Complex Numbers Arihant
Transcript of Complex Numbers Arihant
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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
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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| |
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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]
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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+
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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<
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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
6 | Chapter 1 Complex Numbers
Analytical & Descriptive Questions
True/False
Fill in the Blanks
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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+ = +
8 | Chapter 1 Complex Numbers
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
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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
Chapter 1 Complex Numbers | 9
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= + + +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
10 | Chapter 1 Complex Numbers
y
(z)r
O
r
(z)
x
O A (z1
B (z )2
2
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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=
Chapter 1 Complex Numbers | 11
12
C1
C2
AZ2
Z1B
xx'
y'
(3,4)
z2
z1z3
/3
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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= + = + =
12 | Chapter 1 Complex Numbers
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
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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<
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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 +
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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
Chapter 1 Complex Numbers | 15
AD(1+i)
CB
M
( )2 i
D z(1+i)1A(z)3
BC
M
(2i) z 2
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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+ = +
+
16 | Chapter 1 Complex Numbers
x-axis
z1
z2
z3
O
P(2,0)
y-axis
P(1, 0)
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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
Chapter 1 Complex Numbers | 17
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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 =
On squaring both sides, we get
( ) ( )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
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+
| | 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 ( / )
On squaring both sides, we get
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
Chapter 1 Complex Numbers | 19
O
A(z)1
B(z)2
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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
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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= + ( )
Chapter 1 Complex Numbers | 21
x
z (2, 3)1
z3
z2 z4
z0O
(1, 0)
y
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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)