Preview 43 Years Chapterwise Topicwise Solved Papers (20211979) IIT JEE Mathematics by Amit M Agarwal (2022)

Chapterwise Topicwise

Solved Papers
2021-1979

IITJEE
JEE Main & Advanced

Mathematics
Amit M Agarwal

Arihant Prakashan (Series), Meerut


Arihant Prakashan (Series), Meerut
All Rights Reserved

© Author

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CONTENTS
1. Complex Numbers

1-28

2. Theory of Equations

29-50

3. Sequences and Series

51-75

4. Permutations and Combinations

76-88

5. Binomial Theorem

89-102

6. Probability

103-134

7. Matrices and Determinants

135-170

8. Functions

171-187

9. Limit, Continuity and Differentiability

188-238

10. Application of Derivatives

239-277

11. Indefinite Integration

278-293

12. Definite Integration

294-328

13. Area

329-353

14. Differential Equations

354-378

15. Straight Line and Pair of Straight Lines

379-404


16. Circle

405-439

17. Parabola

440-457

18. Ellipse

458-473

19. Hyperbola

474-485

20. Trigonometrical Ratios and Identities

486-498

21. Trigonometrical Equations

499-514

22. Inverse Circular Functions

515-525

23. Properties of Triangles

526-547

24. Vectors

548-579

25. 3D Geometry

580-605

26. Miscellaneous

606-632

JEE Advanced Solved Paper 2021

1-16

SYLLABUS
JEE MAIN
UNIT I Sets, Relations and Functions

UNIT V Mathematical Induction

Sets and their representation, Union, intersection and
complement of sets and their algebraic properties, Power
set, Relation, Types of relations, equivalence relations,
functions, one-one, into and onto functions, composition
of functions.

Principle of Mathematical Induction and its simple
applications.

UNIT VI Binomial Theorem and its Simple
Applications

UNIT II Complex Numbers and
Quadratic Equations

Binomial theorem for a positive integral index, general
term and middle term, properties of Binomial coefficients
and simple applications.

Complex numbers as ordered pairs of reals,
Representation of complex numbers in the form a+ib and
their representation in a plane, Argand diagram, algebra
of complex numbers, modulus and argument (or
amplitude) of a complex number, square root of a

complex number, triangle inequality, Quadratic
equations in real and complex number system and their
solutions. Relation between roots and co-efficients,
nature of roots, formation of quadratic equations with
given roots.

UNIT III Matrices and Determinants
Matrices, algebra of matrices, types of matrices,
determinants and matrices of order two and three.
Properties of determinants, evaluation of deter-minants,
area of triangles using determinants. Adjoint and
evaluation of inverse of a square matrix using
determinants and elementary transformations, Test of
consistency and solution of simultaneous linear
equations in two or three variables using determinants
and matrices.

UNIT VII Sequences and Series
Arithmetic and Geometric progressions, insertion of
arithmetic, geometric means between two given
numbers. Relation between AM and GM Sum upto n
terms of special series: ∑ n, ∑ n2, ∑n3. Arithmetico Geometric progression.

UNIT VIII Limit, Continuity and Differentiability
Real valued functions, algebra of functions, polynomials,
rational, trigonometric, logarithmic and exponential
functions, inverse functions. Graphs of simple functions.
Limits, continuity and differenti-ability. Differentiation of
the sum, difference, product and quotient of two
functions. Differentiation of trigonometric, inverse

trigonometric, logarithmic, exponential, composite and
implicit functions; derivatives of order upto two. Rolle's
and Lagrange's Mean Value Theorems. Applications of
derivatives: Rate of change of quantities, monotonic increasing and decreasing functions, Maxima and minima
of functions of one variable, tangents and normals.

UNIT IV Permutations and Combinations

UNIT IX Integral Calculus

Fundamental principle of counting, permutation as an
arrangement and combination as selection, Meaning of
P(n,r) and C (n,r), simple applications.

Integral as an anti - derivative. Fundamental integrals
involving algebraic, trigonometric, exponential and
logarithmic functions. Integration by substitution, by


parts and by partial fractions. Integration using
trigonometric identities. Evaluation of simple integrals of
the type
dx

dx

,

x2 ± a2

x ±a

dx
,
ax 2 + bx + c
(px + q) dx
ax 2 + bx + c

,

2

2

dx

dx
,
a2 – x2

dx

,

,

2

a – x2
(px + q) dx

,

ax 2 + bx + c

ax 2 + bx + c

circle when the end points of a diameter are given, points
of intersection of a line and a circle with the centre at the
origin and condition for a line to be tangent to a circle,
equation of the tangent. Sections of cones, equations of
conic sections (parabola, ellipse and hyperbola) in
standard forms, condition for y=mx + c to be a tangent
and point (s) of tangency.

UNIT XII Three Dimensional Geometry
,

2

2

a ± x dx

and

2

2

x – a dx

Integral as limit of a sum. Fundamental Theorem of
Calculus. Properties of definite integrals. Evaluation of
definite integrals, determining areas of the regions
bounded by simple curves in standard form.

UNIT X Differential Equations
Ordinary differential equations, their order and degree.
Formation of differential equations. Solution of
differential equations by the method of separation of
variables, solution of homogeneous and linear differential
equations
of the type

UNIT XI Coordinate Geometry
Cartesian system of rectangular coordinates in a plane,
distance formula, section formula, locus and its equation,
translation of axes, slope of a line, parallel and
dy lines,
perpendicular
of a line on the coordinate
+ p(x)yintercepts
= q(x)
dx
axes.
Straight lines
Various forms of equations of a line, intersection of lines,
angles between two lines, conditions for concurrence of
three lines, distance of a point from a line, equations of

internal and external bisectors of angles between two
lines, coordinates of centroid, orthocentre and
circumcentre of a triangle, equation of family of lines
passing through the point of intersection of two lines.
Circles, Conic sections
Standard form of equation of a circle, general form of the
equation of a circle, its radius and centre, equation of a

Coordinates of a point in space, distance between two
points, section formula, direction ratios and direction
cosines, angle between two intersecting lines. Skew lines,
the shortest distance between them and its equation.
Equations of a line and a plane in different forms,
intersection of a line and a plane, coplanar lines.

UNIT XIII Vector Algebra
Vectors and scalars, addition of vectors, components of a
vector in two dimensions and three dimensional space,
scalar and vector products, scalar and vector triple
product.

UNIT XIV Statistics and Probability
Measures of Dispersion: Calculation of mean, median,
mode of grouped and ungrouped data. Calculation of
standard deviation, variance and mean deviation for
grouped and ungrouped data.
Probability: Probability of an event, addition and
multiplication theorems of probability, Baye's theorem,
probability distribution of a random variate, Bernoulli
trials and Binomial distribution.

UNIT XV Trigonometry
Trigonometrical identities and equations.
Trigonometrical functions. Inverse trigonometrical
functions and their properties. Heights and Distances.

UNIT XVI Mathematical Reasoning
Statements, logical operations And, or, implies, implied
by, if and only if. Understanding of tautology,
contradiction, converse and contra positive.


JEE ADVANCED
Algebra
Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and
principal argument, triangle inequality, cube roots of unity, geometric interpretations.
Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations
with given roots, symmetric functions of roots.
Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite
arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural
numbers.
Logarithms and their Properties, Permutations and combinations, Binomial theorem for a positive integral index,
properties of binomial coefficients.

Matrices
Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product
of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of
order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their
properties, solutions of simultaneous linear equations in two or three variables.

Probability
Addition and multiplication rules of probability, conditional probability, independence of events, computation of
probability of events using permutations and combinations.

Trigonometry
Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving
multiple and sub-multiple angles, general solution of trigonometric equations.
Relations between sides and angles of a triangle, sine rule, cosine rule, half-angle formula and the area of a triangle,
inverse trigonometric functions (principal value only).

Analytical Geometry
Two Dimensions Cartesian oordinates, distance between two points, section formulae, shift of origin.
Equation of a straight line in various forms, angle between two lines, distance of a point from a line. Lines through
the point of intersection of two given lines, equation of the bisector of the angle between two lines, concurrency of
lines, centroid, orthocentre, incentre and circumcentre of a triangle.
Equation of a circle in various forms, equations of tangent, normal and chord.
Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the
points of intersection of two circles and those of a circle and a straight line.
Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric
equations, equations of tangent and normal.


Locus Problems, Three Dimensions Direction cosines and direction ratios, equation of a straight line in space,
equation of a plane, distance of a point from a plane.

Differential Calculus
Real valued functions of a real variable, into, onto and one-to-one functions, sum, difference, product and
quotient of two functions, composite functions, absolute value, polynomial, rational, trigonometric, exponential
and logarithmic functions.
Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two

functions, l'Hospital rule of evaluation of limits of functions.
Even and odd functions, inverse of a function, continuity of composite functions, intermediate value property of
continuous functions.
Derivative of a function, derivative of the sum, difference, product and quotient of two functions, chain rule,
derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions.
Derivatives of implicit functions, derivatives up to order two, geometrical interpretation of the derivative, tangents
and normals, increasing and decreasing functions, maximum and minimum values of a function, applications of
Rolle's Theorem and Lagrange's Mean Value Theorem.

Integral Calculus
Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals
and their properties, application of the Fundamental Theorem of Integral Calculus.
Integration by parts, integration by the methods of substitution and partial fractions, application of definite
integrals to the determination of areas involving simple curves.
Formation of ordinary differential equations, solution of homogeneous differential equations, variables separable
method, linear first order differential equations.

Vectors
Addition of vectors, scalar multiplication, scalar products, dot and cross products, scalar triple products and their
geometrical interpretations.


1
Complex Numbers
Topic 1 Complex Number in Iota Form
Objective Questions I (Only one correct option)
2z − n
1. Let z ∈ C with Im (z ) = 10 and it satisfies
= 2i − 1
2z + n

for some natural number n, then (2019 Main, 12 April II)
(a) n = 20 and Re(z ) = − 10 (b) n = 40 and Re(z ) = 10
(c) n = 40 and Re(z ) = − 10 (d) n = 20 and Re(z ) = 10


α + i
2. All the points in the set S = 
: α ∈ R (i = −1 ) lie

α − i
on a
(2019 Main, 9 April I)
(a) circle whose radius is 2.
(b) straight line whose slope is −1.
(c) circle whose radius is 1.
(d) straight line whose slope is 1.

3. Let z ∈ C be such that|z|< 1. If ω =

5 + 3z
, then
5(1 − z )

3

1
x + iy
(i = −1 ), where x and y are real

3 

27
(2019 Main, 11 Jan I)
numbers, then y − x equals


(c) – 85

(d) – 91


π  3 + 2i sin θ
, π :
is purely imaginary 
 2  1 − 2i sin θ


5. Let A = θ ∈  −


Then, the sum of the elements in A is (2019 Main, 9 Jan I)

(a)

3π
4

(b)

5π
6

(c) π

(b)

6i
7. If 4

–3 i
3i

20

3

 3
−1  1 
(c) sin −1 
 (d) sin 

 3
4



π
6

1
–1 = x + iy, then

i

(1998, 2M)

(a) x = 3, y = 1 (b) x = 1, y = 1 (c) x = 0, y = 3 (d) x = 0, y = 0
13

(a) i

(b) 5 Re (ω) > 1
(d) 5 Re(ω) > 4

(b) 85

π
3

2 + 3i sin θ
is purely imaginary, is
1 − 2i sin θ
(2016 Main)

∑ (i n + i n + 1 ), where i =

−1, equals

n =1

4. Let  −2 − i =
(a) 91

(a)

8. The value of sum

(2019 Main, 9 April II)

(a) 4 Im(ω) > 5
(c) 5 Im (ω) < 1

6. A value of θ for which

(d)

2π
3

(b) i − 1

(1998, 2M)

(c) − i

(d) 0

n

 1 + i
 = 1, is
1 − i

9. The smallest positive integer n for which 
(a) 8
(c) 12

(b) 16
(d) None of these

(1980, 2M)

Objective Question II
(One or more than one correct option)
10. Let a , b, x and y be real numbers such that a − b = 1 and
y ≠ 0. If the complex number z = x + iy satisfies
 az + b
Im 
 = y, then which of the following is(are)
 z+1
possible value(s) of x?

(2017 Adv.)

(a) 1 − 1 + y2

(b) − 1 − 1 − y2

(c) 1 + 1 + y2

(d) − 1 +

1 − y2

Topic 2 Conjugate and Modulus of a Complex Number
Objective Questions I (Only one correct option)
1. The equation|z − i| = |z − 1|, i = −1, represents
1
(2019 Main, 12 April I)
2
(b) the line passing through the origin with slope 1
(c) a circle of radius 1
(d) the line passing through the origin with slope − 1
(a) a circle of radius

2. If a > 0 and z =
equal to
1 3
(a)
− i
5 5
1 3
(c) − + i
5 5

(1 + i )2
2
, has magnitude
, then z is
a−i
5
(2019 Main, 10 April I)

1
(b) − −
5
3
(d) − −
5

3
i
5
1
i
5


2 Complex Numbers
3. Let z1 and z2 be two complex numbers satisfying| z1 | = 9
and | z2 − 3 − 4i | = 4. Then, the minimum value of
(2019 Main, 12 Jan II)
| z1 − z2|is
(a) 1

(b) 2

(c)

(d) 0

2

z −α
4. If
(α ∈ R) is a purely imaginary number and
z+α
(2019 Main, 12 Jan I)
|z| = 2, then a value of α is
(a) 2

1
(b)
2

(c) 1

(where i = − 1).
Then,| z |is equal to
34
3

(b)

(2019 Main, 11 Jan II)

5
3

41
4

(c)

(d)

5
4

6. A complex number z is said to be unimodular, if z ≠ 1.
z – 2 z2
is
If z1 and z2 are complex numbers such that 1
2 – z1z2
unimodular and z2 is not unimodular.
Then, the point z 1 lies on a

(2015 Main)

(a) straight line parallel to X-axis

(d) circle of radius 2

7. If z is a complex number such that |z| ≥ 2, then the
1
2

(c) is strictly greater than 5/2

(x − x0 )2 + ( y − y0 )2 = r 2 and (x − x0 )2 + ( y − y0 )2 = 4r 2,
respectively.
If z0 = x0 + iy0 satisfies the equation 2|z0|2 = r 2 + 2, then

(2013 Adv.)
|α |is equal to
(c)

1
7

(d)

1
3

9. Let z be a complex number such that the imaginary part

of z is non-zero and a = z + z + 1 is real. Then, a cannot
take the value
(2012)
2

(b)

1
3

(c)

1
2

(d)

3
4

10. Let z = x + iy be a complex number where, x and y are
integers. Then, the area of the rectangle whose vertices
are the root of the equation zz3 + zz3 = 350, is
(2009)
(a) 48
(c) 40

 1 
1
2
(c)
(d)
⋅
2
z
+
1
|z + 1|2

 |z + 1|

1
|z + 1|

2

14. For all complex numbers z1 , z2 satisfying |z1| = 12 and
|z2 − 3 − 4i| = 5, the minimum value of|z1 − z2|is
(a) 0
(c) 7

(b) 2
(d) 17

(2002, 1M)

15. If z1 , z2 and z3 are complex numbers such that
1
1
1
+ = 1, then |z1 + z2 + z3|is
|z1| = |z2| = |z3| =  +
z1 z2 z3
(a) equal to 1
(c) greater than 3

(b) less than 1
(d) equal to 3

(2000, 2M)

(a) n1 = n2 + 1
(c) n1 = n2

(b) n1 = n2 − 1
(d) n1 > 0, n2 > 0

complex
numbers
sin x + i cos 2x
cos x − i sin 2x are conjugate to each other, for
(b) x = 0
(d) no value of x

and

(1988, 2M)

vertices of a parallelogram taken in order, if and only if

8. Let complex numbers α and 1 /α lies on circles

(a) − 1

(b)

18. The points z1 , z2, z3 and z4 in the complex plane are the

(d) is strictly greater than 3/2 but less than 5/2

1
2

(a) 0

z −1

(where, z ≠ − 1), then Re (w) is
z+1
(2003, 1M)

(a) x = nπ
(c) x = (n + 1/2) π

(b) lies in the interval (1, 2)

(b)

13. If|z| = 1 and w =

17. The
(2014 Main)

(a) is equal to 5/2

1
2

(b)| z | = 1 and z ≠ 1
(d) None of these

(1 + i )n 1 + (1 + i3 )n1 + (1 + i5 )n 2 + (1 + i7 )n 2 , here
(1996, 2M)
i = −1 is a real number, if and only if

(c) circle of radius 2

(a)

(a)| z | = 1, z ≠ 2
(c) z = z

16. For positive integers n1 , n2 the value of expression

(b) straight line parallel toY -axis

minimum value of z +

 w − wz 
condition that 
 is purely real, then the set of
 1−z 
values of z is
(2006, 3M)

(d) 2

5. Let z be a complex number such that | z | + z = 3 + i

(a)

12. If w = α + iβ, where β ≠ 0 and z ≠ 1, satisfies the

(b) 32
(d) 80

11. If|z|= 1 and z ≠ ± 1, then all the values of

(a) a line not passing through the origin
(b)|z|= 2
(c) the X-axis
(d) the Y-axis

(a) z1 + z4 = z2 + z3
(c) z1 + z2 = z3 + z4

(b) z1 + z3 = z2 + z4 (1983, 1M)
(d) None of these

19. If z = x + iy and w = (1 − iz ) / (z − i ), then |w| = 1 implies
that, in the complex plane

(1983, 1M)

(a) z lies on the imaginary axis (b) z lies on the real axis
(c) z lies on the unit circle
(d) None of these

20. The inequality |z − 4| < |z − 2| represents the region
given by

(1982, 2M)

(a) Re (z ) ≥ 0
(c) Re (z ) > 0

(b) Re (z ) < 0
(d) None of these

5

5

 3 i
 3 i
21. If z = 
+  +
−  , then
2
2


 2 2
(a) Re (z ) = 0
(c) Re (z ) > 0, Im (z ) > 0

(1982, 2M)

(b) Im (z ) = 0
(d) Re (z ) > 0, Im (z ) < 0

22. The complex numbers z = x + iy which satisfy the
z
lie on
1 − z2
(2007, 3M)

z − 5i 
 = 1, lie on

equation
z + 5i 
(a) the X-axis
(b) the straight line y = 5
(c) a circle passing through the origin
(d) None of the above

(1981, 2M)


Complex Numbers 3
29. Let z be any point in A ∩ B ∩ C and let w be any point

Objective Questions II
(One or more than one correct option)

satisfying |w − 2 − i| < 3.
between

23. Let S be the set of all complex numbers z satisfying

(a) − 6 and 3
(c) − 6 and 6

| z 2 + z + 1| = 1. Then which of the following statements
is/are TRUE?
(2020 Adv.)

1 1
≤ for all z ∈S

2 2
1 1
(c) z +
≥ for all z ∈S
2 2

(a) z +

solutions z = x + iy (x, y ∈ R, i = − 1 ) of the equation
sz + tz + r = 0, where z = x − iy. Then, which of the
following statement(s) is (are) TRUE?
(2018 Adv.)

(a) If L has exactly one element, then| s|≠ |t |
(b) If|s|=|t |, then L has infinitely many elements
(c) The number of elements in L ∩ {z :| z − 1 + i| = 5} is at most 2
(d) If L has more than one element, then L has infinitely many
elements

25. Let z1 and z2 be complex numbers such that z1 ≠ z2 and

30. Let z be any point in A ∩ B ∩ C.
The|z + 1 − i|2 + |z − 5 − i|2 lies between
(a) 25 and 29
(c) 35 and 39
(a) 0
(c) 2

that |z1| = |z2| = 1 and Re (z1z2) = 0, then the pair of
complex numbers w1 = a + ic and w2 = b + id satisfies

(1985, 2M)

Passage Based Problems

32. Match the statements of Column I with those of
Column II.
Here, z takes values in the complex plane and Im (z )
and Re (z ) denote respectively, the imaginary part and
(2010)
the real part of z

A.

The set of points z satisfying
| z − i| z|| = | z + i | z|| is
contained in or equal to

p.

an ellipse with
eccentricity 4/5

B.

The set of points z satisfying
| z + 4| + | z − 4| = 0 is
contained in or equal to

q.

the set of points z
satisfying Im ( z) = 0

C.

If| w| = 2 , then the set of
1
points z = w − is contained
w
in or equal to

r.

the set of points z
satisfying|Im( z) |≤ 1

D.

If| w| = 1, then the set of points s.
1
z = w + is contained in or
t.
w
equal to

Passage I

27. min|1 − 3i − z|is equal to
z ∈s

2− 3
(a)
2

2+ 3
(b)
2

3− 3
(c)
2

3+ 3
(d)
2

28. Area of S is equal to
(a)

10 π
3

(b)

20 π
3

(c)

16 π

3

(d)

32 π
3

Column II

Column I

Read the following passages and answer the questions
that follow.
Let A, B, C be three sets of complex number as defined
below
A = { z : lm (z ) ≥ 1}
B = { z :|z − 2 − i| = 3}
(2008, 12M)
C = { z : Re((1 − i )z ) = 2 }

(b) 1
(d) ∞

Match the Columns

(b) real and positive
(d) purely imaginary

26. If z1 = a + ib and z2 = c + id are complex numbers such

(b) 30 and 34
(d) 40 and 44

31. The number of elements in the set A ∩ B ∩ C is

|z1| = |z2|. If z1 has positive real part and z2 has negative
z + z2
imaginary part, then 1
may be
(1986, 2M)
z1 − z2

(b)|w2| = 1
(d) None of these

(b) − 3 and 6
(d) − 3 and 9

Let S = S1 ∩ S 2 ∩ S3 , where


z − 1 + 3 i 
S1 = { z ∈ C :|z | < 4}, S 2 = z ∈ C : lm 
 > 0
 1− 3i 


and S3 : { z ∈ C : Re z > 0}
(2008)

(d) The set S has exactly four elements

(a)|w1| = 1
(c) Re (w1 w2 ) = 0

lies

Passage II

(b)|z|≤ 2 for all z ∈S

24. Let s, t, r be non-zero complex numbers and L be the set of

(a) zero
(c) real and negative

Then, |z | − |w| + 3

the set of points
satisfying|Re( z)|≤ 2
the set of points z
satisfying| z| ≤ 3

Fill in the Blanks
33. If α , β, γ are the cube roots of p, p < 0, then for any x, y
and z then

xα + yβ + zγ
= ... .
xβ + yγ + zα

(1990, 2M)

34. For any two complex numbers z1 , z2 and any real
numbers a and b,|az1 − bz2|2+ |bz1 + az2|2 = K .
(1988, 2M)


4 Complex Numbers


 x
 x
sin  2 + cos  2 − i tan (x)

 is real,
35. If the expression

 x 
1 + 2 i sin  2 


then the set of all possible values of x is… .
(1987, 2M)

41. If z1 and z2 are two complex numbers such that
|z1| < 1 < |z2|, then prove that

1 − z1z2
< 1.

z1 − z2

(2003, 2M)

42. For complex numbers z and w, prove that

| z |2 w − |w|2 z = z − w, if and only if z = w or z w = 1.
(1999, 10M)

43. Find all non-zero complex numbers z satisfying

True/False

z = iz 2.

36. If three complex numbers are in AP. Then, they lie on a
circle in the complex plane

(1985 M)

37. If the complex numbers, z1 , z2 and z3 represent the
vertices of an equilateral triangle
| z1 | = | z2| = | z3 |, then z1 + z2 + z3 = 0.

such

that

(1984, 1M)

38. For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, we
write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. Then, for all complex
1−z
numbers z with 1 ∩ z, we have
∩ 0.
(1981, 2M)
1+ z

Analytical & Descriptive Questions
39. Find the centre and radius of the circle formed by all the

points represented by z = x + iy satisfying the relation
z − α 


= k (k ≠ 1), where α and β are the constant
z −β
complex numbers given by α = α 1 + iα 2, β = β1 + iβ 2.
(2004, 2M)

40. Prove that there exists no complex number z such that
|z| < 1 /3 and

n

∑

44. If

(1996, 2M)

iz + z − z + i = 0,
3

2

then

show

that

|z| = 1.

(1995, 5M)

45. A relation R on the set of complex numbers is defined
z1 − z2
is real.
z1 + z2
Show that R is an equivalence relation.

by z1 R z2, if and only if

(1982, 2M)

46. Find the real values of x and y for which the following
equation is satisfied
(1 + i ) x − 2i (2 − 3i ) y + i
+

= i.
3+ i
3−i

47. Express

( 1980, 2M)

1
in the form A + iB.
(1 − cos θ ) + 2i sin θ
(1979, 3M)

48. If x + iy =

a + ib
a 2 + b2
, prove that (x2 + y2)2 = 2
c + id
c + d2
(1978, 2M)

Integer & Numerical Answer Type Question
49. If z is any complex number satisfying | z − 3 − 2i | ≤ 2,
then the maximum value of|2z − 6 + 5i |is …… (2011)

a rz r = 1, where|a r|< 2.

r =1

(2003, 2M)

Topic 3 Argument of a Complex Number
Objective Questions I (Only one correct option)
1. If z1 , z2 are complex numbers such that Re(z1 ) = |z1 − 1|,
Re(z2) = |z2 − 1| and arg(z1 − z2) =

π
, then Im(z1 + z2) is
6

equal to
(a)

(2020 Main, 3 Sep II)

3
2

(b)

1
3

(c)

2
3

(d) 2 3

2. Let z1 and z2 be any two non-zero complex numbers such
that 3|z1| = 4|z2|. If z =
(a) |z| =

1 17
2 2

3z1 2z2
+
, then
2z2 3z1
(2019 Main, 10 Jan I)

(d) |z| =

5
2

(a) − θ

(b)

π
−θ
2

(c) θ

(2000, 2M)

(b) − π
(d) π /2

5. Let z and w be two complex numbers such that|z| ≤ 1,
|w| ≤ 1 and|z + i w|= |z − iw| = 2 , then z equals
(1995, 2M)

(a) 1 or i
(c) 1 or −1

(b) i or −i
(d) i or −1

6. Let z and w be two non-zero complex numbers

such that |z| = |w| and arg (z ) + arg (w) = π, then z
(1995, 2M)
equals
(b) −w
(d) −w

7. If z1 and z2 are two non-zero complex numbers such

3. If z is a complex number of unit modulus and argument
 1 + z
θ, then arg 
 is equal to
1 + z

(a) π
(c) − π/2

(a) w
(c) w

(b) Im(z ) = 0

(c) Re(z) = 0

4. If arg (z ) < 0 , then arg (−z ) − arg (z ) equals

(2013 Main)

(d) π − θ

that|z1 + z2| = |z1| + |z2|, then arg (z1 ) − arg (z2) is equal
to
(a) − π
(c) 0

π
(b) −
2
π
(d)
2

(1987, 2M)

Complex Numbers 5
8. If a , b, c and u , v, w are the complex numbers

10. Let z1 and z2 be two distinct complex numbers and let

representing the vertices of two triangles such that
c = (1 − r ) a + rb and w = (1 − r ) u + rv, where r is a
(1985, 2M)
complex number, then the two triangles

z = (1 − t ) z1 + tz2 for some real number t with 0 < t < 1. If
arg (w) denotes the principal argument of a non-zero
(2010)
complex number w, then

(a) have the same area
(c) are congruent

(a) |z − z1| + |z − z2|= |z1 − z2|(b) arg (z − z1 ) = arg (z − z2 )

(b) are similar
(d) None of these

(c)

Objective Questions II
(One or more than one correct option)
principal argument with − π < arg(z ) ≤ π . Then, which of
the following statement(s) is (are) FALSE ? (2018 Adv.)

π
, where i =
4

z − z1

z2 − z1

z2 − z1

=0

(d) arg (z − z1 ) = arg (z2 − z1 )

Match the Columns

9. For a non-zero complex number z, let arg(z ) denote the

(a) arg (−1 − i ) =

z − z1

11. Match the conditions/expressions in Column I with
statement in Column II (z ≠ 0 is a complex number)
Column II

Column I

−1

Re ( z) = 0
π
arg ( z) =
4

A.

(b) The
function
defined
by
f : R → (− π, π],
f (t ) = arg (−1 + it ) for all t ∈ R, is continuous at all
points of R, where i = −1.
(c) For any two non-zero complex numbers z1 and z2,
z 
arg  1  − arg (z1 ) + arg (z2 ) is an integer multiple of
 z2 

B.

p.

Re ( z2 ) = 0

q.

Im ( z2 ) = 0

r.

Re ( z2 ) = Im ( z2 )

Analytical & Descriptive Questions
12. |z| ≤ 1,|w|≤ 1, then show that
|z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)2

2π.
(d) For any three given distinct complex numbers z1 , z2 and
z3 , the locus of the point z satisfying the condition
 (z − z1 ) (z2 − z3 ) 
arg 
 = π, lies on a straight line.
 (z − z3 ) (z2 − z1 ) 

(1995, 5M)

13. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex
number such that the argument of (z − z1 ) / (z − z2) is
(1991, 4M)
π /4, then prove that|z − 7 − 9i| = 3 2.

Topic 4 Rotation of a Complex Number
Objective Questions I (Only one correct option)
1. Let S be the set of all complex numbers z satisfying
| z − 2 + i | ≥ 5. If the complex number z0 is such that

 1
1
is the maximum of the set 

: z ∈ S , then
| z0 − 1 |
|
z
−
1
|


the principal argument of
(a)

π
4

(b)

3π
4
5

4 − z0 − z0
is
z0 − z0 + 2 i
(c) −

π
2

(2019 Adv.)

(d)

π
2

5

 3 i
 3 i
+  +
−  . If R(z ) and I (z )
2
2
 2
 2

2. Let z = 

respectively denote the real and imaginary parts of z,
then
(2019 Main, 10 Jan II)
(a) R (z ) > 0 and I (z ) > 0
(c) R (z ) < 0 and I (z ) > 0

(b) I (z ) = 0
(d) R (z ) = − 3

3. A particle P starts from the point z0 = 1 + 2 i, where
i = −1. It moves first horizontally away from origin by

5 units and then vertically away from origin by 3 units
to reach a point z1. From z1 the particle moves 2 units
in the direction of the vector $i + $j and then it moves
π
in anti-clockwise direction on a
through an angle
2
circle with centre at origin, to reach a point z2. The point
(2008, 3M)
z2 is given by
(a) 6 + 7i

(b) −7 + 6i

(c) 7 + 6i

(d) − 6 + 7i

4. A man walks a distance of 3 units from the origin
towards the North-East (N 45° E) direction. From there,
he walks a distance of 4 units towards the North-West
(N 45° W) direction to reach a point P. Then, the
position of P in the Argand plane is
(2007, 3M)
(a) 3ei π/ 4 + 4 i
(b) (3 − 4 i ) ei π / 4
(c) (4 + 3 i )ei π / 4
(d) (3 + 4 i ) ei π / 4

5. The shaded region, where P = (−1, 0), Q = (−1 + 2 , 2 )

R = (−1 + 2 , − 2 ), S = (1, 0) is represented by (2005, 1M)
π
4
π
(b)|z + 1|< 2,| arg (z + 1)|<
2
π
(c)|z + 1|> 2,| arg (z + 1)|>
4
π
(d)|z − 1|< 2,| arg (z + 1)|>
2

Y

(a)|z + 1|> 2,| arg (z + 1)|<

Q
X′

P

S

O

X

R

Y′

π
is a fixed angle. If P = (cos θ ,sin θ ) and
2
Q = {cos(α − θ ),sin(α − θ )}, then Q is obtained from P by

6. If 0 < α <

(2002, 2M)

(a) clockwise rotation around origin through an angle α
(b) anti-clockwise rotation around origin through an angleα
(c) reflection in the line through origin with slope tan α
α
(d) reflection in the line through origin with slope tan
2


6 Complex Numbers
7. The complex numbers z1 , z2 and z3

satisfying
z1 − z3 1 − i 3
are the vertices of a triangle which is
=
z2 − z3
2
(2001, 1M)

(a) of area zero
(c) equilateral

(b) right angled isosceles
(d) obtuse angled isosceles

(2016 Adv.)

1
1 
and centre 
(a) the circle with radius
, 0 for
 2a 
2a
a > 0, b ≠ 0
1 
1
and centre  −
(b) the circle with radius −
, 0 for
 2a 
2a
a < 0, b ≠ 0
(c) the X-axis for a ≠ 0, b = 0
(d) the Y-axis for a = 0, b ≠ 0

3+i
and P = {W n: n = 1, 2, 3,... }.
2

1

Further H 1 = z ∈ C : Re (z ) > 
2

1
−


and H 2 = z ∈ C : Re (z ) <
, where C is the set of all

2 
complex numbers. If z1 ∈ P ∩ H 1, z2 ∈ P ∩ H 2 and O
represents the origin, then ∠ z1Oz2 is equal to

9. Let W =

π
6

(c)

14. Let bz + bz = c, b ≠ 0, be a line in the complex plane,

(1997C, 5M)

15. Let z1 and z2 be the roots of the equation z + pz + q = 0,
2



1
Suppose S = z ∈ C : z =
, t ∈ R, t ≠ 0, where
a
+
i
bt


i = − 1. If z = x + iy and z ∈ S, then (x, y) lies on

(b)

circle |z − 1| = 2 is 2 + 3i. Find the other vertices of
(2005, 4M)
square.

that c = z1b + z2b.

8. Let a , b ∈ R and a 2 + b2 ≠ 0.

π
2

13. If one of the vertices of the square circumscribing the

where b is the complex conjugate of b. If a point z1 is the
reflection of the point z2 through the line, then show

Objective Questions II
(One or more than one correct option)

(a)

Analytical & Descriptive Questions

2π
3

(2013 JEE Adv.)

(d)

5π
6

Fill in the Blanks
10. Suppose z1 , z2, z3 are the vertices of an equilateral
triangle inscribed in the circle |z| = 2. If z1 = 1 + i 3,
then z2 = K, z3 = … .
(1994, 2M)

11. ABCD is a rhombus. Its diagonals AC and BD intersect
at the point M and satisfy BD = 2 AC. If the points D and
M represent the complex numbers 1 + i and 2 − i
respectively, then A represents the complex number
…or…
(1993, 2M)

where the coefficients p and q may be complex numbers.
Let A and B represent z1 and z2 in the complex plane. If
∠ AOB = α ≠ 0 and OA = OB, where O is the origin prove
α
that p2 = 4q cos 2  .
 2
(1997, 5M)

16. Complex numbers z1 , z2, z3 are the vertices A , B, C
respectively of an isosceles right angled triangle with
right angle at C. Show that
(z1 − z2)2 = 2(z1 − z3 ) (z3 − z2).
(1986, 2 1 M)
2

17. Show that the area of the triangle on the argand

diagram formed by the complex number z , iz and z + iz
1
is |z|2.
2
(1986, 2 1 M)
2

18. Prove that the complex numbers z1 , z2 and the origin
form an equilateral triangle only if z12 + z22 − z1z2 = 0.
(1983, 2M)

19. Let the complex numbers z1 , z2 and z3 be the vertices of

an equilateral triangle. Let z0 be the circumcentre of the
triangle. Then, prove that z12 + z22 + z32 = 3z02. (1981, 4M)

Integer & Numerical Answer Type Questions
20. For a complex number z, let Re(z ) denote the real part of
z. Let S be the set of all complex numbers z satisfying
z 4 − |z|4 = 4 i z 2, where i = −1. Then the minimum
possible value of|z1 − z2|2, where z1 , z2 ∈ S with Re(z1 ) > 0
and Re(z2) < 0 is …… .
(2020 Adv.)
kπ 
 kπ 
 + i sin   , where
 7
 7
i = −1. The value of the expression

21. For any integer k, let α k = cos 
12

∑|α k + 1 − α k|

k =1

12. If a and b are real numbers between 0 and 1 such that
the points z1 = a + i , z2 = 1 + bi and z3 = 0 form an
equilateral triangle, then a = K and b = K . (1990, 2M)

3

∑|α 4k − 1 − α 4k − 2|

k =1

is
(2016 Adv.)


Complex Numbers 7

Topic 5 De-Moivre’s Theorem, Cube Roots and nth Roots of Unity
Objective Questions I (Only one correct option)

9. Let z1 and z2 be nth roots of unity which subtend a right
angled at the origin, then n must be of the form
(2001, 1M)
(where, k is an integer)

3

2π
2π 

+ i cos

 1 + sin
9
9  is
1. The value of 
 1 + sin 2π − i cos 2π 


9
9 

(a) 4k + 1

[2020 Main, 2 Sep I]

1
(a) − ( 3 − i )
2
1
(c) ( 3 − i )
2

1
(b) − (1 − i 3 )
2
1
(d) (1 − i 3 )
2

π
, then
2

3. If z =
to

(2019 Main, 10 April II)

(a) 1

(c) −1

(d) 0

(d)

5. Let z = cos θ + i sin θ . Then, the value of

π
3

∑ Im (z

θ = 2° is

2m − 1

) at

6. The minimum value of |a + bω + cω |, where a, b and c
are all not equal integers and ω (≠ 1) is a cube root of
unity, is
(2005, 1M)

(c) 1

(d) 0

7. If ω (≠ 1) be a cube root of unity and (1 + ω 2)n = (1 + ω 4 )n,
then the least positive value of n is
(a) 2

(b) 3

(c) 5

(2004, 1M)

(d) 6

1
3
, then value of the determinant
+i
2
2
1
1
1
1 −1 − ω 2 ω 2 is
(2002, 1M)
ω2
ω
1

8. Let ω = −

(a) 3 ω

(a) – 1

(b) 3 ω (ω − 1) (c) 3 ω2

3

(d) −128 ω2

B

are

respectively

(b) 1, 1



∑ sin

(c) 1, 0

(d) 3 ω (1 − ω)

(d) −1, 1

2 πk
2 πk 

– i cos
 is
7
7 

(b) 0

(c) – i

(1998, 2M)

(d) i

Match the Columns
2 kπ 
 2 kπ 
 ; k = 1, 2, …9.
 + i sin 
 10 
 10 

14. Let zk = cos 

Column II

P.

For each zk, there exists a z j such that
zk ⋅ z j = 1

(i)

True

Q.

There exists a k ∈ { 1, 2, … , 9 } such that
z1 ⋅ z = zk has no solution z in the set of
complex numbers

(ii)

False

R.

|1 − z1||1 − z2| … |1 − z9|
equal
10
9
2 kπ 
1 − ∑ cos 
 equals
 10 
k =1

(iii)

1

(iv)

2

S.

2

1
2

and

A

k =1

(2009)

1
(b)
3 sin 2°
1
(d)
4 sin 2°

(b)

(d) −i

(1998, 2M)

(c) 128 ω2

Column I

m =1

1
(a)
sin 2°
1
(c)
2 sin 2°

3

12. If ω (≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω,

(2019 Main, 9 Jan II)

(c) 0

(b) −128 ω

6

15

(a) 3

(a) 128 ω

13. The value of

z = 3 + 6iz081 − 3iz093 , then arg z is equal to
π
(b)
6

(c) i

(1995, 2M)

4. Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If
π
(a)
4

365

(1999, 2M)

(b) −1 + i 3

3

(a) 0, 1

(2019 Main, 8 April II)

(b) (−1 + 2i )

 1 i 3
+ 3 − +

2 
 2

(1 + ω − ω 2)7 is equal to

then

3 i
+ (i = −1 ), then (1 + iz + z5 + iz 8 )9 is equal
2
2
9

(d) 4k

11. If ω is an imaginary cube root of unity, then

(b) zw =

(c) zw = i

i 3

2 

334

is equal to

1− i
2
− 1+ i
(d) zw =
2

(a) zw = − i

 1
 2

(c) 4k + 3

10. If i = −1, then 4 + 5  − +
(a) 1 − i

2. If z and w are two complex numbers such that | zw| = 1
and arg(z ) − arg(w) =

(b) 4k + 2

(2011)

Codes
P

(a) (i)
(b) (ii)
(c) (i)
(d) (ii)

Q
(ii)
(i)
(ii)
(i)

R
(iv)
(iii)
(iii)
(iv)

S
(iii)
(iv)
(iv)
(iii)

Fill in the Blanks
2π
2π
+ i sin . Then
3
3
the number of distinct complex number z satisfying

15. Let ω be the complex number cos

ω
ω2
z+1
2
ω
1 = 0 is equal to ... .
z+ω
ω2
1
z+ω

(2010)


8 Complex Numbers
16. The value of the expression

19. If 1, a1 , a 2, ... , a n − 1 are the n roots of unity, then show

1 ( 2 − ω ) (2 − ω 2) + 2(3 − ω ) (3 − ω 2) + ...
+ (n − 1) ⋅ (n − ω ) (n − ω 2) ,
where, ω is an imaginary cube root of unity, is….

that (1 − a1 ) (1 − a 2) (1 − a3 ) K (1 − a n − 1 ) = n

(1984, 2M)

20. It is given that n is an odd integer greater than 3, but n

(1996, 2M)

True/False

is not a multiple of 3. Prove that x3 + x2 + x is a factor of
(1980, 3M)
(x + 1)n − xn − 1 .

21. If x = a + b, y = aα + bβ, z = aβ + bα , where α , β are

17. The cube roots of unity when represented on Argand
diagram form the vertices of an equilateral triangle.

complex cube
xyz = a3 + b3 .

roots

of

unity,

then

show

that

(1979, 3M)

(1988, 1M)

Integer & Numerical Answer Type Questions

Analytical & Descriptive Questions
18. Let a complex number α , α ≠ 1, be a root of the equation
z p + q − z p − zq + 1 = 0
where, p and q are distinct primes. Show that either
1 + α + α 2 + ... + α p − 1 = 0
or
1 + α + α 2 + ... + α q − 1 = 0
but not both together.

(2002, 5M)

22. Let ω ≠ 1 be a cube root of unity. Then the minimum of
the set {| a + bω + cω 2|2 : a , b, c distinct non-zero
integers} equals ............
(2019 Adv.)

23. Let ω = eiπ /3 and a , b, c, x, y, z be non-zero complex
numbers such that a + b + c = x, a + bω + cω 2 = y,
a + bω 2 + cω = z.
| x|2 + | y|2 + | z |2
Then, the value of
(2011)
is ……
| a |2 + | b|2 + | c|2

Answers
Topic 1
1. (c)
5. (d)
9. (d)

2. (c)
6. (d)
10. (b, d)

3. (b)
7. (d)

4. (a)
8. (b)

1
cot (θ / 2 )
−i
θ


1 + 3 cos2 (θ / 2 )
2 1 + 3 cos2 

2

49. 5

Topic 3
1. (d)

Topic 2
1. (b)

47. A + iB =

2. (b)

3. (d)

4. (d)

5. (c)
9. (a, b, d)

2. (*)
6. (d)
10. (a, c, d)

3. (c)

4. (a)

7. (c)

8. (b)

11. A → q ; B → p

5. (b)

6. (c)

7. (b)

8. (c)

9. (d)

10. (a)

11. (d)

12. (b)

13. (a)

14. (b)

15. (a)

16. (d)

1. (c)

2. (b)

3. (d)

4. (d)

17. (d)

18. (b)

19. (b)

20. (d)

5. (a)

6. (d)

7. (c)

8. (a,c,d)

21. (b)

22. (a)

23. (b,c)

24. (a, c, d)

25. (a,d)

26. (a, b, c)

27. (c)

28. (b)

29. (d)

30. (c)

31. (b)

32. A → q, r ; B → p; C → p, s, t ; D → q, r, s, t
33. ω

2

34. (a + b )(| z1| + | z 2| )
2

2

2

2

−1

35. x = 2nπ + 2α , α = tan k, where k ∈(1, 2 ) or x = 2nπ
36. False

37. True

38. True
k (α − β )
α − k 2β

39. Centre =
, Radius = 
2
2
1 −k
 1 −k 

Topic 4

9. (c, d)
10. z 2 = − 2, z 3 = 1 − i 3
i
3i
11. 3 − or 1 −
12. a = b = 2 ± 3
2
2
13. z 2 = − 3 i , z 3 = (1 − 3 ) + i and z 4 = (1 + 3 ) − i
20. (8)

21. (4)

Topic 5
1. (a)

2. (a)

3. (c)

3. (a)

5. (d)

6. (c)

7. (b)

8. (b)

9. (d)

10. (c)

11. (d)

12. (b)

14. (c)

15. (1)

13. (d)



3 i
43. z = i , ±
– 
2
2


 n (n + 1 )
16. 
 −n

2 

46. (x = 3 and y = −1)

23. (3)

2

17. True

22. (3)


Hints & Solutions
Topic 1 Complex Number in Iota Form
1. Let z = x + 10i, as Im (z ) = 10 (given).
Since z satisfies,
2z − n
= 2i − 1, n ∈ N ,

2z + n
∴ (2x + 20i − n ) = (2i − 1) (2x + 20i + n )
⇒ (2x − n ) + 20i = (− 2x − n − 40) + (4x + 2n − 20)i
On comparing real and imaginary parts, we get
2x − n = − 2x − n − 40 and 20 = 4x + 2n − 20
⇒
4x = − 40 and 4x + 2n = 40
⇒
x = − 10 and − 40 + 2n = 40 ⇒ n = 40
So,
n = 40 and x = Re (z ) = − 10
α+i
2. Let x + iy =
α −i
(α + i )2 (α 2 − 1) + (2α )i α 2 − 1  2α 
⇒ x + iy = 2
=
+
= 2
i
α +1
α2 + 1
α + 1  α 2 + 1

3

 3 + 2i sin θ   1 + 2 i sin θ 
 ×

 1 − 2i sin θ   1 + 2 i sin θ 

5. Let z = 

(rationalising the denominator)
=

2α
α2 −1
and y = 2
α +1
α2 + 1
2

2
 α 2 − 1
 2α 
Now, x + y =  2
 + 2

 α + 1
 α + 1
2

=

2

α 4 + 1 − 2α 2 + 4α 2 (α 2 + 1)2
= 2
=1

(α 2 + 1)2
(α + 1)2

⇒
x2 + y2 = 1
Which is an equation of circle with centre (0, 0) and
radius 1 unit.

3 − 4 sin θ + 8i sin θ
1 + 4 sin 2 θ
2

[Q a 2 − b2 = (a + b)(a − b) and i 2 = − 1]
 3 − 4 sin θ   8 sin θ 
=
 +
i
 1 + 4 sin 2 θ   1 + 4 sin 2 θ 

On comparing real and imaginary parts, we get
x=

2

As z is purely imaginary, so real part of z = 0
3 − 4 sin 2 θ
= 0 ⇒ 3 − 4 sin 2 θ = 0
∴
1 + 4 sin 2 θ
3

⇒
sin 2 θ =
4
3
⇒
sin θ = ±
2
Y
1
√3/2


α + i
So, S = 
; α ∈ R lies on a circle with radius 1.

α − i
X′

3. Given complex number

5 ω − 5 ω z = 5 + 3z
(3 + 5 ω )z = 5 ω − 5
…(i)
|3 + 5 ω||z| = |5 ω − 5|
[applying modulus both sides and |z1z2| = |z1||z2|]
Q
|z| < 1
[from Eq. (i)]
∴ |3 + 5 ω| > |5 ω − 5|

3

⇒
ω + > |ω − 1|
5

2
3

Let ω = x + iy, then  x +  + y2 > (x − 1)2 + y2

5
9
6
2
2
⇒
x +
+ x > x + 1 − 2x
25 5
1
16x 16
⇒
>
⇒ x > ⇒ 5x > 1
5
5
25
⇒
5 Re( ω ) > 1

y=sin θ

–π/2 –π/3
O π/3

5 + 3z
ω=
5(1 − z )

⇒
⇒
⇒

3

1   –1
x + iy 

=  − 2 − i =
(6 + i )


27
3   3
1
x + iy
⇒
=−
(216 + 108i + 18i 2 + i3 )

27
27
1
=−
(198 + 107i )
27
[Q (a + b)3 = a3 + b3 + 3a 2b + 3ab2, i 2 = − 1, i3 = − i]
On equating real and imaginary part, we get
x = − 198 and y = − 107
⇒ y − x = − 107 + 198 = 91

4. We have,

−1

2π/3

π

X

–√3/2

Y′

 π π 2π 
θ ∈ − , ,

 3 3 3 
2π

.
Sum of values of θ =
3
2 + 3i sin θ
6. Let z =
is purely imaginary. Then, we have
1 − 2i sin θ
Re(z ) = 0
2 + 3i sin θ
Now, consider z =
1 − 2i sin θ
(2 + 3i sin θ) (1 + 2i sin θ )
=
(1 − 2i sin θ ) (1 + 2i sin θ)
⇒

=

2 + 4i sin θ + 3i sin θ + 6i 2 sin 2 θ
12 − (2i sin θ) 2


10 Complex Numbers
=

2 + 7i sin θ − 6 sin 2 θ
1 + 4 sin 2 θ

=

2 − 6 sin 2 θ
7 sin θ
+i
1 + 4 sin 2 θ
1 + 4 sin 2 θ

Re(z ) = 0
2 − 6 sin 2 θ
= 0 ⇒ 2 = 6 sin 2 θ
1 + 4 sin 2 θ
1
sin 2 θ =
3
1
sin θ = ±
3
1
−1 
−1  1 
θ = sin  ±
 = ± sin  


 3
3

Q
∴
⇒
⇒

⇒

7. Given,

 6i
4
20

 6i
−3 i  4
 20

⇒

−3 i
3i
3

1
−1  = x + iy
i

1
−1
i

1
−1  = x + i y
i

⇒

x = 0, y = 0

13

8.

∑ (i

+i

n

n=1

n+1

13

) = ∑ i (1 + i ) = (1 + i )
n

n=1

1. Let the complex number z = x + iy
| z − i | =| z − 1|

Also given,
⇒

| x + iy − i | = | x + iy − 1|

⇒

x2 + ( y − 1)2 = (x − 1)2 + y2
[Q| z | = (Re(z ))2 + (Im(z ))2 ]

On squaring both sides, we get
x2 + y2 − 2 y + 1 = x2 + y2 − 2x + 1
⇒ y = x, which represents a line passing through the
origin with slope 1.
(1 + i )2
a−i
(1 − 1 + 2i ) (a + i )
=
a2 + 1
2i (a + i ) −2 + 2ai
=
=
a2 + 1
a2 + 1

2. The given complex number z =

x + iy = 0 [Q C 2 and C3 are identical]

⇒

Topic 2 Conjugate and Modulus of

Complex Number

13

∑

i

n

n =1

 i − (1 − i13 ) 
= (1 + i ) (i + i 2 + i3 + K + i13 ) = (1 + i ) 


 1−i
 i (1 − i ) 
= (1 + i ) 
 = (1 + i ) i = i − 1
 1−i 
Alternate Solution

4 + 4a 2
2
2
2
⇒
=
=

2
5
5
(a 2 + 1)2
1+ a
4
2
= ⇒ a 2 + 1 = 10 ⇒ a 2 = 9 ⇒ a = 3 [Qa > 0]
⇒
1 + a2 5
–2 + 6i
[From Eq. (i)]
∴
z=
10
1 3
 −2 + 6 i   1 3 
So, z = 
 =  − + i ⇒ z = − − i
 10   5 5 
5 5

+ (i 2 + i3 + K + i14 ) = i + i 2 = i − 1
n

 1 + i

 =1
1 − i

n

 1 + i 1 + i
×
⇒ 
 =1
 1 − i 1 + i

n

⇒

 2i 
  =1
 2

⇒
in = 1
The smallest positive integer n for which i n = 1 is 4.
∴
n =4
az + b ax + b + aiy (ax + b + aiy)((x + 1) − iy)
10.
=
=
z+1
(x + 1) + iy
(x + 1)2 + y2
∴

 az + b − (ax + b) y + ay(x + 1)
Im 
=
 z+1
(x + 1)2 + y2

⇒

(a − b) y
=y
(x + 1)2 + y2

Q
∴

a − b =1
(x + 1)2 + y2 = 1

∴

[Qif z = x + iy, then z = x − iy]

3. Clearly|z1|= 9, represents a circle having centre C1 (0, 0)

n=1

9. Since,

x = − 1 ± 1 − y2

[given]

⇒

13

∑ (i n + i n + 1 ) = (i + i 2 + K + i13 )

…(i)

z = 2 /5

Q

Since, sum of any four consecutive powers of iota is zero.
∴

[Q i 2 = − 1]

and radius r1 = 9.
and |z2 − 3 − 4i|= 4 represents a circle having centre
C 2(3, 4) and radius r2 = 4.
The minimum value of |z1 − z2| is equals to minimum
distance between circles|z1|= 9 and|z2 − 3 − 4i|= 4.
Q

C1C 2 = (3 − 0)2 + (4 − 0)2 = 9 + 16 = 25 = 5

and
|r1 − r2|=|9 − 4|= 5 ⇒ C1C 2 =|r1 − r2|

∴ Circles touches each other internally.
Hence,
|z1 − z2|min = 0
z −α
4. Since, the complex number
(α ∈ R) is purely
z+α
imaginary number, therefore
z −α
z −α
[Qα ∈ R]
+
=0
z+α z+α
⇒

zz − αz + αz − α 2 + zz − αz + αz − α 2 = 0

⇒

2 z

⇒

α2 = z

⇒

α=±2

2

− 2 α2 = 0
2

=4

2

[Qzz = z ]
[| z | = 2 given]


Complex Numbers 11
5. We have,|z | + z = 3 + i
∴

x2 + y2 + x + iy = 3 + i

⇒

(x + x + y ) + iy = 3 + i

⇒

x + x2 + y2 = 3 and y = 1

2

⇒

∴
⇒

= |z1|2 − z1z2 − z2z1 + |z2|2
Here, (x − x0 ) + ( y − y0 ) = r 2
2

|z − z0|2 = r 2 and |z − z0|2 = 4r 2
Since, α and
∴
⇒

(α − z0 ) (α − z0 ) = r 2

⇒

|α|2 − z0α − z0α + |z0|2 = r 2

⇒ (z1 − 2z2)(z1 − 2z2) = (2 − z1z2) (2 − z1z2)

[zz = |z|2 ]

⇒ |z1|2+4|z2|2−2z1z2 − 2z1z2

⇒


 1
1

2
 − z0   − z0  = 4 r

 α
α

⇒

z
z
1
− 0 − 0 + |z0|2 = 4r 2
α
|α|2 α
|α|2 = α ⋅ α

Since,

= 4+|z1|2|z2|2−2z1z2 − 2z1z2 ⇒ (|z2|2−1)(|z1|2−4) = 0
Q
|z2|≠ 1
∴
|z1|= 2
Let
z1 = x + iy ⇒ x2 + y2 = (2)2
∴ Point z1 lies on a circle of radius 2.

⇒

z ⋅α

z
1
− 0
− 0 ⋅ α + |z0|2 = 4r 2
|α|2 |α|2 |α|2
1 − z0α − z0α + |α|2|z0|2 = 4r 2|α|2

⇒

(|α|2 − 1) + |z0|2 (1 − |α|2 ) = r 2 (1 − 4|α|2 )

(0, 0) and radius is 2.
1
is distance of z, which lie on circle
Minimum z +
2
| z | = 2 from (−1 / 2, 0).
1
 1 
= Distance of  − , 0 from (−2, 0)
∴ Minimum z +
 2 
2

⇒

(|α|2 − 1) (1 − |z0|2 ) = r 2(1 − 4|α|2 )

⇒


r 2 + 2
(|α|2 − 1) 1 −
 = r 2(1 − 4|α|2 )
2 

|z0|2 =

Given,

2

1
3
3
 −1

+ 2 + 0 =
 +0= = 
 2

2
2
2

⇒

|α|2 − 1 = − 2 + 8|α|2 ⇒ 7|α|2 = 1
|α| = 1 / 7

∴

9.
A
1 , (0,0)
(– —
)
2 0

(2,0)

1
= AD
2
1
Hence, minimum value of z +
lies in the interval
2
(1, 2).

Geometrically Min z +

PLAN If ax 2 + bx + c = 0 has roots α, β, then
α, β =

X

Y′

r2 + 2

2

 − r 2
(|α|2 − 1) ⋅ 
 = r 2(1 − 4|α|2 )
 2 

⇒

Y

D
(–2,0)

…(ii)

On subtracting Eqs. (i) and (ii), we get

7. |z| ≥ 2 is the region on or outside circle whose centre is

X′

…(i)

1
− z0 = 4 r 2
α

and

z1 − 2z2
= 1 ⇒ |z1 − 2z2|2 =|2 − z1z2|2
2 − z1z2

2

1
lies on first and second respectively.
α
2
1
− z0 = 4 r 2
|α − z0|2 = r 2 and
α

2

Given, z2 is not unimodular i.e.|z2|≠ 1
z − 2 z2
is unimodular.
and 1
2 − z1z2


=  −2 +


2

(x − x0 )2 + ( y − y0 )2 = 4r 2 can be written as,

and

PLAN If z is unimodular, then| z| = 1. Also, use property of modulus
i.e. z z =| z|2

⇒

|z|2 = z ⋅ z
|z1 − z2|2 = (z1 − z2) (z1 − z2)

and

x2 + 1 = 9 − 6x + x2
4
6x = 8 ⇒ x =
3
4
z= +i
3
5
16
25
|z | =
+1=
⇒ |z | =
9
9
3

⇒

PLAN Intersection of circles, the basic concept is to solve the
equations simultaneously and using properties of modulus of
complex numbers.

Formula used

2

x2 + 1 = 3 − x

Now,

6.

8.

z = x + iy

Let

− b ± b 2 − 4ac
2a

For roots to be real b 2 − 4ac ≥ 0.

Description of Situation
z = x + iy is non-zero.
⇒

y ≠0

As imaginary part of

Method I Let z = x + iy
∴

a = (x + iy)2 + (x + iy) + 1

⇒

(x2 − y2 + x + 1 − a ) + i (2xy + y) = 0

⇒

(x2 − y2 + x + 1 − a ) + iy (2x + 1) = 0,

…(i)


12 Complex Numbers
It is purely real, if y (2x + 1) = 0

On squaring both sides, we get
1 + |w|2 − 2|w| Re (w) = 1 + |w|2 + 2|w| Re (w)

but imaginary part of z, i.e. y is non-zero.

[using|z1 ± z2|2 = |z1|2 + |z2|2 ± 2|z1||z2| Re (z1z2)]

⇒

2x + 1 = 0 or x = − 1 / 2
1
1
From Eq. (i),
− y2 − + 1 − a = 0
4
2
3
2 3
a =− y +
⇒ a<
⇒
4
4
Method II

4|w|Re|w| = 0

⇒

Re (w) = 0

14. We know, |z1 − z2| = |z1 − (z2 − 3 − 4i ) − (3 + 4i )|
≥ |z1| − |z2 − 3 − 4i | − |3 + 4i|

Here, z 2 + z + (1 − a ) = 0

∴

z=

− 1 ± 1 − 4 (1 − a )
2 ×1

⇒

z=

− 1 ± 4a − 3
2

For z do not have real roots, 4 a − 3 < 0 ⇒

≥ 12 − 5 − 5
∴

Alternate Solution

a<

3
4

Clearly from the figure|z1 − z2|is minimum when z1 , z2
lie along the diameter.
Y

2

2

2

,
(3

x2 − y2 = 7
x2 = 16, y2 = 9 ⇒ x = ± 4, y = ± 3

Y′

∴

|z1 − z2| ≥ C 2B − C 2A ≥ 12 − 10 = 2
|z1| = |z2| = |z3| = 1
|z1| = 1

Now,

11. Let z = cos θ + i sin θ
z
cos θ + i sin θ
=
1 − z 2 1 − (cos 2 θ + i sin 2 θ )
cos θ + i sin θ
=
2 sin 2 θ − 2i sin θ cos θ
cos θ + i sin θ

i
=
=
− 2i sin θ (cos θ + i sin θ ) 2 sin θ
z
lies on the imaginary axis i.e. Y-axis.
Hence,
1 − z2

⇒

Alternate Solution
z
z
1
Let E =
which is an imaginary.
=
=
1 − z 2 zz − z 2 z − z
w − wz
12. Let z1 =
be purely real ⇒ z1 = z1
1−z
w − wz w − wz
=
∴
1− z
1−z
w − wz − wz + wz ⋅ z = w − zw − wz + wz ⋅ z

⇒
⇒ (w − w ) + (w − w)| z |2 = 0
(w − w ) (1 − | z |2 ) = 0
⇒
| z |2 = 1 [as w − w ≠ 0, since β ≠ 0]

⇒

| z | = 1 and z ≠ 1
z −1
13. Since,|z| = 1 and w =
z+1
|1 + w|
1+ w
⇒ z − 1 = wz + w ⇒ z =
⇒ |z| =
|1 − w|
1−w
|1 − w| = |1 + w|

[ Q|z| = 1]

X

12

C2

15. Given,

∴ Area of rectangle = 8 × 6 = 48

Z1

Z2

4)

C1

X′

... (ii)

From Eqs. (i) and (ii),

⇒

B

A

2 (x + y ) (x − y ) = 350
(x2 + y2) (x2 − y2) = 175
2

Since, x, y ∈ I, the only possible case which gives
integral solution, is
... (i)
x2 + y2 = 25

⇒

[using|z1 − z2| ≥ |z1| − |z2|]

|z1 − z2| ≥ 2

zz (z 2 + z 2) = 350

10. Since,
⇒
⇒

⇒

⇒

|z1|2 = 1

⇒

z1z1 = 1

Similarly,

z2z2 = 1, z3 z3 = 1

1 + 1 + 1 
= 1
z1 z2 z3

Again now,

⇒ | z1 + z2 + z3 |= 1 ⇒ |z1 + z2 + z3|= 1
⇒
|z1 + z2 + z3| = 1

16. (1 + i )n1 + (1 − i )n1 + (1 + i )n2 + (1 − i )n2
= [n1 C 0 +

n1

C1i +

+ [n1 C 0 −
+ [n2 C 0 +
+ [n2 C 0 –
= 2 [n1 C 0 +

n1

= 2 [ C0 −

n1

n1

C 2i 2 +

C3 i3 + K ]

C1 i + n1 C 2i 2 − n1 C3 i3 + ... ]
C1i + n2C 2i 2 + n2C3 i3 + K ]
n2
C1 i + n2C 2i 2 – n2C3 i3 + .. ]
n1

n2

C2 i2 +

n1

C 4i 4 + K ]

+ 2 [n2 C 0 +
n1

n1

C2 +

n1

n2

C 2i 2 +

n2

C 4i 4 + K ]

C4 − K ] + 2 [ C0 −
n2

n2

+

C2
n2
C 4 −... ]

This is a real number irrespective of the values of n1 and
n2.
Alternate Solution
{(1 + i )n1 + (1 − i )n1 } + {(1 + i )n2 + (1 − i )n2 }
⇒ A real number for all n1 and n2 ∈ R.
[Q z + z = 2 Re (z ) ⇒ (1 + i )n1 + (1 − i )n1
is real number for all n ∈ R]


Complex Numbers 13
17. Since, (sin x + i cos 2x) = cos x − i sin 2x
⇒

sin x − i cos 2x = cos x − i sin 2x

⇒

sin x = cos x and cos 2x = sin 2x

⇒

tan x = 1 and tan 2x = 1

⇒ x = π / 4 and x = π / 8 which is not possible at same
time.

Alternate Solution
We know that,
z + z = 2 Re(z )
5
5
 3 i
 3 i
If
z=
+  +
−  , then
2
2
 2
 2
z is purely real. i.e. Im (z ) = 0
z − 5i 
 = 1 ⇒ |z − 5i| = |z + 5i|
z + 5i 

22. Given, 

Hence, no solution exists.

[Q if|z − z1| = |z − z2|, then it is a perpendicular
bisector of z1 and z2 ]

18. Since, z1 , z2, z3 , z4 are the vertices of parallelogram.
D(z4)

C(z3)

Y

(0, 5)
O

X′

X

(0, –5)
Y′

A(z1)

B(z2)

∴ Perpendicular bisector of (0, 5) and (0, – 5) is X-axis.

∴ Mid-point of AC = mid-point of BD

z1 + z3 z2 + z4
=
⇒
2
2
⇒
z1 + z3 = z2 + z4

23. It is given that the complex number Z, satisfying
|z 2 + z + 1| = 1

 1 − iz 
= 1
z−i 

19. Since,|w| = 1 ⇒ 
⇒

⇒

|z − i| = |1 − iz|

|z − i | = |z + i |


z +


⇒
Q

|z1 − z2| ≥ ||z1| − |z2||

∴

1
1
3

 3
− −
z +  − −  ≥ z +




2
4
2
4

2

[Q |1 − iz | = | − i || z + i | = | z + i |]

∴It is a perpendicular bisector of (0, 1) and (0, − 1)
i.e. X-axis. Thus, z lies on the real axis.

2

2

⇒

z+

20. Given,|z − 4| < |z − 2|
Since, |z − z1| > |z − z2| represents the region on right
side of perpendicular bisector of z1 and z2.
∴
|z − 2| > |z − 4|
⇒
Re (z ) > 3 and Im (z ) ∈ R
Y

X′

O

(2, 0) (3, 0) (4, 0)

5

 3 i
 3 i
+  +
− 
2
2
 2

 2

X

⇒

1
3
− ≤1
2
4
2

⇒

−

1
1
7
≤ z+
≤
4
2
4

 −1 + i 3 
3+i
= −i 
 = − iω

2
2


 −1 − i 3 
3−i
=i
 = iω 2
2
2


z = (− iω )5 + (iω 2)5 = − iω 2 + iω
= i(ω − ω 2) = i (i 3 ) = − 3
Re(z ) < 0 and lm (z ) = 0

0≤ z+

1
7
≤
2
4

⇒

0≤ z+

1
7

≤
2
2

Q

z+

1
7
≤
2
2

5


−1 + i 3
−1 − i 3 
and ω 2 =

Q ω =
2
2



∴

−1 ≤ z +

2

21. Given, z = 

and

1
3
− ≤1
2
4
2

⇒

⇒

Y′

Now,

2

1
3
 + =1

2
4

Q
∴

|z1 + z2| ≥ ||z1| − |z2||
1
1
z + ≥ |z| −
2
2
1
1
7
≤ z+ ≤
2
2
2

⇒

|z| −

⇒

7
1
7
≤ |z| − ≤
2
2

2
1− 7
7+1
≤ |z| ≤
2
2
1+ 7
|z| ≤
2
|z| ≤ 2

⇒
⇒
∴

−

{Q|z| ≥ 0}
… (i)


14 Complex Numbers

z +


Q

2

1≤ z+

2

⇒

|w2| = b2 + d 2 = a 2 + b2 = 1

1
3
1
3
+ ≥1
+
⇒ z+
2
4
2
4

1
1
z+
≥
2
4

|w1| = a 2 + c2 = a 2 + b2 = 1

Now,

2

2

⇒

Also, given w1 = a + ic and w2 = b + id

2

1
3
1
3
+
 + ≤ z+
2
4
2
4

⇒

1 1
z+ ≥
2 2

… (ii)

and Re(w1 w2) = ab + cd = (bλ )b + c(− λc) [from Eq. (i)]
= λ (b2 − c2) = 0

27. min|1 − 3 i − z|= perpendicular distance of point (1, − 3)
Z ∈S

From Eqs. (i) and (ii), we get
1
1
7
≤ z+ ≤
2
2
2

from the line

3x + y = 0 ⇒

28. Since, S = S1 ∩ S 2 ∩ S3

24. We have,
sz + tz + r = 0
On taking conjugate
sz + tz + r = 0
On solving Eqs. (i) and (ii), we get
rt − rs
z= 2
|s| − |t|2

Y

…(i)
…(ii)

150°

X′

For unique solutions of z
|s|2 − |t|2 ≠ 0 ⇒ |s| ≠ |t|
It is true
(b) If|s| = |t|, then rt − rs may or may not be zero.
So, z may have no solutions.
∴ L may be an empty set.
It is false.
(c) If elements of set L represents line, then this line
and given circle intersect at maximum two point.
Hence, it is true.
(d) In this case locus of z is a line, so L has infinite
elements. Hence, it is true.

O

25. Given,|z1| = |z2|

y = –3√x

∴

S=

1 2
1
5π 20π
r θ = × 42 ×
=
2
2
3
6

|w − (2 + i )| < 3 ⇒ |w| − |2 + i| < 3

29. Since,
⇒

−3 + 5 < |w| < 3 + 5

⇒

−3 − 5 < − |w| < 3 − 5

…(i)

|z − (2 + i )| = 3

Also,

−3 + 5 ≤ |z| ≤ 3 + 5

⇒
∴

…(ii)

−3 < |z| − |w| + 3 < 9
= (x + 1)2 + ( y − 1)2 + (x − 5)2 + ( y − 1)2
= 2(x2 + y2 − 4x − 2 y) + 28
= 2(4) + 28 = 36

[Q|z1|2 = |z2|2 ]

[Q x2 + y2 − 4x − 2 y = 4]

31. Let z = x + iy
Set A corresponds to the region y ≥ 1

…(i)

Set B consists of points lying on the circle, centre at
(2, 1) and radius 3.
i.e.
…(ii)
x2 + y 2 − 4 x − 2 y = 4

As, we know z − z = 2i Im (z )
∴

∴

30. |z + 1 − i| + |z − 5 − i|2

|z |2 + (z2 z1 − z1 z2) − |z2|2
= 1
|z1 − z2|2
z2z1 − z1z2
|z1 − z2|2

Y′

Clearly, the shaded region represents the area of sector

2

z1 + z2 z1 − z2 z1z1 − z1z2 + z2z1 − z2 z2
=
×
z1 − z2 z1 − z2
|z1 − z2|2

=

X
(4, 0)

(a)

Now,

| 3 − 3| 3 − 3
=
2
3+1

z2z1 − z1z2 = 2i Im (z2z1 )
z1 + z2 2i Im (z2 z1 )
=
z1 − z2
|z1 − z2|2

Set C consists of points lying on the x + y = 2

…(iii)

Y

which is purely imaginary or zero.

26. Since, z1 = a + ib and

P

z2 = c + id

⇒ |z1|2 = a 2 + b2 = 1 and |z2|2 = c2 + d 2 = 1

(0,√2)

…(i)

[Q|z1|=|z2| = 1]
Also, Re (z1z2) = 0 ⇒ ac + bd = 0
a
d
⇒
=− =λ
b
c
From Eqs. (i) and (ii), b2λ2 + b2 = c2 + λ2c2
⇒
b2 = c2 and a 2 = d 2

X′

(√2,0)

(2,1)

y=1

X

[say]…(ii)
Y′

Clearly, there is only one point of intersection of the line
x + y = 2 and circle x2 + y2 − 4x − 2 y = 4.

Complex Numbers 15
z = x + iy

32. A. Let

y x 2 + y2 = 0

⇒ we get
⇒

y = 0 ⇒ Im (z ) = 0

B. We have
2ae = 8, 2a = 10 ⇒ 10e = 8
4
16

e = ⇒ b2 = 25 1 −  = 9

5
25

⇒

C. Let
∴

z = 2 (cos θ + i sin θ ) −
= 2 (cos θ + i sin θ ) −
=

⇒

2

⇒

x2
y2
+
=1
9 / 4 25 / 4
e= 1−

⇒
Let
and
Then,
and

9 /4
4
=
25 / 4 5
1
cos θ + i sin θ

x = 2 cos θ , y = 0

ω 2 (x + yω + z ω 2)

ω 2 (xω + yω 2 + z )
ω 2 (x + yω + zω 2)
= ω2
x + yω + zω 2

34. |az1 − bz2| + |bz1 + az2|

2

= [a 2|z1|2 + b2|z2|2 − 2ab Re (z1z2)]
+ [b2|z1|2 + a 2|z2|2 + 2ab Re (z1z2)]
= (a 2 + b2) (|z1|2 + |z2|2 )

35.

... (i)

x

 

2 x
2 x
 tan + 1 1 − tan  + 1 + tan  = 0

 
2
2 
2

xα + yβ + z γ
x( p)1/3 + y( p)1/3 ω + z ( p)1/3 ω 2
33.
=
xβ + yγ + zα x( p)1/3 ω 2 + y( p)1/3 ω3 + z ( p)1/3 ω

2

x = 2 nπ

x
x 
x

2x
2 x
sin + cos   cos − sin  + cos = 0




2
2
2
2
2

x
On dividing by cos3 , we get
2

= 2 cos θ

=

x
=0
2

1
(cos θ − i sin θ )
2

z = x + iy = cos θ + i sin θ +

⇒

sin

or

w = cos θ + i sin θ

Then,

x 
x
x 
x
2x

2 x
sin + cos   cos − sin  + cos  = 0
2 
2
2 
2
2
2

1
2 (cos θ + i sin θ )

2

⇒

D. Let

⇒ sin

3
5
cos θ + i sin θ
2
2

 2x
 2 y
  +   =1
 3

5

∴

2 sin

⇒

z = x + iy
3
5
x = cos θ and y = sin θ
2
2

Let

x
x
x
x
x
sin + cos  cos x + 2 sin cos = 0
2
2
2
2
2

⇒

∴

x2 y2
+
=1
25 9
w = 2 (cos θ + i sin θ )

∴

Since, it is real, so imaginary part will be zero.
x
x
x
−2 sin sin + cos  − tan x = 0
∴
2
2
2

x
x

sin + cos  − i tan x

2
2
∈R
x

1 + 2 i sin
2
x
x
x


sin + cos − i tan x 1 − 2i sin 



2
2
2
=
x
1 + 4 sin 2
2

tan3

x
x
− tan − 2 = 0
2
2
x
tan = t
2
f (t ) = t3 − t − 2

f (1) = − 2 < 0
f (2) = 4 > 0

Thus, f (t ) changes sign from negative to positive in the
interval (1, 2).
∴ Let t = k be the root for which
f (k) = 0 and k ∈(1, 2)
x
∴
t = k or tan = k = tan α
2
⇒
x/2 = nπ + α
x = 2nπ + 2α , α = tan −1 k, where k ∈ (1, 2)
⇒

or x = 2nπ


36. Since, z1 , z2, z3 are in AP.
⇒
2z2 = z1 + z3
i.e. points are collinear, thus do not lie on circle. Hence,
it is a false statement.

37. Since, z1 , z2, z3 are vertices of equilateral triangle and
|z1| = |z2| = |z3|
⇒ z1 , z2, z3 lie on a circle with centre at origin.
⇒ Circumcentre = Centroid
z + z2 + z3

⇒
0= 1
3
∴
z1 + z2 + z3 = 0

38. Let z = x + iy ⇒ 1 ∩ z gives 1 ∩ x + iy
or
Given,
⇒

1 ≤ x and 0 ≤ y
1−z
1 − x − iy
∩0 ⇒
∩0
1+ z
1 + x + iy
(1 − x − iy) (1 + x − iy)
∩ 0 + 0i
(1 + x + iy) (1 + x − iy)

…(i)


16 Complex Numbers
⇒

1 − x2 − y2
2iy

−
∩ 0 + 0i
2
2
(1 + x) + y
(1 + x)2 + y2

⇒

⇒

x2 + y2 ≥ 1 and −2 y ≤ 0

⇒

⇒

or x2 + y2 ≥ 1 and y ≥ 0 which is true by Eq. (i).

39. As we know,|z| = z ⋅ z
|z − α|2
= k2
|z − β|2

⇒

1 − |z1|2 − |z2|2+ | z1|2|z2|2 < 0

2

2


1 − z1z2
< 1
 z1 − z2 

⇒

⇒ |z| − αz − αz + |α| = k (|z| − βz − β z + |β| )
2

2

2

Hence proved.

42. Given,|z|2 w − |w|2 z = z − w

⇒ |z|2 (1 − k2) − (α − k2β )z − (α − β k2) z
+ (|α|2 − k2|β|2 ) = 0
|α|2 − k2|β|2
(α − k2β )
(α − β k2)
z+
z−
= 0 …(i)
2
2

(1 − k2)
(1 − k )
(1 − k )

⇒

zz w − ww z = z − w

[Q |z|2 = zz ] …(i)

Taking modulus of both sides, we get
|zw||z − w| = |z − w|

On comparing with equation of circle,

⇒

|zw||z − w| = |z − w|

|z| + az + az + b = 0
whose centre is (− a ) and radius = |a|2 − b

⇒

| zw|| z − w | = |z − w |

2

∴ Centre for Eq. (i) =

α −kβ
1 − k2
2

 α − k β  α − k β  αα − k ββ
and radius = 
 −
 
1 − k2
 1 − k2   1 − k2 
2

2

2

∴
⇒

1
3
|a1z + a 2z 2 + a3 z3 + K + a nz n| = 1
|a1z| + |a 2z 2| + |a3 z3| + K + |a nz n| ≥ 1

⇒

2{(|z| + |z| + |z| + K + |z|n } > 1

|z| <

…(i)

[using|z1 + z2| ≤ |z1| + |z2|]

⇒
⇒
⇒
⇒
⇒

[using|a r| < 2]

3

2|z|(1 − |z|n )
>1
1 − |z|

[using sum of n terms of GP]

2|z| − 2|z|n + 1 > 1 − |z|
3|z| > 1 + 2|z|n + 1
1 2
|z| > + |z|n + 1
3 3
1
|z| > , which contradicts
3

|zw| − 1 = 0

⇒

|z − w| = 0

or

|zw| = 1

⇒

z − w=0

or

|z w|= 1

⇒

z=w

or

|zw| = 1




z1
 |z1|

using z = |z |


2
2 

…(ii)

(1 − z1z2)(1 − z1z2) < (z1 − z2)(z1 − z2) [using|z|2 = zz ]

1 iφ
e
r

On putting these values in Eq. (i), we get
1
1
 1
r 2 ei φ  − 2 (reiθ ) = reiθ − eiφ
r
 r
r
1 iθ
1 iφ
iφ
iθ
re − e = re − e
⇒
r
r

1  iφ 
1  iθ

⇒
=
+
r
+
e
r
 e





r
r

NOTE

…(i)

[say]

z = reiθ and w =

Let

⇒

r =1

|1 − z1z2| < |z1 − z2|
⇒
On squaring both sides, we get,

or

…(ii)

n


1 − z1z2
< 1
 z1 − z2 

|z − w| = 0

Therefore,

|z| < 1 / 3 and ∑ a rz r = 1

Then, to prove

|z − w|(|zw| − 1) = 0

⇒

∴ There exists no complex number z such that

41. Given,|z1| < 1 and |z2| > 1

⇒
⇒

Then,|zw| = 1 or|z||w| = 1
1
|z| =
=r
⇒
|w|

40. Given, a1z + a 2z 2 + K + a nz n = 1

2

[∴ |z| = |z| ]

Now, suppose z ≠ w

k(α − β ) 

=
2
1−k 

and

…(iii)

∴ Eq. (iii) is true whenever Eq. (ii) is true.

(z − α )(z − α ) = k2(z − β ) (z − β )

⇒ |z|2 −

1 + |z1|2|z2|2 <|z1|2 + |z2|2

⇒
(1 − |z1|2 )(1 − |z2|2 ) < 0
which is true by Eq. (i) as|z1| < 1 and|z2| > 1
∴
(1 − |z1|2 ) > 0 and (1 − |z2|2 ) < 0

2

Given,

1 − z1z2 − z1z2 + z1z1z2z2 < z1z1 − z1z2 − z2z1 + z2z2

eiφ = eiθ ⇒ φ = θ
1
z = reiθ and w = eiθ
r
1
zw = reiθ . e−iθ = 1
r

‘If and only if ’ means we have to prove the relation in
both directions.

Conversely
Assuming that z = w or z w = 1
If
z = w, then
LHS = zz w − w wz = |z|2⋅z − |w|2⋅z
= |z|2⋅z − |z|2⋅z = 0
and
RHS = z − w = 0
If
zw = 1, then zw = 1 and
LHS = zz w − ww z = z ⋅ 1 − w ⋅ 1
= z − w = z − w = 0 = RHS
Hence proved.