1.(A) ii
ei 2 iei2 2e22.(A)
3.(D) and 2
19 19
and 7
2 7142 The required equation is x2 x 1 0
4.(D) 6 3i
i2 3
3i 4i20
1 12 60 i
6 0i 3 4i i _20 12 60i 0 x iy
x y 0
5.(A) Let arg z
, where 0 arg
z arg
z arg z
6.(C) z3z
Taking mod of both sides, we get : 3 3 z z z z z or 10 0 z z0 Now, z3z z4zz z2 If z the above equation becomes1
4 2 1 1
z z 1, i
So, the solutions of the equation z3 arez z 0, 1, i
7.(C) z 1 z 4 3 z 4 3
z 1 3 3 or z 1 6
8.(C) The required number =
1 1 1 2 1 2 2 2 1 i i i i 1 1 i 9.(B) For the series
1
12
. . . .
10
102
2
2 1 r T r r r r
10 2 2 2 1 1 1 10 10 1 r . . . . r r
10 11 21 10 11 10 385 55 10 450 6 2
2
2
450 1 1 10 10 900 900 cos . . . . cos cos 2 0 Complex Numbers
HWT - 1
10.(A) Let z x iy
2 2 2 1 1 1 1 1 1 1 1 1 x x y i x y y x x yi z z x yi x y
2 2 2 2 1 2 1 x y i y x y According to the question x2y2 or1 0 x2y2 1 z2
z 1
1.(A) Let z x iy z2
x2y2
i
2xy
2 0Re z x2y20 or y x
2.(C) The cube roots of –1 are 1, , 2 Now,
1
2 1 3.(B) x 5 2 4 5 4i
x5
2x210x25 16 or x210x410 Now,
4 3 2 2 2 2 2 2 9 35 4 10 41 10 41 4 10 41 160 0 0 4 0 160 160 x x x x x x x x x x x x x x 4.(A) We can compare two real numbers only b = d = 0
5.(D) 1 20
7 7 7 7 2 2 2 2 14 2 1 2 2 128 6.(C)
10 23
2
4 4 sin sin 5 1 4 4 2 sin sin 7.(B) For the given series
2
2
2 1 1 1 1 1 3 3 r T r r . r r r r r r r 3 2 3 3 r r r Required sum
2 2 1 2 2 3 2 1 1 1 2 1 3 1 1 3 3 3 4 6 2 4 n r n n n n n . n n n n r r r . n
8.(C) If all of the coefficients in an equation are real, the complex roots exist in pairs.
Roots of given quadratic equation are 2i 3 and 2i 3.
Sum of roots a
2i 3
2i 3
4 a and product of roots4 b
2i 3
2i 3
79.(D) Let 1 1 1 to x or1 x x 1 x x3 or x2 x 1 0 x , 2 10.(D) 1 3 2 2 3 4 1 3 i i i Argument 1 2 3 2 2 3 3 tan or 4 3
Complex Numbers
HWT - 2
1.(D) According to the question, z lies on negative y-axis.
2 arg z 2.(A) 2 2 3i
1 3i
3.(C)
2 2 2 1 1 1 1 1 n n n i i i i or
2 1 ni 2 is the smallest positive integer.
4.(B)
2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2cos i sin . cos cos i sin cos cos i sin
i sin i cos i cos i sin
i cos i sin cos
2 2 2 2 2 2 2 2 2 2 i i i
cos cos i sin
e
i ie
e i cos cos i sin
4 4 4 1 4 4 i i cos i sin ie e cos i sin i sin i cos 5.(D) i i2 i3 i4 i 1 i 1 0Similarly, i5 i6 i7 i8 i i2 i3 i4 0 Required sum = 0
6.(ACD) x2 x 1
x
x2
If P(x) is divisible by x2 x 1, P
P
2 0 P
g
3 h 3 g
1 h
1 0 . . . .(i) and P
2 g 6 2h
6 g
1 2h
1 0 . . . .(ii)Solving (i) and (ii) we get :
1 1 0 g h g
1 andh
1 g
1 h1 0 0 0 7.(A) Let
2 1 2 2 1 1 1 x yi z z x iy iz y xi
2 2 2 2 2 2 2 1 2 2 2 2 1 1 2 2 1 2 1 2 1 1 x y i x y x y x y xy i x x y y x y y y x According to the question
2 2 2 2 2 2 2 2 2 3 5 5 8 3 0 2 1 x y x y x y y x x y y 8.(C) 3 3 3 r i r r r x cos i sin e 1 1 3 9 3 9 1 2 i . . . i . . . x . x . . . . e e i 2 e i 9.(B)
x1
3 8 0
x1
3 8 x 1 2, 2,22 or x 1 1 2, ,1 2 2 10.(D) z 1 1 z or 2 1 0 z z z , 2
99 99 99 99 1 1 1 1 2 1 z z Complex Numbers
HWT - 3
1.(A) log1 2 z 2 log1 2 z z 2 z
Distance of z from (2, 0) is less than distance of z from (0, 0) Re z
12.(D) 6 6 1 1 2 2 2 2 7 7 7 7 k k k k k k
sin i cos i cos i sin
6 7 1 i k k i e i
[sum of 6 seventh roots of unity except 1] 0 1i i
(Sum of nth roots of unity = 0)
3.(B)
2 2 2 2 2 a b abi a ib z a ib a b According to the questiontan 22ab2 a b 4.(B) Let z x iy z2
x2y2
2xyi
2 2 2 0 0 Re z x y or y x 2 2 2 2 z x y or x2y24 x 2 and y 2So, the solutions of the system are z 2 2i
5.(C) xn has roots1 1, a , a , . . ., a1 2 n1 xn 1
x1
xa1
xa2
. . . x
an1
1
2
1
1 1 n n x x a x a . . . x a x or
1 x x2. . .xn1
xa1
xa2
. . . x
an1
Put x = 1 in above equation to get :n
1 a1
1a2
. . .
1an1
6.(C) Let z1 anda bi z2 c di
1 2 z z a c b d i
1 2
0 Im z z adbc or0 addc0 or d a
c
0 ac or d = 0 In both cases we find that z1z27.(C) x iy
1 i 1 2 i
1 3 i
2
2
2
2
2 1 1 2 1 3 1 1 2 1 3 x iy i i i i i i 2 5 10 100. . 8.(C)
2 2 1 1 2 p q r p q r i i i a b c a b c p22 q22 r22 2 pq qr pr p22 q22 r22 2pqr c a b p22 q22 r22 2i ab bc ac abc r p q a b c a b c a b c Complex Numbers
HWT - 4
9.(B)
cosi sin
cos3i sin3
cos5i sin5
. . . . cos
2n1
i sin
2n1
ei . ei3 . ei5. . . ei2n1 2
1 3 5 2 1 2 2 1 i n i . . . . n e e cos n i sin n
2 1cos n and sin n
2 0 2 2 r n (r is an integer) 10.(AD) aei, bei and cei 1 i i i a b c e e e b c a
cos
cos
cos
1 and sin
sin
sin
01.(D) 2 2 1 2 3 2 2 1 2 2 2 1 2 1 2 3 i i i i i i i i i i
2.(D) We cannot compare two complex (non-real) numbers
3.(B) 1 1a 2 2 . . .nan 1 1a 2a2 . . . nan 1 a1 1 a1 . . . n an
1 a1 2 a2 . . . n an 1 2 . . . n 1
4.(D) According to the question
sin xcos x and cos x2 sin x2 or tan x1 and tan x2 1 Which is not possible for any value of x.
5.(D) 3 i
a ib c id
ac bd
i adbc
. . . .(i)1 1 1 1
1 b d
b d a c bc ad
tan tan tan tan
bd a c ac bd ac 1 1 3 tan [From (i)] 6 n , n Z 6.(C) Multiplicative inverse of 3 4 4 5 8 31 4 5 3 4 25 i i i i i
7.(B) Let z1 z2 r `and arg z
1 arg z
2 z1r cos
i sin
and z2r cos
i sin
z1z28.(D)
6 6 6 6 2 2 2 2 1 1
6 6 2 2 2 64 64 128 9.(A) x and 3 y 52
x3
2y2 or5 2 2 6 4 0 x y x 10.(A) For given seriesTr r 1 r 12 r2 r 1
2
1 1 1 2 1 1 1 6 2 n n r r r n n n n n T r r n
2 2
3 n n Complex Numbers
HWT - 5
1.(C) x 2 i 3
x2
2x24x 4 3 x24x 7 0
4 3 2 2
2
2
2
2
2 5 7 41 2 4 7 3 4 7 5 4 7 6 2 0 3 0 5 0 6 6 f x x x x x x x x x x x x x x x 2.(B) Determinant = 2 2 2 2 2 4 2 4 1 1 1 3 0 0 1 1 1 1 1 1 (Applying R1R1R2R3)
2
3 1 3 3 1 3.(D) Let z x iy, y0
2 2 2 2 2 1 2 1 1 2 az z x y xyi x yi x y x xyy i As a is real 2xy y 0 y
2x 1
0 1 2 x 2 2 2 1 2 1 3 2 1 1 2 2 4 ax y x y y As y is non-zero a cannot be 3 4. 4.(D)
2 2 1 1 1 1 1 0 2 2 1 z z z w w z z Re w z as z 1 5.(B)
12
n 1 4
n
n
2 n n2n 3 is the least value of n.
6.(C)
2 2 2 2 a b c a b c . a b c
2
2
2 2 2
a b c . a b c a b c ab bc ca 1
2
2
2 2 a b b c c a Minimum value occurs when two integers are equal and third is 1 more than them.
2 2 1
2 1 2
abc or a b c2 1
7.(A) eiz eir cos i sin eir cosr cos er sin
8.(A) We have z
z 1
z z 1 1 z z 1 Hence, the minimum value of z is 1.z 19.(B) Let akibkrk
coski sink
1 2 3 1 2 3 n i . . . n r . r . r . . . r e A iB
1 2 3 n
B tan . . . A or 1 2 3 . . . k tan 1 B A or 1 1 1 n k k k b B tan tan a A
10.(D) 8 8 2 2 1 2 2 8 8 16 16 16 1 2 2 8 8 16 16 16cos i sin cos i sin . cos cos i sin cos i sin . cos
8 8 16 8 8 16 16 16 16 1 16 16 16 i i i icos cos i sin
e
e e
e cos cos i sin
Complex Numbers
HWT - 6
1.(B)
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 i isin i cos i cos i sin
e e sin i cos i cos i sin
8 4 2 1 i e e 2.(A) x2 x 1
x
x2
According to the question
n1
n and1 0
2 n1 2 n 1 0 n6k1
3.(B) x22x cos 1
x cos
2sin20 x cos 0 and sin 0
n and cos x 10r1 2
2 1
n n
x x cos n
4.(D) Conjugate complex number of
2
2 2 2 1 2 1 2 i i i i 2 2 11 22 11 3 4 25 25 i i i i 5.(D) Let z x iy
25 1 i i x iy w wz x iy
2 2 1 x y i y x x y
2 2 1 1 1 x y x y y x i x y x y x y x y According to the question
1x
yx
y xy
0 or
yxxxyx2
yxyy20 or
1 x2y2
0 or x2y21 z and1 z1 6.(B) Let z . Thenx iy x2y21
2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 x x y xy i z x iy y x y x y z x y xyi x y xyi We can see that
2 0 1 z Re z 2 1 z z lies on y-axis.
7.(B) The locus of point z satisfying
3arg z is a ray starting from origin (excluding origin) and going in the I quadrant with making angle of
3
with +ve x-axis.
8.(A) z11, z2 1, z3i, z4 i
2 2 2 2 2 2 2 2 1 2 3 4 1 1 1 1 1 1 0 z z z z i i Complex Numbers
HWT - 7
9.(B)
30 30 2 3 30 2 3 30 1 1 k k k k . . . . . . . . .
1 2 3 30 15 31 0 . . . 0 1 10.(D) z2m1ei2m1
2 1
1 3 29 k m mIm z sin sin . . . sin
2
15 15 30 1
2 4 2
sin . sin sin
sin sin sin
1.(B) 2 and 2 1 2.(B) Let zki z2 k2 Im z
2 0 3.(B) 1 2 1 3 1 2 i i z i (lies in II quadrant) 4.(A)
x iy
1 3 a ib x iy
aib
3a3b i3 3a bi2 3a
b2
3 2
2 3
3 3 a ab i a b b 3 2 3 xa ab and y3a b b2 3 or x y
a2 3b2
3a2 b2
4a2 4b2 a b 5.(C) f
3k 1 3k6k3
2 3 1 3 2 1 0 f k f k Range of f n
0 3, 6.(D) 2 2 2 2 2 1 a b c a b c c a b b c a 7.(CD)
2 2 1 1 1 1 1 n n n n i i i i i
2 1 1 2 1 1 2 2 2 2 n n n n n i i i i i i . i i 8.(B) z 1 i 3 22
2 3 3 3 arg z arg z 9.(D)
sini cos
4i4
cosi sin
41. ei4 cos4i sin410.(B) According to the question0 1
z 1
1 i Let z x iy Then0
x 1
iy 1 i or 0 y
1 x i
1 x = 1 and 0 y 1Complex Numbers
HWT - 8
1.(A) 3 3 1 4 4 2 2 i zr cos i sin r 2 2 1 1 r z z zz r 1 z z arg zz 2.(B)
1i 5 1 i 5
2 5
ei 4 5
2 5
ei4
5
5 4
5 4
5 4 2 4 2 4 2 2 4 i i e e cos 1 4 2 2 8 2 3.(A) 2 2 2 2 2 1 1 1 1 1 x x x x x x x x (ApplyingR1R1R2R3)
2
2 2
2
1 1 x x x x x x x 3 2 2 2 2 3 x x x x x x x 4.(B) Let us take z1cos A i sin A, z 2cos Bi sin B, z3cos Ci sin C
z1z2z30 3 3 3 1 2 3 31 2 3 3 3 i A B C iA iB iC z z z z z z e .e . e e But z13z23z33ei A3 ei B3 ei C3 3 3 3 3ei A B C ei Aei Bei C
Equating real parts no both sides we get: 3cos A
B C
3cos
3 cos A3 cos B3 cos C35.(B) ei811 2 3 4 5 i8 11 i16 11 i22 11 i32 11 i40 11 e e e e e Also 2345 8 11 i16 11 i24 11 i32 11 i40 11 e e e e e We can re-write 2 3 4 5 i8 11 i6 11 i2 11 i10 11 i18 11 e e e e e and 2345ei1411ei611ei2011ei1211ei411
We can see that are 10 eleventh roots of unity. 1 is the eleventh root of unity which is, 2, 3, 4, 5, , 2, 3, 4, 5 remaining.
2 3 4 5 2 3 4 5
1 (sum of nth roots of unity is zero).0
2 3 4 5
12 Re
6.(C) 6 7 1 6 7 4 4 i x
4x6
2 49 or 16x236 48 x 49 or 16x248x850 Now, 16x316x211x173x
16x248x85
2 16x248x85
3 x
0 2 0 3 3 7.(B) 3i32ai2
1 a i
5
1 a 3
i 5 2a
As the above number is real 1 ora 3 0 a 28.(B) s 1 232. . . n n1 2 3 2 3 n s . . . n
2 1
1 1 n n s . . . . . n
1 1 1 1 n n n
1 2 1 1 1 1 1 n n n s
2
2
1 1 1 1 1 1 1 1 1 1 n n n n n 1 n 9.(A) 2 1 1 2 1 2 2 4 3 4 3 3 0 z z z z zarg arg arg arg
z z z z z 10.(B)
2 2 2 2 2 a b abi a ib a ib a b Let a ib r a ib and a ib arg a ib a ib i re a ib and a ib tan i log a ib
22ab2 22ab2tan i log r i tan
a b a b
1.(C) Let z x iy 2 1 1 z m z 2 2 2 2z1 m z1 or
2x1
2 2y 2m2
x1
2y2 or 4x24y24x 1 m2x2y22x1 2 4m or m for above to be the equation of a circle.2
2.(A) 6 6 6 2 6 2 1 3 1 3 2 2 1 1 2 2 1 3 1 3 2 i i i i 3.(D) 2 30 2 2 3 2 1 tan z z log z or 2 2 2 2 3 1 3 1 3 z z z or 2 2 z 2 z 3 3 z 3 or 2 z 2 z 6 0 or
2 z 3
z 2
0 3 2 z or z 2 z 2 4.(A) Let z x iy 8
8
6 6 x y i z i z x yi
2 2 6 8 6 8 6 x x y y i x y xy x y According to the question x2y26x8y0
5.(B) aei2, bei2, cei2, dei2 abcd 1 ei e i 2cos
abcd 6.(D) inin1in2in3in1 i i2 i3 in1 i 1 i 0 7.(D) x iy a ib c id . . . .(i) Replace i by to get :i a ib x iy c id . . . .(ii)Multiply (i) and (ii) to get : 2 2 2 2 2 2 a b x y c d or
2 2 2 2 2 2 2 a b x y c d Complex Numbers
HWT - 10
8.(D) 8 8 1 1 2 2 2 2 9 9 9 9 r r r r r r
sin i cos i cos i sin
2 8 9 1 i r r i e i
[Sum of 8 9th roots of unity excluding 1]0 1
i i
(Sum of nth roots of unity = 0)
9.(D) z3 z 0 z3 z
Taking mod of both sides we get : 3 3 z z z z 0 1, 0 z z = 0 Now, z3 z z4 zz z2 If z then above equation becomes1
4 1
z then above equation becomes
5 solutions of the equation z3 z 0
10.(A) 1 1 2 1 2 3 2 2 1 2 3 i i n n n n n z e e 1 1 1 1 1 1 2 3 5 5 7 2 1 2 3 1 2 i . . . . n n n z . z . . . z e 1 1 2 3 2 3 i n e As n 1 6 2 3 1 2 i i n z . z . . . . z e e