Complex Numbers Arihant

8/11/2019 Complex Numbers Arihant

1/22

1. (a)z z

z z

=12

1 locus of point representing z is the

perpendicular bisector of line joining z1and z2.

(b) | | Re ( ) | | , | | Im( ) | |z z z z z z(c) If a and b are real numbers and z z1 2, are complex

numbers, then

| | | |az bz bz az 1 22

1 2

2

+ + = + +( )(| | | | )a b z z 2 2

1

2

2

2

2. If z z z1 2 3, , are three distinct complex numbers a b c, , are

three positive real numbers such that

a

z z

b

z z

c

z z| | | | | |2 3 3 1 1 2 =

=

, then

a

z z

b

z z

c

z z

2

2 3

2

3 1

2

1 2

0

+

+

=

3. If a b c ii i i, , ( , , )= 1 2 3 are complex numbers and

z

a b c

a b c

a b c

=1 1 1

2 2 2

3 3 3

, then z

a b c

a b c

a b c

=1 1 1

2 2 2

3 3 3

4. If cos cos cos sin + + = + + =sin sin 0, then(a) cos ( ) cos ( ) cos ( )2 2 2 0 + + =(b) sin ( ) sin ( ) sin ( )2 2 2 0 + + =(c) cos ( ) cos ( ) cos ( ) + + + + + = 0(d) sin ( ) sin ( ) sin ( ) + + + + + = 0(e) sin sin sin /2 2 2 3 2 + + =

(f) cos cos cos /2 2 2 3 2 + + = (g) sin ( ) sin ( ) sin ( )3 3 3 + + = + +3sin ( ) (h) cos ( ) cos ( ) cos ( )3 3 3 + + = + +3cos ( )

5. If cos cos cos ,A B C+ + = 0 sin sin sinA B C+ + = 0, then

(a) A B B C C A = = =2

3

2

3

2

3

, ,

(b) 2 0 2 0A B C B C A = =, ,2 0C A B =6. If l A m B n C cos cos cos+ + = 0,

l A m B n C sin sin sin+ + = 0, then(a) l A m B n C 3 3 33 3 3cos cos cos+ +

= + +3lmn A B C cos ( )(b) l A m B n C 3 3 33 3 3sin sin sin+ +

= + +3lmn A B C sin ( )

7. Distance If z1and z2are two complex numbers, then the

distance between z z1 2, is | |z z1 2 .8. The equation of the line[in para metric form] joiningz1and

z2is given by z tz t z= + 1 21( ) .Where tis a real parameter.

The equation of the line[non-parametric form] joining z

andz2is given by

z z

z z

z z

z z

=

1

2 1

1

2 1

or

z z

z z

z z

1

1

1

01 1

2 2

=

9. The general equation of a lineisaz az b+ + = 0, where bia real number.

10. Complex slope of a straight line If A z B z( ), ( )1 2 are two

points in the argand plane, then the complex slope ()of the

straight lineABis given by =

z z

z z

1 2

1 2

11. Parallel and perpendicular lines Two lines having

complex slopes 1and 2which are(a) parallel iff 1 2=(b) perpendicular iff

1 2= or 1 2 0+ =

12. An gle between two lines If A z B z C z D z( ), ( ), ( ), (1 2 3 4

are four points in the argand plane, then the angle between

the linesABandCDis given by =

arg

z z

z z

1 2

3 4

13. Collinear points z z z1 2 3, , are collinear iff

(a)

z z

z z

z z

1 1

2 2

3 3

1

1

1

0=

(b) z z

z z

3 1

2 1

is purely real

(c) arg ( )z z2 1 =arg ( )z z3 1(d) Three complex numberz z z1 2 3, , are in AP.

14. Reflection of a line Reflection of the line in the real axis isaz az + = 0Reflection of the line az az + = 0 in the imaginary axis isalso az az + = 0

Complex Numbers

1


2/22

15. Length of the perpendicular from a point to a line The

length of the per pendicular from a point z1 to the line

az az b+ + = 0is given

byaz az b

a

1 1

2

+ +| |

16. Section formulae

(a) Two pointsPandQhave affixesz1andz2respectively in

the argand plane, then the affix of a pointRdividingPQ

in the ratio m n:

(i) internally ismz nz

m n

2 1++

(ii) externally ismz nz

m n

2 1

(b) IfRis the mid point ofPQ, then affix ofRisz z1 2

2

+.

17. Area of the triangle with vertices z z1 2, and z3 is

( )| |z z ziz

2 3 12

14

sq unit.

18. Equilateral triangle The triangle whose ver tices are the

pointsz z z1 2 3, , on the argand plane is an equilateral triangleiff

z z z z z z z z z12

22

32

1 2 2 3 3 1+ + = + +

or1 1 1

01 2 2 3 3 1z z z z z z

+

+

=

19. If the complex numbers z z z1 2 3, , be the ver tices of an

equilateral triangle and if z0 be the circumcentre of the

triangle, then z z z z12

22

32

023+ + =

20. Centroid The centroid of the tri angle (in the argand

plane), formed byz z z1 2 3, , is given by1

31 2 3( )z z z+ +

21. Incentre The incentre of the triangle (in the argand plane),

formed by z z z1 2 3, , isaz bz cz

a b c

1 2 3+ ++ +

where

a z z b z z c z z = = = | |, | |, | |2 3 3 1 1 222. Excentres The excentres of the triangle (in the argand

plane), formed by z z z1 2 3, , are

given by,

(i)I az bz cz

a b c1

1 2 3= + + + +

(ii)I az bz cz

a b c2

1 2 3= + +

(iii) I az bz cz

a b c3

1 2 3= + +

,

where a z z b z z c z z = = = | |, | |, | |2 3 3 1 1 223. Circumcentre The circumcentre of the tri angle (in the

Argand plane), formed by

z z z1 2 3, , is given by z z z z

z z z

1 1 2 3

1 2 3

( )

( )

24. Orthocentre The orthocentre of the tri angle (in the argand

plane), formed byz z z1 2 3, , is given by

+

z z z z z z

z z z z

12

2 3 12

2 3

1 2 1 2

( ) | | ( )

( )

25. Parallelogram z z z z1 2 3 4, , , are vertices of a

parallelogram iff z z z z1 3 2 4+ = +26. Square If z z z z1 2 3 4, , , are the ver tices of a square in tha

order, then

(a) z z z z 1 3 2 4+ = +(b) | | | | | | | |z z z z z z z z

1 2 2 3 3 4 4 1 = = = (c) | | | |z z z z1 3 2 4 =

(d)( )

( )

z z

z z

1 3

2 4

is purely imaginary.

Circle

27. The equa tion of a circle having centre z0and radius r is

| |z z r =0 or zz z z z z z z r + =0 0 0 02 0

28. The equation of the circle described on the line segmen

joining z1andz2as diameter is

( ) ( ) ( )( )z z z z z z z z + =1 2 2 2 0or | | | | | |z z z z z z + = 1

22

21 2

2

29. The gen eral equation of a circle iszz az az b+ + + = 0where bis a real number

The centre of the circle is a and its radius is aa b

30. | | | | ( )z z z z k k R + = 12

22 will represent a cir cle if

k z z 12

1 22| |

31. Concyclic points Four points z z z1 2 3, , and z4 are

concyclic if and only if( )( )

( )( )

z z z z

z z z z

1 3 2 4

1 4 2 3

is purely real.

32. Locus of z(in the argand plane) such that

(a) | | | |z z z z k + =1 2 is an ellipse, if k z z> | |1 2 is astraight line, if k z z= | |1 2

(b) | | | |z z z z k =1 2 is an ellipse, if k z z< | |1 2 is astraight line, if k z z= | |1 2(c) | | | | | |z z z z z z + = 1

22

21 2

2is a circle (with z1and

z2as ends of a diameter)

(d) z z

z zk k

= 12

1( ) is a circle

(e) argz z

z za

=1

2

fixed angle, is a circle.

33. The least value of | | | |z a z b + is | |a b .

34. IfPrepresents a complex number z in the argand diagram

and OP is rotated through an angle and Q is the newposition of P, then the complex number represented byQis

z i + (cos sin ).35. amp amp( ) ( )z n zn = 36. Product of the nth roots of any complex number z is

z n( ) 1 1.

37. If zz

a a+ =1 , is positive real number, then

+ +

+ +a az

a a2 24

2

4

2| |

2 | Chapter 1 Complex Numbers


3/22

1. The smallest positive integer nfor which1

11

+

=

i

i

n

, is(1980, 2M)

(a) 8 (b) 16

(c) 12 (d) None of these

2. The complex numbers z x iy= + which satisfy the equationz i

z i

+

=5

51, lie on (1981, 2M)

(a) thex-axis

(b) the straight line y= 5(c) a circle passing through the origin

(d) None of the above

3. If z i i= +

+

3

2 2

3

2 2

5 5

, then

(1982, 2M)

(a) Re ( )z = 0 (b) Im ( )z = 0(c) Re ( ) , Im ( )z z> >0 0 (d) Re ( ) , Im ( )z z> 0 (d) None of these

5. If z x iy= + and w iz z i= ( ) / ( )1 , then | |w = 1 impliesthat, in the complex plane (1983, 1M)

(a)zlies on the imaginary axis

(b)zlies on the real axis

(c)zlies on the unit circle

(d) None of the above

6. The pointsz z z z1 2 3 4, , , in the complex plane are the ver ti ces

of a parallelogram taken in order, if and only if (1983, 1M)

(a) z z z z1 4 2 3+ = + (b) z z z z1 3 2 4+ = +(c) z z z z1 2 3 4+ = + (d) None of these7. If a b c, , and u v w, , are the complex numbers representing the

verti ces of two triangles such that c r a rb= +( )1 andw r u rv= +( )1 , where r is a complex number, then thetwo triangles (1985, 2M)

(a) have the same area (b) are similar

(c) are congruent (d) None of these

8. The value of k

ki

k

=

1

6 2

7

2

7sin cos

is

(1987, 2M)

(a) 1 (b) 0 (c) i (d) i9. If z1 and z2 are two non-zero complex numbers such that

| | | | | |z z z z1 2 1 2+ = + , then arg ( ) ( )z z1 2 arg is equal to(1987, 2M)

(a) (b) 2

(c) 0 (d)2

10. The complex numbers sin cosx i x+ 2 and cos sinx i x 2are conju gate to each other, for (1988, 2M)

(a)x n= (b)x= 0(c)x n= +( / )1 2 (d) no value ofx

11. If ( ) 1 is a cube root of unity and ( )1 7+ = + A B , thenAandBare respectively (1995, 2M

(a) 0, 1 (b) 1, 1

(c) 1, 0 (d) 1, 1

12. Let z and w be two non-zero complex numbers such tha| | | |z w= and arg ( )z + arg ( )w = , thenzequals (1995, 2M(a) w (b) w(c) w (d) w

13. Letzand wbe two complex numbers such that | |z 1, | |w and | |z iw+ = | |z iw = 2, thenzequals (1995, 2M)(a) 1 or i (b) ior i(c) 1 or 1 (d) ior 1

14. For positive integers n n1 2, the value of expression

( ) ( ) ( ) ( )1 1 1 11 1 2 23 5 7+ + + + + + +i i i in n n n , here i= 1 is areal number, if and only if (1996, 2M

(a) n n1 2 1= + (b) n n1 2 1=

(c) n n1 2= (d) n n1 20 0> >,15. If is an imaginary cube root of unity, then ( )1 2 7+ i

equal to (1998, 2M

(a) 128 (b) 128 (c) 128 2 (d) 128 2

16. The value of sum n

n ni i=

++1

131( ), where i= 1 equals

(1998, 2M

(a) i (b) i 1(c) i (d) 0

17. If

6 3 1

4 3 1

20 3

i i

i

i

x iy

,= + then

(1998, 2M

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

18. If i= 1, then 4 5 12

3

23

1

2

3

2

334 365

+ +

+ +

i i is

equal to (1999, 2M

(a) 1 3i (b) +1 3i(c) i 3 (d) i 3

19. If arg ( )z < 0, then arg ( ) ( ) z zarg equals (2000, 2M(a) (b) (c) / 2 (d) / 2

20. If z z1 2, and z3 are complex numbers such tha

| | | | | |z z zz z z

1 2 31 2 3

1 1 1= = = + +

= 1, then | |z z z1 2 3+ + is

(2000, 2M

(a) equal to 1 (b) less than 1

(c) greater than 3 (d) equal to 3

21. Let z1 and z2 be nth roots of unity which sub tend a righ

angled at the origin, then nmust be of the form (where kis an

integer) (2001, 1M

(a) 4 1k+ (b) 4 2k+ (c) 4 3k+ (d) 4k

Chapter 1 Complex Numbers | 3

Objective Questions I[Only one correct option]


4/22

22. The complex numbers z z1 2, and z3 satisfying

z z

z z

i1 3

2 3

1 3

2

= are the vertices of a tri angle which is(2001, 1M)

(a) of area zero

(b) right-angled isosceles

(c) equilateral

(d) obtuse-angled isosceles

23. Let = +12

3

2i , then value of the determinant

1 1 1

1 1

1

2 2

2

is

(2002, 1M)

(a) 3 (b) 3 1 ( )(c) 3 2 (d) 3 1 ( )

24. For all complex numbers z z1 2, satisfying | |z1 12= and| |z i2 3 4 5 = , the minimum value of | |z z1 2 is

(2002, 1M)

(a) 0 (b) 2(c) 7 (d) 17

25. If | |z = 1and w zz

= +

1

1 (wherez 1), then Re ( )w is

(2003, 1M)

(a) 0 (b)1

12| |z+

(c)1

1

1

12z z+

+| | (d)

2

12| |z+

26. If ( )1 be a cube root of unity and ( ) ( )1 12 4+ = + n n,then the least positive value of nis (2004, 1M)

(a) 2 (b) 3

(c) 5 (d) 6

27. The minimum value of | |,a b c+ + 2 where a, band careall not equal integers and ( ) 1 is a cube root of unity, is

(2005, 1M)

(a) 3 (b)1

2

(c) 1 (d) 0

28. The shaded region, where P Q= = +( , ), ( , )1 0 1 2 2 R= + ( , )1 2 2 , S= ( , )1 0 is represented by (2005, 1M)

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


5/22

1. Ifz a ib1= + andz c id2= + are complex numbers such that| | | |z z1 2 1= = and Re ( )z z1 2 0= , then the pair of complexnumbers w a ic1= + and w b id 2= + satisfies (1985, 2M)(a) | |w1 1= (b) | |w2 1=(c) Re ( )w w1 2 0= (d) None of these

2. Let z1 and z2 be complex numbers such that z z1 2 and| | | |z z1 2= . If z1 has positive real part and z2 has negative

imaginary part, thenz z

z z

1 2

1 2

+

may be (1986, 2M)

(a) zero (b) real and positive

(c) real and negative (d) purely imaginary

3. Let z1 and z2 be two distinct complex numbers and le

z t z tz= +( )1 1 2 for some real number twith 0 1<


6/22

1. If the expression

sin cos tan ( )

sin

x xi x

i x

2 2

1 2

2

+

+

is real, then

the set of all possible values ofxis . (1987, 2M)

2. For any two complex numbersz z1 2, and any real numbers a

and b, | | | |az bz bz az 1 22

1 22 + + = K . (1988, 2M)

3. If aand bare real numbers between 0 and 1 such that the

points z a i z bi1 2 1= + = +, and z3 0= form an equilateraltriangle, then a= Kand b= K . (1990, 2M)

4. ABCDis a rhombus. Its diagonalsACandBDintersect at the

point M and satisfy BD AC= 2 . If the points D and Mrepresent the complex numbers 1+iand 2i respectivelythenArepresents the complex number or . (1993, 2M

5. Suppose z z z1 2 3, , are the vertices of an equilateral triangle

inscribed in the circle | | .z = 2If z i1 1 3= + , then z2= Kz3= . (1994, 2M

6. The value of the ex pression

1 ( ) ( ) ( ) ( ) ...2 2 2 3 32 2 + + + ( ) ( ) ( )n n n 1 2 ,

whereis an imaginary cube root of unity, is . (1996, 2M

1. For complex numbers z x iy1 1 1= + and z x iy2 2 2= + , wewrite z z1 2 , ifx x1 2 and y y1 2 . Then, for all complex

numberszwith 1 z, we have 11

0+

zz (1981, 2M)

2. If the complex numbers, z z1 2, and z3represent the vertices

of an equilateral tri angle such that | | | | | |,z z z1 2 3= = thenz z z1 2 3 0+ + = . (1984, 1M

3. The cube roots of unity when represented on Argand diagram

form the verti ces of an equi lateral triangle. (1988, 1M

1. Ifx iy a ib

c id+ = +

+, prove that

( )x y a b

c d

2 2 22 2

2 2+ = +

+

(1978, 2M)

2. If x a b= + , y a b= + , z a b= + , where , arecomplex cube roots of unity, show thatxyz a b= +3 3.

(1979, 3M)

3. Express1

1 2( cos ) sin + iin the formA iB+ .

(1979, 3M)

4. It is given that nis an odd integer greater than 3, but nis not a

multiple of 3. Prove that x x x3 2+ + is a factor of( )x xn n+ 1 1. (1980, 3M)

5. Find the real values of x and y for which the following

equation is satisfied

( ) ( )

.1 2

3

2 3

3

+ +

+ +

=i x ii

i y i

ii

( 1980, 2M)

6. Let the complex numbers z z1 2, and z3be the vertices of an

equilateral triangle. Let z0 be the circumcentre of the

triangle. Then prove that z z z z12

22

32

023+ + = . (1981, 4M)

7. A relation Ron the set of complex numbers is defined by

z R z1 2, if and only ifz z

z z

1 2

1 2

+

is real.

Show thatRis an equivalence relation. (1982, 2M

8. Prove that the complex numbersz z1 2, and the origin form an

equilateral triangle only if z z z z12

22

1 2 0+ = . (1983, 2M

9. If 1, a a an1 2 1, , ....., are the nroots of unity, then show that( ) ( ) ( ) ( )1 1 1 11 2 3 1 =a a a a nnK (1984, 2M

10. Show that the area of the triangle on the Argand diagram

formed by the complex number z iz, and z iz+ is 12

2| |z

(1986, 21

2M

11. Complex numbers z z z1 2 3, , are the ver ti ces A B C, ,

re spectively of an isosceles right angled tri angle with righ

angle at C. Show that ( ) ( ) ( )z z z z z z1 22

1 3 3 22 = (1986, 2

1

2M

12. Letz i1 10 6= + andz i2 4 6= + . Ifzis any complex number

such that the argument of ( ) / ( )z z z z 1 2 is / 4 , thenprove that | |z i =7 9 3 2. (1991, 4M

13. If iz z z i3 2 0+ + = , then show that | |z = 1. (1995, 5M


Analytical & Descriptive Questions

True/False

Fill in the Blanks


7/22

14. | | , | | ,z w 1 1 show that

| | (| | | | ) ( )z w z w z w + 2 2 2arg arg (1995, 5M)15. Find all non-zero complex numberszsatisfying z iz= 2.

(1996, 2M)

16. Let z1and z2be the roots of the equation z pz q2 0+ + = ,

where the coefficientspand qmay be complex numbers.

Let A and B represent z1 and z2 in the complex plane. If

= AOB 0 and OA OB= , where O is the origin prove

that p q2 242

=

cos

. (1997, 5M)

17. Let bz bz c+ = , b 0, be a line in the complex plane, wherebis the complex conju gate of b. If a point z1is the reflection

of the pointz2through the line, then show that c z b z b= +1 2 .(1997C, 5M)

18. For complex numbers z and w, prove that

| | | | ,z w w z z w2 2 = if and only if z w= or z w = 1.

(1999, 10M)19. Let a complex number , 1, be a root of the equation

z z zp q p q+ + =1 0

wherep,qare distinct primes. Show that either

1 02 1+ + + + = ..... p

or 1 02 1+ + + + = ..... q

but not both together. (2002, 5M

20. Ifz1andz2are two complex numbers such that | | |z z1 21< 1 2 represents the region on right side ofperpendicular bisector of z1and z2.

| | | |z z > 2 4 Re( )z > 3 and Im( )z R

Hence, option (d) is correct.

5. Since, | |w = 1

1

iz

z i= = 1 1| | | |z i iz

| | | |z i z i = +( | |Q 1 iz= +| | | |i z i = +| |z i

It is a perpendicular bisector of ( , )0 1 and ( , )0 1 ie,x-axis

Thus,zlies on real axis.

6. Since,z z z z1 2 3 4, , , are the vertices of parallelogram.

Mid point of AC=mid point ofBD

z z z z1 3 2 42 2

+ = +

z z z z1 3 2 4+ = +


Hints & So utions

(0,5)

(0,5)

x

y

Ox'

y'

y

O (2,0) (3,0) (4,0) xx'

y'

A(z )1 B(z )2

D(z )4 C(z )3


9/22

7. Since a b c, , and u v w, , are the ver tices of two triangles.

Also, c r a rb= +( )1 and w r u rv= +( )1 (i)

Considera ub vc w

111

ApplyingR R r R rR3 3 1 21 + {( ) }

=

ab

c r a rb( )1

uv

w r u rv r r ( ) ( )1

11

1 1

=

ab

uv

0 0

110

[form Eq. (i)]

= 0Hence, two triangles are similar.

8. k

ki

k

=

1

6 2

7

2

7sin cos

= +

=

i k i kk 1

6 2

7

2

7cos sin

=

=

i ei k

k

27

1

6

= + +i e e ei i i{ /2 4 6 7 / 7 / 7

+ + +e e ei i i8 10 12 / 7 / 7 / 7 }

=

i e e

e

ii

i

212

2 7

1

1

/ 7 / 7

/( )

=

i e e

e

i i

i

2 14

21

/ 7 / 7

/ 7, (Q e i14 1 / 7= )

=

=i ee

ii

i

2

2

1

1

/ 7

/ 7

9. Given, | | | | | |z z z z1 2 1 2+ = +

On squaring both sides, we get

| | | | | || | cos ( )z z z z z z12

22

1 2 1 22+ + arg arg

= +| | | |z z12

22+ 2 1 2| | | |z z

2 1 2 1 2| | | | cos ( )z z z zarg arg = 2 1 2| | | |z z cos ( )arg argz z1 2 1 = arg arg( ) ( )z z1 2 0 =

10. Since, (sin cos ) cos sinx i x x i x+ = 2 2

sin cos cos sinx i x x i x = 2 2 sin cosx x= and cos sin2 2x x= tanx= 1and tan 2 1x= x= /4andx= /8which is not possible at same time.Hence, no solution exist.

11. ( ) ( ) ( )1 1 17 6+ = + +

= + ( ) ( )1 2 6

= +1 A B+ =1 + A B= =1 1,

12. Since, | | | |z w= and arg ( ) arg ( )z w=

Let w rei= , then w re i=

z re re ei i i= = ( )

= re i= w

13. Given, | | | |z iw z iw

+ = =2

| ( ) | | ( ) |z iw z iw = = 2 | ( ) | | ( ) |.z iw z iw = zlies on the perpendicular bisector of the line joining iwand iw. Since iwis the mirror image of iwin thex-axisthe locus ofzis thex-axis.

Let z x iy= + and y= 0.Now, | |z x + 1 0 12 2 1 1x .

zmay take values given in (c).Alternate Solution

| | | | | |z iw z iw+ += +| | | |z w

+ =1 1 2 | |z iw+ 2 | |z iw+ = 2 holds when

arg zarg iw= 0

arg ziw

= 0

ziw

is purely real

zw

is purely imaginary

Similarly, when | |z iw = 2,

thenz

w is purely imaginary

Now, given relation

| | | |z iw z iw+ = = 2put w i= , we get

| | | |z i z i+ = + =2 2 2 | |z =1 2 z= 1 ( | |Q z 1put w i= ,we get

| | | |z i z i = =2 2 2 | |z+ =1 2 z= 1 ( | |Q z 1z= 1or 1is the one correct option given.

14. ( ) ( ) ( ) ( )1 1 1 11 1 2 2+ + + + + i i i in n n n

= + + + +[ ]n n n nC C i C i C i1 1 1 10 1 22

33

K

+ + +[ ...]n n n nC C i C i C i1 1 1 10 1 22

33

+ + + + +[ ]n n n nC C i C i C i2 2 2 20 1 22

33

K

+ + +[ ..n n n nC C i C i C i2 2 2 20 1 22

33



10/22

= + + +2 1 1 10 22

44[ ]n n nC C i C i K

+ + + +2 2 2 20 22

44[ ]n n nC C i C i K

= + + 2 21 1 1 2 20 2 4 0 2[ ] [n n n n nC C C C C K

+ n C2 4 ...]

This is a real number irrespective of the values of n1and n2.Alternate Solution

{( ) ( ) } {( ) ( ) }1 1 1 11 1 2 2+ + + + + i i i in n n n

a real number for all n1and n R2 .[Qz z z+ = 2Re( ) ( ) ( )1 11 1+ + i in n is real number forall n R ]Hence, option (d) is the best option.

15. ( ) ( )1 2 7 2 2 7+ = ( )Q1 02+ + =

= = = ( ) ( )2 2 1282 7 7 14 2

16. n

n n

n

n

n

ni i i i i i=

+

= =+ = + = +

1

131

1

13

1

13

1 1( ) ( ) ( )

= + + + + + = +( ) ( ) ( )1 12 3 13i i i i i iK i ii

( )1

1

= + = +( )1 1i i iAlternate Solution

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

n

n ni i i i i=

++ = + + +1

131 2 13( ) ( )K

+ + + +( )i i i2 3 14K = +i i2 = i 1

17. Given,6420

333

11

i ii

ix iy

= +

= +3

6

420

1

1

1

1i

i

i i x i y

x iy+ = 0 (QC2and C3are identical) x y= =0 0, .

18. If in a complex number a ib+ ,the ratio a b: is 1 3: ,then italways convert the complex number in .

Since, = +12

3

2i

4 5 12

3

23

1

2

3

2

334 365

+ +

+ +

i i

= + +4 5 3334 365

= + + 4 5 33 111 3 121 2

( ) ( ) = + +4 5 3 2 (Q 3 1= ) = + + + +1 3 2 3 3 2

= + + + + = + + 1 2 3 1 1 2 3 02 ( )(Q 1 02+ + = )

= + + =1 1 3 3( )i i.19. Since, arg ( )z < 0

arg ( )z =

z r i= + cos ( ) sin ( ) = r i(cos sin )

And = z r i[cos sin ] = + r i[cos ( ) sin ( )] arg ( ) = z Thus, arg arg( ) ( ) z z

= = ( )Alternate Solution

Reason : arg ( ) z zarg

=

= =arg argz

z( )1

and also arg z z arg ( )

=

= =arg arg

z

z( )1

20. Given, | | | | | |z z z1 2 3 1= = =

Now, | |z1 1= | |z1

2 1=

z z1 1 1=Similarly, z z z z2 2 3 31 1= =,

Again now,1 1 1

11 2 3z z z

+ +

=

| |z z z1 2 3 1+ + =

| |z z z1 2 3 1+ + = | |z z z1 2 3 1+ + =

21. Since, argz

z

1

2 2=

zz

i i1

2 2 2= + =cos sin

zz

in

n

n1

2

= ( ) in = 1 ( | | | |Q z z2 1 1= =

n k= 4Alternate Solution

Since, argz

z

2

1 2=

zz

z

ze

i2

1

2

1

2=

zz

i2

1

= ( | | | | )Q z z1 2 1= =

zz in

n2

1 =

z1and z2are nth roots of unity z zn n1 2 1= =

zz

n

2

1

1

=

in = 1 n k= 4 , where k is an integer


y

(z)r

O

r

(z)

x

O A (z1

B (z )2

2


11/22

22. z z

z z

i i i

i

1 3

2 3

1 3

2

1 3 1 3

2 1 3

= = ++

( )( )

( )

= +

1 3

2 1 3

2i

i( )

=+

=+

4

2 1 3

2

1 3( ) ( )i i

z zz z

ii2 3

1 3

1 3

2 3 3

= + = +cos sin

z zz z

2 3

1 3

1

= and arg z z

z z

2 3

1 3 3

=

Hence, the triangle is an equilateral.

Alternate Solution

z zz z

i1 3

2 3

1 3

2

=

z zz z i

i2 3

1 3

2

1 3

1 3

2

=

= +

= +cos sin 3 3i

arg z zz z

2 3

1 3 3

=

and also

z z

z z

2 3

1 3

1

=

Therefore, triangle is equilateral.

23. Let = 1 1 1

1 1

1

2 2

2

ApplyingR R R R R R2 2 1 3 3 1 ;

=

100

12

1

111

2

2

2

= ( )( ) ( )2 1 12 2 2 = + + +2 2 2 13 2 4 2 ( )

= 3 32 = 3 1 ( ) ( )Q 4 =24. We know, | | | ( ) ( )|z z z z i i1 2 1 2 3 4 3 4 = +

+| | | | | |z z i i1 2 3 4 3 4 12 5 5(using | | | | | | )z z z z1 2 1 2

| |z z1 2 2 Alternate Solution

Clearly from the figure | |z z1 2 is minimum when z z1 2, liealong the diameter.

| |z z C B C A1 2 2 2 =12 10 2

25. Since, | |z= 1and w zz

= +

1

1

z wz w = +1 z ww

= +

1

1

| | | || |

z w

w= +

1

1

| | | |1 1 = +w w (Q| |z= 1)On squaring both sides, we get

1 2 1 22 2+ = + +| | | | ( ) | | | | ( )w w w w w wRe Re[using | | | | | | | |z z z z z1 2

21

22

212 = + | | ( )z z z2 1 2Re ]

4 0| | | |w wRe = Re ( )w = 026. Given, ( ) ( )1 12 4+ = + n n

( ) ( ) = n n2 (Q3 1= and 1 02+ + = ) n = 1 n= 3 is the least positive value of n.

27. Let z a b c= + +| | 2

z a b c2 2 2

= + +| | = + + ( )a b c ab bc ca2 2 2

or z a b b c c a2 2 2 21

2= + + {( ) ( ) ( ) } (i)

Since, a b c, , are all integers but not all simultaneously equal

If a b= then a c and b cBecause, difference of integers =integer.( )b c 2 1{as minimum difference of two consecutiveintegers is ( )}1 also ( )c a 2 1

and we have taken a b a b= =( )2 0From Eq. (i),

z2 + +1

2

0 1 1( ) z2 1

Hence, minimum value of | |z is 1 .

28. Since, | | | | | |PQ PS PR= = = 2 Shaded part represents the external part of circle havingcentre ( , )1 0 and radius 2.As we know equation of circle having centre z0and radius r

is | |z z r =0 | ( )|z i + >1 0 2 | |z+ >1 2Also, argument of z+ 1with respect to positive direction ofx-axis is /4.

arg ( )z+ 14

(i)

and argument of z+ 1in anticlockwise direction is / 4 + / ( )4 1arg z (ii)From Eqs. (i) and (ii), we get

| ( ) | /arg z+ 1 4

29. Let z w wz

z1

1=

, be purely real.

z z1 1=


12

C1

C2

AZ2

Z1B

xx'

y'

(3,4)

z2

z1z3

/3


12/22

w wzz

w wz

z

= 1 1

w wz wz wz z w zw + = + wz wz z ( ) ( ) | |w w w w z + =2 0 ( ) ( | | )w w z =1 02

| |z2

1= (as, w w 0, since 0) | |z = 1and z 1.Therefore, (b) is the answer.

30. Let OA= 3, so that the complexnumber associated with A is

3 4ei / .Ifzis the complex number

associated withP, then

z e

ee

ii

i

i

= = 30 3

4

3

4

3

4

4

2

/

/

/

3 9 124 4z e iei i = / /

z i ei= +( ) ./3 4 4

31. Let z i= +cos sin

zz

i

i1 1 2 22 = +

+cos sin

(cos sin )

= +

cos sin

sin sin cos

i

i2 22

= + +

cos sin

sin (cos sin )

i

i i2 = i

2sin

Hence,z

z1 2lies on the imaginary axis ie,y-axis.

Alternate Solution

LetE z

z

z

zz z z z=

=

=

11

2 2which is imaginary.

32.

z2 6 2 45 5 2 45 = + + ( cos , sin )= = +( , )7 6 7 6iBy rotation about (0, 0),

z

z ei2

2

2

= /

z z e

i

2 2

2

=

z i i i i2 7 62 2

7 6= + +

= +( ) cos sin ( )( ) = +6 7i

33. Given that z i ei= + =cos sin

Im( ) Im( )z em i mmm

2 1 2 1

1

15

1

15

===

= =

Im ( )ei mm

2 1

1

15

= + + + +sin sin sin ... sin 3 5 29

=

+

sin sin

sin

29

2

15 2

2

2

2

= =

sin ( )sin ( )

sin sin

15 15 1

4 2

34. Since,zz z z( )2 2 350+ =

2 3502 2 2 2( )( )x y x y+ = ( )( )x y x y2 2 2 2 175+ =Since, x y I, , the only possible case which gives integrasolution, is

x y2 2 25+ = ... (i)

x y

2 2 7

=... (ii

From Eqs. (i) and (ii), we get

x y2 216 9= =;

x= 4; y= 3

Area of rectangle = 8 6= 48

Objective Questions II1. Since, z a ib1= + and z c id2= +

| |z a b12 2 2 1= + = and | |z c d2

2 2 2 1= + = (i)

( | | | | )Q z z1 2 1= =Also, Re ( )z z1 2 0= ac bd + = 0

ab

d

c= = (say)(ii)

From Eqs. (i) and (ii), b b c c2 2 2 2 2 2 + = +

b c2 2= and a d2 2=

Also, given w a ic1= + and w b id 2= +

Now, | |w a c a b12 2 2 2 1= + = + =

| |w b d a b22 2 2 2 1= + = + =


90

1

z2

Imaginary axis

Real axis

z' (7,6)2

13

5(6,2)

z(1

,2)

0

p

4

3

Oy'

x' x/4

i/43eA


13/22

and Re( ) ( ) ( )w w ab cd b b c c1 2 = + = + [from Eq. (i)]

= =( )b c2 2 0

Therefore, (a), (b), (c) are the correct answers.

2. Given, | | | |,z z1 2=

Now, z zz z

z zz z

1 2

1 2

1 2

1 2+

= +

z z z z z z z z

z z

1 1 1 2 2 1 2 2

1 22| |

= +

| | ( ) | |

| |

z z z z z z

z z

12

2 1 1 2 22

1 22

=

z z z z

z z

2 1 1 2

1 22| |

(Q| | | | )z z12

22=

As, we knowz z i = 2 Im( )z z z z z i z z2 1 1 2 2 12 = Im ( )

z zz z

i z zz z

1 2

1 2

2 1

1 22

2+ = Im ( )

| |

Which is purely imaginary or zero.

Therefore, (a) and (d) are correct answers.

3. Given, z t z t z

t t= +

+( )

( )

1

1

1 2

Clearly,zdividesz1andz2in the ratio of t t: ( )1 , 0 1<


14/22

Clearly, there is only one point of intersection of the line

x y+ = 2and circlex y x y2 2 4 2 4+ =

2. | | | |z i z i+ + 1 52 2

= + + + + ( ) ( ) ( ) ( )x y x y1 1 5 12 2 2 2

= + +2 4 2 282 2( )x y x y

= +2 4 28( ) (Qx y x y2 2 4 2 4+ = ) = 36

3. Since, | ( )|w i +


15/22

2. | | | |az bz bz az 1 22

1 22 + +

= + [ | | | | Re ( )]a z b z ab z z 2 12 2

22

1 22

+ + +[ | | | | Re ( )]b z a z ab z z 2 12 2

22

1 22

= + +( ) (| | | | )a b z z 2 2 12

22

3. Since, z z1 2, andz3form an equilateral triangle.

z z z z z z z z z12

22

32

1 2 2 3 3 1+ + = + +

( ) ( ) ( ) ( ) ( )a i ib a i ib+ + + + = + + + +2 2 21 0 1 0 0

a ia b ib a i ab b2 21 2 1 2 1 + + + = + + ( )

( ) ( ) ( ) ( )a b i a b a b i ab2 2 2 1 + + = + +

a b a b2 2 =

and 2 1( )a b ab+ = + (a b= or a b+ = 1)and 2 1( )a b ab+ = +If a b= , 2 2 12( )a a= +

a a2 4 1 0 + =

a=

= 4 16 4

22 3

If a b+ = 1,2 1 1= +a a( )

a a2 1 0 + =

a= 1 1 4

2, but aand b R

Only solution when a b= a b= = 2 3

4. Given, D i M i= + = ( ), ( )1 2

and diagonals of a rhombus bisect each other.LetB a ib +( ),therefore

a+ =12

2,b+ = 1

21

a+ =1 4, b+ = 1 2 a= 3, b= 3 B i ( )3 3

Again, DM= + = + =( ) ( )2 1 1 1 1 4 52 2

But BD DM= 2 BD= 2 5

and 2AC BD= 2 2 5AC= AC= 5

and AC AM= 2 5 2= AMAM= 52

Now, let coordinate ofAbe ( )x iy+ .But in a rhombus AD AB= , therefore we haveAD AB2 2= ( ) ( ) ( ) ( )x y x y + = + +1 1 3 32 2 2 2

x x y y x x2 2 21 2 1 2 9 6+ + + = + + + +y y2 9 6 4 8 16x y =

x y =2 4 x y= +2 4 (i)

Again, AM= 52

AM2 54

=

( ) ( )x y + + =2 1 54

2 2

( ) ( )2 2 1 54

2 2y y+ + + = [from Eq. (i)

5 10 5 54

2y y+ + =

20 40 15 02

y y+ + = 4 8 3 02y y+ + = ( ) ( )2 1 2 3 0y y+ + = 2 1 0y+ = ,2 3 0y+ =

y= 12

, y= 32

On putting these values in Eq. (i),

x=

+2 1

24,x=

+2 3

24

x= 3,x= 1

Therefore,Ais either 32

i or 1 32

i

Alternate Solution

Since,Mis the centre of Rhombus.

By rotatingDaboutMthrough an angle of / 2 , we gepossible position ofA.

z zz z

z z

z ze i3 2

1 2

3 2

1 2

2

=

| |

| |

/

z ii

i32

1 2

1

2

+

= ( ) ( )

z i i i3 21

22 1= ( ) ( )

= ( ) ( )2 12

2i i

=

+ +( ),

4 2 2

2

4 2 2

2

i i i i

= 1 32

32

i i

,

Ais either 1 32

i or 3

2

i

.

5. z i r i1 1 3= + = +(cos sin ) (let)

r rcos , sin = =1 3 r= 2 and = / 3


AD(1+i)

CB

M

( )2 i

D z(1+i)1A(z)3

BC

M

(2i) z 2


16/22

So, z1 2 3 3= +(cos / (sin / )) Since, | | | |z z2 3 2= = (given)

Now, the triangle z z1 2, and z3being an equilateral and the

sides z z1 2and z z1 3make an angle 2 3 / at the centre.

Therefore, = + =POz23

2

3

and = + + =POz33

2

3

2

3

5

3

Therefore, z i2 2 2 1 0 2= + = + = (cos sin ) ( ) and z i i i3 2

5

3

5

32

1

2

3

21 3= + =

=

cos sin

Alternate Solution

Whenever vertices of an equilateral triangle having centroid

is given its vertices are of the from z z z, , 2.

If one of the vertex isz i1 1 3= + ,then other two verticesare ( ), ( )z z1 1

2 .

( ) ( )1 3 1 32

+ +i i , ( ) ( )1 3 1 32

+ i i

+( )1 32

, + +( ( ) )1 3 2 32

2 2i i

2, + = ( )2 2 3

2

1 3i

i

z2 2= and z i3 1 3=

6. Here, T r r r r= + + [( ) ] [( ) ]1 12

= + + + +r r r[( ) ( ) ( ) ]1 12 2 3 = + + + +r r r[( ) ( ) ]1 1 12

= + + + + +r r r r [ ]2 1 2 1 1 = + +r r r3 23 3

Therefore, the sum of the given series

= + +=

r

n

r r r1

13 23 3

( )

( )

=

+

+ ( )( ) ( ) ( ) ( )( )n n n n n n n1

23

1 2 1

63

1

2

2

= + +

( ) ( )( )( ) ( )

n n n n n

11

4

2 1

2

3

2

= + +14

1 1 2 2 1 6( ) ( ) [( ) ( ) ]n n n n n

= + +14

1 3 42( ) [ ]n n n n

True / False

1. Let z x iy= + 1zgives 1 +x iyor 1xand 0 y (i)

Given,1

10

+

zz

11

0 + +

x iyx iy

( ) ( )( ) ( )

1 1

1 10 0

+ + + +

+x iy x iyx iy x iy

i

11

2

10 0

2 2

2 2 2 2

+ +

+ +

+x yx y

iy

x yi

( ) ( )

x y2 2 1+ and 2 0yor x y2 2 1+ and y 0which is true by Eq. (i).

2. Since, z z z1 2 3, , are vertices of equilateral triangle and

| | | | | |z z z1 2 3= =z z z1 2 3, , lie on a circle with centre at origin. Circumcentre = Centroid

03

1 2 3= + +z z z

z z z1 2 3 0+ + =

3. Since, cube root of unity are 1 2, , given by,

A B C( , ), , , ,1 01

2

3

2

1

2

3

2

AB BC CA= = = 3.

Hence, cube roots of unity form an equilateral triangle.

Analytical & Descriptive Questions

1. Since, ( )x iy a ib

c id+ = +

+2

| | | || |

x iy a ib

c id+ = +

+2 Q

z

z

z

z

1

2

1

2

=

| |

| |

( )x y a b

c d

2 22 2

2 2+ = +

+

( )x y a bc d

2 2 22 2

2 2+ = +

+


x-axis

z1

z2

z3

O

P(2,0)

y-axis

P(1, 0)


17/22

2. Since, , are the complex cube roots of unity.

We take = and = 2.

Now, xyz a b a b a b= + + +( )( )( ) = + + + +( )[ ( ) ]a b a ab b2 2 2 2 = + + + + ( )[ ( ) ( ) ( )]a b a ab b2 2 2 4 2 2

= + +( )( )a b a ab b2 2 (Q1 0 12 3+ + = = and )= +a b3 3

3. Now,1

1 2

1

22

42 2

2( cos ) sin sin sin cos +

=+ i i

=+

1

22 2

22

22

2

22sin sin cos

sin cos

sin c

i

i

i os2

=

+

sin cos

sin sin cos

2

22

2 2 2 4 2

2 2

i

=

+

sin cos

sin cos

2

22

22

1 32

2

i

A iB i+ =+

+

1

2 1 32

2

1 32

2 2cos

cot

cos

4. Since, n is not a multiple of 3, but odd integers and

x x x3 2 0+ + = x= 0 2, ,

Now, whenx= 0 ( )x xn n

+ 1 1 = =1 0 1 0 x= 0 is root of ( )x xn n+ 1 1Again, whenx= ( ) ( )x xn n n n+ = + 1 1 1 1

= = 2 1 0n n

(as nis not a multiple of 3 and odd)

Similarly,x= 2is root of {( ) }x xn n+ 1 1Hence,x= 0, , 2are the roots of ( )x xn n+ 1 1Thus,x x x3 2+ + divides ( )x xn n+ 1 1.

5. ( ) ( )1 2

3

2 3

3

+

+

+ +

=i x i

i

i y i

i

i

( ) ( ) ( ) ( ) ( )1 3 2 3 3 2 3+ + + i i x i i i i y + + =i i i( )3 10

4 2 6 2 9 7 3 1x ix i y iy i+ + + = 10i 4 9 3 0x y+ = and 2 7 3 10x y = x= 3 and y= 1

6. Since, z z z1 2 3, , are the ver tices of an equi lat eral triangle.

Circumcentre ( )z0 = centroidz z z1 2 3

3

+ +

...(i)

Also, for equilateral triangle

z z z z z z z z z12

22

32

1 2 2 3 3 1+ + = + + ... (ii)On squaring Eq. (i), we get

9 202

12

22

32z z z z= + + + ( )z z z z z z1 2 2 3 3 1+ +

= + + +9 202

12

22

32z z z z ( )z z z1

222

32+ + [from Eq. (ii)

3 02 12 22 32z z z z= + +

7. Here, z Rz1 2z z

z z

1 2

1 2

+

is real

(1) Reflexivez Rz1 1z z

z z

1 1

1 2

0+

= (purely real

z Rz1 1is reflexive.

(2) Symmetric z Rz1 2z z

z z

1 2

1 2

+

is real

+

( )z z

z z

2 1

1 2

is real

z Rz2 1 z Rz1 2 z Rz2 1Therefore, it is symmetric.

(3) Transitive z Rz1 2 z z

z z

1 2

1 2

+

is real

and z Rz2 3 z z

z z

2 3

2 3

+

is real

Here, let z x iy z x iy1 1 1 2 2 2= + = +, and z x iy3 3 3= +

z zz z

1 2

1 2

+

is real ( ) ( )( ) ( )

x x i y y

x x i y y

1 2 1 2

1 2 1 2

+ + + +

is real

{( ) ( )}{( ) ( )}( ) (

x x i y y x x i y y

x x y y

1 2 1 2 1 2 1 2

1 22

1

+ + ++ + + 2

2)

( ) ( ) ( ) ( )y y x x x x y y1 2 1 2 1 2 1 2 0 + + = 2 2 02 1 2 1x y y x =

x

y

x

y

1

1

2

2

= ... (i)

Similarly, z Rz2 3 x

y

x

y

2

2

3

3

= ... (ii)

From Eqs. (i) and (ii), we havex

y

x

y

1

1

3

3

=

z Rz1 3Thus, z Rz1 2and z Rz2 3z Rz1 3. (transitive).Hence,Ris an equivalence relation.

8. Since,z z1 2, and origin form an equilateral tri angle.

Q if from an equilateral triangle, thenz z z

z1 2 3

12

, ,

+ + = + +

z z z z z z z z22

32

1 2 2 3 3 1

z z z z z z12

22 2

1 2 2 10 0 0+ + = + +

z z z z12

22

1 2+ =

z z z z12

22

1 2 0+ =

9. Since, 1 1 2 1, , , ......,a a an are nth roots of unity.

( ) ( ) ( ) ( ) .... ( )x x x a x a x an n = 1 1 1 2 1



18/22

xx

x a x a x an

n

= 1

11 2 1( ) ( ) .....( )

x x x xn n + + + + +1 2 2 1.....

= ( ) ( ) ..... ( )x a x a x an1 2 1

Q

x

x x x x

nn n

= + + + +

1

1 11 2

.....

On puttingx= 1, we get 1 1+ + .......ntimes = ( ) ( ) ..... ( )1 1 11 2 1a a an

( ) ( )..... ( )1 1 11 2 1 =a a a nn10. We have, iz zei= / 2. This implies that iz is the vector

obtained by rotating vector z in anti-clockwise direction

through 90. Therefore, OA AB . So,

Area of OAB= 12

OA OB

= =12

1

2

2| | | | | | .z iz z

11. Since, is right angled isosceles .Rotatingz2aboutz3in anti-clockwise direction through anangle of / ,2 we get

z z

z z

z z

z zei2 3

1 3

2 3

1 3

2

=

| |

| |

/

where, | |z z2 3 = | |z z1 3 ( ) ( )z z i z z2 3 1 3 =


( ) ( )z z z z2 32

1 32 =

z z z z z z z z22

32

2 3 12

32

1 32 2+ = +

z z z z z z z z12

22

1 2 1 3 2 32 2 2+ = + 2 232

1 2z z z

( ) {( ) ( )}z z z z z z z z z1 22

1 3 32

2 3 1 22 = +

( ) ( )( )z z z z z z1 22

1 3 3 22 =

12. Since,z i z i1 210 6 4 6= + = +,

and argz z

z z

=1

2 4

represents locus of zis a circle shown

as from the figure whose centre is (7, y) and = AOB 90clearly, OC= 9 .

OD= + =6 3 9 Centre = ( , )7 9 and radius = =6

23 2

Equation of circle is | ( )|z i + =7 9 3 2

13. Given, iz z z i3 2 0+ + =

iz i z z i3 2 2 0 + = (Q i2 1=

iz z i z i2 1 0( ) ( ) =

( )( )iz z i2 1 0 =

z i = 0 or iz2 1 0 =

z i= or zi

i21= =

If z i= , then | | | |z i= = 1.If z i2 = , then | | | |z i2 1= =

| |z2 1= | |z = 1

14. Let z r i= +1 1 1(cos sin ) and w r i= +2 2 2(cos sin )

We have, | | , | | ,z r w r= =1 2 arg ( )z = 1and arg ( )w = 2Given, | | , | |z w


19/22

+

| | sinr r1 2

2 1 2

2

42

( , )Q r r1 2 1

and |sin | | |, R

Therefore, | | | |z w r r +

2

1 22 1 2

2

42

+ | | | |r r1 22

1 22

| | (| | | | ) ( )z w z w z w + 2 2 2arg arg

Alternate Solution

| | | | | | | | | |cos ( )z w z w z w z w = + 2 2 2 2 arg arg

= + +| | | | | | | | | | | |z w z w z w2 2 2 2

2| || |cos ( )z w z warg arg

= +

(| | | | ) | | | | sinz w z w z w2 22 2

2

arg arg(i)

| | (| | | | )z w z w z w +

2 22

24 1 1

arg arg

(Q sin ) | | (| | | | ) ( )z w z w z w + 2 2 2arg arg

15. Let z x iy= + .Given, z iz= 2

( ) ( )x iy i x i y+ = + 2

x iy i x y i xy = +( )2 2 2

x iy xy i x y = + 2 2 2( )

Note It is a compound equation, therefore, we can generate fromit more than one primary equations.

On equating real and imaginary parts, we get

x xy= 2 and = y x y2 2

x xy+ =2 0 and x y y2 2 0 + =

x y( )1 2 0+ = x= 0 or y= 1 2/Whenx= 0, x y y2 2 0 + =

0 02 + =y y

y y( )1 0 = y= 0 or y= 1When y= 1 2/ ,x y y2 2 0 + =

x2 14

1

20 = x2 3

4=

x= 32

Therefore, z i= +0 0, 0 32 2

+ i i;

z i i= , 32 2

( )Qz 0

16. Since, z z p1 2+ = andz z q1 2=

Now,z

z

z

zi1

2

1

2

= +| || |

(cos sin )

zz

i1

2 1= +cos sin ( | | | | )Qz z1 2=

Applying componendo and dividendo, we get

z z

z z

i

i

1 2

1 2

1

1

+

= + ++

cos sin

cos sin

= + +2 2 2 2 2

2 2 2

2

2

cos ( / ) sin ( )cos ( / )

sin ( / ) sin ( /

/

i

i 2 2)cos ( / )

= +

+

2 2 2 2

2 2 2

cos ( / ) [cos ( / ) sin ( / )]

sin ( / )[cos ( / )

i

i i sin ( / )] 2 = = cot ( / ) cot / 2 2

ii

= pz z

i1 2

2cot ( / )


p

z z

2

1 22

2 2( )

cot ( / )

=

pz z z z

2

1 22

1 2

2

42

( )cot ( / )

+ =

p

p q

2

2

2

4 2 = cot ( / ) p p q2 2 2 22 4 2= +cot ( / ) cot ( / )

p q2 2 21 2 4 2( cot / ) cot ( / )+ =

p q2 2 22 4 2cosec ( / ) cot ( / ) =

p q2 24 2= cos /

17. Let Qbez2and its reflection be the pointP z( )1 in the given

line. If O z( ) be any point on the given line then by def inition

ORis right bisector of QP.

OP OQ= or | | |z z z z = 1 2 | | | |z z z z = 1

22

2

( ) ( ) ( ) ( )z z z z z z z z = 1 1 2 2 z z z z z z z z z z( ) ( )1 2 1 2 1 1 2 2 + = Comparing with given linezb zb c+ =

z z

b

z z

b

z z z z

c

1 2 1 2 1 1 2 2 = = = , say


O

A(z)1

B(z)2


20/22

z z b z z b z z z z c1 2 1 2 1 1 2 2 = = =

, , (i)

z b z b z z z z z z1 2 1 1 2 2 1 2+ =

+

= =zz z z c1 2 2

[from Eq. (i)]

18. Given, | | | |z w w z z w2 2 =

zz w ww z z w = [ | |Q z zz2 = ] (i)

Taking modulus of both sides, we get

| | | | | |zw z w z w = | | | | | |zw z w z w = [ =| | | |z z ]

| | | | | |zw z w z w = | | (| | )z w zw =1 0 | |z w = 0 or | |zw =1 0

| |z w = 0 or | |zw = 1 z w = 0 or | |z w= 1

z w= or | |zw = 1Now, suppose z wThen, | |zw = 1or | || |z w = 1

| || |

zw

r= =1 (say)

Let z rei= and wr

ei= 1

On putting these values in Eq. (i), we get

rr

er

re rer

ei i i i22

1 1 1

= ( )

rer

e re

r

ei i i i = 1 1

rr

e rr

ei i+

= +

1 1

e ei i =

=

Therefore, z rei= and wr

ei= 1

zw rer

ei i= = .1 1

Note if and only if means we have to prove the relation inboth directions.

Conversely

Assuming that z w= or z w= 1If z w= ,then

LHS = zz w w wz= | | | |z z w z2 2

= =| | | |z z z z2 2 0

and RHS = =z w 0If zw= 1, thenzw= 1and

LHS= zz w ww z= z w1 1= = z w z w = =0 RHS Hence proved.

Alternate Solution

We have, | | | |z w w z z w2 2 =

| | | |z w w z z w2 2 0 + =

(| | ) (| | )z w w z2 21 1 0+ + =

(| | ) (| | )z w w z2 21 1+ = +

zw

z

w= +

+| |

| |

2

2

1

1

zw

is purely real.

zw

z

w= zw zw= (i)

Again, | | | |z w w z z w2 2 =

z zw w wz z w = z zw w zw( ) ( ) =1 1 0 ( )( )z w zw =1 0 [from Eq. (i)]

z w

= or zw

=1

Therefore, | | | |z w w z z w2 2 = if and only if z w= ozw= 1.

19. Given, z z zp q p q+ + =1 0 (i)

( )( )z zp q =1 1 0

Since, is root of Eq. (i), either p =1 0 or q =1 0

either

p

=11

0

or

q

=11

0 (as 1

either 1 02 1

+ + + + =

... p

or 1 01+ + + = K q

But p =1 0 and q =1 0 cannot occur simultaneouslyaspand qare distinct primes, so neither pdivides qnor q

dividesp, which is the requirement for 1 = = p q.

20. Given, | |z1 1< and | |z2 1> (i)Then, to prove

111 2

1 2


21/22

which is true by Eq. (i) as | |z1 1< and | |z2 1> ( | | )1 01

2 >z

and ( | | )1 022 K (using | |ar < 2)

2 11

1| |( | | )

| |

z z

z

n

> (using sum of nterms of GP)

2 2 11| | | | | |z z zn > + 3 1 2 1| | | |z zn> + +

| | | |z zn> + +13

2

3

1

| | ,z> 13

which contradicts (ii)

There exists no complex numberzsuch that

| |z< 13 and

r

n

rra z

= =

11

22. As we know | |z z z2 =

Given,| |

| |

z

zk

=

2

2

2

( )( ) ( ) ( )z z k z z = 2

| | | | (| | | | )z z z k z z z2 2 2 2 2 + = +

| | ( ) ( ) ( )z k k z k z2 2 2 21

+ =(| | | | ) 2 2 2 0k

| | ( )( )

( )

( )z

k

kz

k

kz2

2

2

2

21 1

+

=| | | |( )

2 2 221

0k

k (i)

On comparing with equation of circle,

| |z az az b2 0+ + + =

whose centre is ( )a and radius = | |a b2

Centre for Eq. (i)

=

k

k

2

21 and radius

=

kk

k

k

k

k

2

2

2

2

2

21 1 1

radius =

k

k

( ) 1 2

23. Here, centre of circle is (1, 0) is also the mid point of

diagonals of square

z z z1 2 02

+ =

z i2 3= , (wherez i0 1 0= +

andz

ze i3

1

21

1

= /

z i i3 1 1 3 2 2= + + ( ) cos sin ,

(Q z i1 2 3= + = +1 1 3i i( )

= + ( )1 3 iz i3 1 3= +( )

and z i4 1 3= + ( )


x

z (2, 3)1

z3

z2 z4

z0O

(1, 0)

y


22/22

1. Let z be a complex number and a be a real

parameter such thatz az a2 2 0+ + = , then(a) locus ofzis an ellipse (b) locus ofzis a circle

(c) are ( )z = 2

3

(d)| | | |z a= 3

2. The complex numberz i= +1 is rotated through anangle 3 2/ in anti-clockwise direction about theorigin and stretched by additional 2 unit, then thenew complex number is

(a) 2 2i (b) 2 2 i(c) 2 2 i (d) None of these

3. The complex numberz1andz2are such thatz z1 2and | | | |.z z z1 2 1= If has positive real part andz2has

negative imaginary part, thenz z

z z1 2

1 2

+

may be

(a) zero (b) real and positive(c) real and negative (d) purely imaginary

4. If z z C z z R z z z1 2 12

22

1 12

223 2 + =, , ( ) and

z z z2 12

223 1( ) , = then the value ofz z1

222+ is

(a) 5 (b) 6(c) 10 (d) 12

5. Consider an ellipse having its foci at A z( )1 andB z( )2in the argand plane. If the eccentricity of the ellipsebe e and it is known that origin is an interior point ofthe ellipse, then

(a) ez z

z z +

+

0 1 2

1 2

,| |

| | | |(b) e

z z

z z

+

0 1 2

1 2

,| |

| | | |

(c) ez z

z z +

0 1 2

1 2

,| |

| | | |(d) cannot be discussed

6. If z a ib1 1 1= + and z a ib2 2 2= + are complexnumbers such that| | ,| |z z1 21 2= = and Re ( ) ,z z1 2 0=then the pair of complex numbers = +1 1 2

2a

iaand

= +2 1 22b ib satisfy(a)| |1 1= (b)| | =2 2(c) Re ( ) =1 2 0 (d) Im ( ) =1 2 0

7. If from a pointPrepresenting the complex numberz1on the curve| | ,z = 2pair of tan gents are drawn to thecurve| | ,z = 1meeting at point Q z( )2 andR z( )3 , then(a) complex number

z z z1 2 33

+ + will lie on the

curve | |z = 1

(b)4 1 1 4 1 1

91 2 3 1 2 3z z z z z z

+ +

+ +

=

(c) argz

z2

3

2

3

=

(d)orthocentre and circumcentre of PQR willconcide

8. One vertex of the triangle of maximum area thacan be inscribed in the curve | |z i =2 2, is 2 2+ iremaining vertices is/are

(a) + +1 2 3i( ) (b) +1 2 3i ( )(c)

+ 1 2 3i( ) (d)

1 2 3i( )

Passage for Q. Nos. 9 to 11

In argand plane| |z represent the distance of a point z

from the origin. In general | |z z1 2 represent thedistance between two points z1 and z2. Also, for a

general moving point z in argrand plane, if arg( )z = , thenz z ei= | | , where e ii .= +cos sin

9. The equation | | | |z z z z + =1 2 10 if z1 3 4= +andz i2 3 4= rep re sents(a) point circle (b) ordered pair ( , )0 0(c) ellipse (d) None of these

10. || | | || ,z z z z t =1 2 where t is real parametealways represents(a) ellipse (b) hyperbola

(c) circle (d) None of these

11. If | ( )| cos arg ,z i z z + =

3 24

then locus o

zis

(a) circle (b) parabola(c) ellipse (d) hyperbola

12. If z z z z1 2 3 4, , , are the roots of the equa tionz z z z4 3 2 1 0+ + + + = , thenMatch the statement of Column I with values oColumn II.

Column I Column II

(A) i

jz= 1

44 is equal to

(p) 0

(B) i

jz= 14 5is equal to (q) 4

(C) i

iz= +

1

42( ) is equal to (r) 1

(D) Least value of[| |]z z1 2+is (Where [ ] represents

greatest integer function)

(s) 11

13. The complex numbers z is simultaneously satisfy

the equationsz

z i

z

z

=

=128

5

3

4

81, , then the

Re ( )z is

14. If| | ,z 3then the least value of zz

+ 1 is 3

,then the

value of is

apter est

Answers

1. (c) 2. (d) 3. (d) 4. (a) 5. (b) 6. (a, b, c, d) 7. (a, b, c, d) 8. (a, c)

9. (d) 10. (b) 11. (b) 12. A r; B q; C s; D p 13. (6) 14. (8)

Documents

Complex Numbers Arihant