Objective Type Questions
Example 16 Fill in the blanks:
(i) The real value of ‘a’ for which 3i3 – 2ai2 + (1 – a)i + 5 is real is ________
...

4
π
(iii) The locus of z satisfying arg (z) =
is _______
...


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84

EXEMPLAR PROBLEMS – MATHEMATICS

(v) The conjugate of the complex number

1− i
is _____
...

(vii) If (2 + i) (2 + 2i) (2 + 3i)
...
8
...
(4 + n2) = ______
...
e
...

1 
π
π  1

+i
(ii) z = z  cos + i sin  = 2 
 = 2 (1+ i)
4
4  2
2

(iii) Let z = x + iy
...
Given that θ =
...

3
π y
tan =
⇒ y = 3 x , where x > 0, y > 0
...

4 n –3
= (−i) 4 n−3 = (−i )4 n (−i) −3 =
(iv) Here (– −1)

1
( −i)3

1
1 i
= = 2 = −i
3
i i
−i
1− i 1 − i 1 − i 1+ i 2 − 2i 1 −1 − 2i
=
×
=
=
= −i
(v)
1 + i 1+ i 1 − i
1+ 1
1− i 2
=

1− i
is i
...
Therefore, if a number lies in the third quadrant, then its image lies in
the second quadrant
...
(2 + ni) = x + iy

...
(2 + ni ) = x + iy = ( x − iy )

i
...
,

(2 – i) (2 – 2i) (2 – 3i)
...
(2)

18/04/18

COMPLEX NUMBERS AND QUADRATIC EQUATIONS

85

Multiplying (1) and (2), we get 5
...
13
...

Example 17 State true or false for the following:
(i) Multiplication of a non-zero complex number by i rotates it through a right
angle in the anti- clockwise direction
...

(iii) If a complex number coincides with its conjugate, then the number must lie on
imaginary axis
...

(vi) If three complex numbers z1, z2 and z3 are in A
...
, then they lie on a circle in
the complex plane
...

Solution
(i) True
...
Then iz = –3 + 2i,
represented by OQ, where if OP is rotated in the anticlockwise direction through
a right angle, it coincides with OQ
...
Because cosθ + isinθ = 0 ⇒ cosθ = 0 and sinθ = 0
...

(iii) False, because x + iy = x – iy ⇒ y = 0 ⇒ number lies on x-axis
...


z1 + z3
⇒ z2 is the midpoint
2
of z1 and z3, which implies that the points z1, z2, z3 are collinear
...
P
...

Column A
2

Column B
4

6

20

(a) The value of 1+i + i + i +
...
+ i20
= 1 – 1 + 1 – 1 +
...

(c) ⇔ (iv), conjugate of 1 + i is 1 – i, which is represented by the point (1, –1) in
the fourth quadrant
...

 2 2

(d) ⇔ (iii), because

(e) ⇔ (vi), If b2 – 4ac < 0 = D < 0, i
...
, square root of D is a imaginary
number, therefore, roots are x =

−b ± Imaginary Number
, i
...
, roots are in
2a

conjugate pairs
...


b = – (5 +

Now D = b2 – 4ac = {– (5 +
Therefore x =

2)x+5

87

2
2 )} – 4
...
5 2 = (5 –

2
2)
...

2

Example 19 What is the value of
Solution i, because

i 4 n +1 − i 4 n −1
?
2

i 4 n +1 − i 4n −1 i 4 ni − i 4 ni − i
=
2
2

1
2
i = i −1 = −2 = i
=
2
2i
2i
Example 20 What is the smallest positive integer n, for which (1 + i)2n = (1 – i)2n?
i−

2n

 1+ i 
Solution n = 2, because (1 + i)2n = (1 – i)2n = 
 =1
 1− i 
⇒

(i)2n = 1 which is possible if n = 2

Example 21 What is the reciprocal of 3 +
Solution Reciprocal of z =
Therefore, reciprocal of 3 +
Example 22 If z1 =

(∴ i4 = 1)

7i

z
z

2

7 i=

3− 7 i
3
7i
=
–
16
16 16

3 + i 3 and z2 =

3 + i , then find the quadrant in which

 z1 
  lies
...


18/04/18

88

EXEMPLAR PROBLEMS – MATHEMATICS

5 + 12i + 5 −12i
Example 23 What is the conjugate of

5 +12i − 5 −12i

?

Solution Let
z=

=

5 +12i + 5 −12i
5 +12i − 5 −12i

×

5 +12i + 5 −12i
5 +12i + 5 −12i

5 +12i + 5 −12i + 2 25 + 144
5 + 12i − 5 +12i

3i
3
3
=
= 0− i
−2
2i
2
3
Therefore, the conjugate of z = 0 + i
2
Example 24 What is the principal value of amplitude of 1 – i ?
=

Solution Let θ be the principle value of amplitude of 1 – i
...
Then z – 2 – 3i = (x – 2) + i (y – 3)
y −3
Let θ be the amplitude of z – 2 – 3i
...
e
...

Example 27 If 1 – i, is a root of the equation x2 + ax + b = 0, where a, b ∈ R, then
find the values of a and b
...

1
(since non real complex roots occur in conjugate pairs)
b
Product of roots, = (1 − i ) (1 + i) ⇒ b = 2
1
Choose the correct options out of given four options in each of the Examples from 28
to 33 (M
...
Q
...
+ i2n is
(A) positive
(C) 0

(B) negative
(D) can not be evaluated

Solution (D), 1 + i2 + i4 + i6 +
...
(–1)n
which can not be evaluated unless n is known
...

Example 30 The area of the triangle on the complex plane formed by the complex
numbers z, – iz and z + iz is:
(A)
(C)

z
z

2

(B)

z

2

2

2

(D) none of these

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90

EXEMPLAR PROBLEMS – MATHEMATICS

Solution (C), Let z = x + iy
...
Therefore,
z + iz = (x – y) + i (x + y)
2

Required area of the triangle =

z
1 2
(x + y2 ) =
2
2

Example 31 The equation z + 1− i = z −1+ i represents a
(A) straight line
(C) parabola

(B) circle
(D) hyperbola

Solution (A), z + 1− i = z −1+ i

z − (−1 + i) = z − (1 − i)

⇒
⇒
⇒

PA = PB, where A denotes the point (–1, 1), B denotes the point (1, –1) and P
denotes the point (x, y)
z lies on the perpendicular bisector of the line joining A and B and perpendicular
bisector is a straight line
...
Hence infinitely many solutions
...
cos  
 10 
 10 

π
π
tan θ = tan   i
...
, θ=
10
 10 

⇒

5
...
For a positive integer n, find the value of (1 – i)n  1− 
 i

n

13

∑ (i n + i n+1 ) , where n∈N
...
Evaluate

n =1

3

3

 1+ i   1 − i 
3
...

 1- i   1+ i 
4
...

2−i
100

 1− i 
5
...


1+ a

...
If (1 + i) z = (1 – i) z , then show that z = – i z
...
If z = x + iy , then show that z z + 2 (z + z ) + b = 0, where b ∈ R, represents
a circle
...
If a = cos θ + i sinθ, find the value of

z +2
is 4, then show that the locus of the point representing
z −1
z in the complex plane is a circle
...
If the real part of

 z −1  π
10
...

11
...


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92

EXEMPLAR PROBLEMS – MATHEMATICS

Long Answer Type
12
...

13
...
where z = x + iy
14
...
If

z −2
= 2 represents a circle
...

z −3

z −1
is a purely imaginary number (z ≠ – 1), then find the value of z
...
z1 and z2 are two complex numbers such that z1 = z2 and arg (z1) + arg (z2) =
π, then show that z1 = − z2
...

z1 +1
18
...
If z1 = 1 (z1 ≠ –1) and z2 =

 z1 
z 
arg   + arg  2 
...
If z1 = z2 =
...
+ zn =

1
1
1
1
+
+
+
...

z1 z2 z3
zn

20
...
Solve the system of equations Re (z2) = 0, z = 2
...
Find the complex number satisfying the equation z + 2 |(z + 1)| + i = 0
...
Write the complex number z =

1− i
in polar form
...
If z and w are two complex numbers such that zw =1 and arg (z) – arg (w) =

π
, then show that z w = – i
...
Fill in the blanks of the following
(i) For any two complex numbers z 1, z 2 and any real numbers a, b,
2

2

az1 − bz2 + bz1 + az2 =
...

(1 − i)3
is equal to
...
upto 1000 terms is
...

(vi) If z1 and z2 are complex numbers such that z1 + z2 is a real number,
then z2 =
...

(viii) If z + 4 ≤ 3 , then the greatest and least values of z + 1 are
...

(ix) If

z −2 π
= , then the locus of z is
...

6

26
...

(ii) Multiplication of a non zero complex number by – i rotates the point about
origin through a right angle in the anti-clockwise direction
...

(iv) The locus represented by z − 1 = z − i is a line perpendicular to the join of
(1, 0) and (0, 1)
...

(vi) The inequality z − 4 < z − 2 represents the region given by x > 3
...

(viii) 2 is not a complex number
...
Match the statements of Column A and Column B
...

(c) If z + 2 = z − 2 , then

(iii)

2π
3

locus of z is
(d) If z + 2i = z − 2i , then

(iv) Perpendicular bisector of segment

locus of z is

joining (0, – 2) and (0, 2)
...


1 + 2i
lies in
1− i

(h) Reciprocal of 1 – i lies in
28
...
If z1 = z2 , is it necessary that z1 = z2?
(a 2 + 1) 2
30
...
Find z if z = 4 and arg (z) =
32
...

6

(2 + i )
(3 + i)

33
...

34
...

z + 5i

Choose the correct answer from the given four options indicated against each of the
Exercises from 35 to 50 (M
...
Q)
35
...
The real value of α for which the expression
(A)

( n +1)

π
2

(C) n π
37
...
The value of (z + 3) ( z + 3) is equivalent to
(A)

z+3

2

2

(C) z + 3

(B)

z −3

(D) None of these

x

 1+ i 
39
...
A real value of x satisfies the equation 
 = α − iβ (α, β ∈ R )
3 + 4ix 

if α2 + β2 =
(A) 1
(B) – 1
(C) 2
(D) – 2
41
...
arg (z2)

(C)

z1 + z2 = z1 + z2

(D)

z1 + z2 ≥ z1 − z2

42
...
Let x, y ∈ R, then x + iy is a non real complex number if:
(A) x = 0
(B) y = 0
44
...


(A)

z2 > z

2

(C)

z2 < z

2

(D) y ≠ 0

(B) b2 + c2 = 0
(D) a2 + b2 = c2 + d2

45
...
If z is a complex number, then

(D) –1 + 2 i

i+z
= 1 lies on
i−z

(B) the x-axis
(D) the line x + y = 1
...
The real value of θ for which the expression

(A) nπ +

π
4

π
2
49
...
If f (z) =

(A)
(C)

z
2

2z

97

1 + i cos θ
is a real number is:
1 − 2 i cos θ

(B)

nπ+ (−1) n

π
4

(D) none of these
...


18/04/18



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