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Jee 2014 Booklet3 Hwt Solutions Complex Numbers

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(1)

1.(A) ii

 

ei 2 iei2 2e2

2.(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 4i20

 

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) z3z

Taking mod of both sides, we get : 3 3 zzzzz  or 10 0 z   z0 Now, z3zz4zzz2 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

12

. . . .

10

102

2

2 1 r T  r rr  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

(2)

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 x2y2  or1 0 x2y2 1 z2

z 1

1.(A) Let z x iyz2

x2y2

i

 

2xy

 

2 0

Re z   x2y20 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

x5

2x210x25 16 or x210x410 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 xxx   x x xx x xx  xx  xx    

4.(A) We can compare two real numbers only  b = d = 0

5.(D) 1  20 

 

  

 

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 Trr  . 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 2i 3 and 2i 3.

Sum of roots  a

2i 3

 

 2i 3

4  a  and product of roots4  b

2i 3



2i 3

7

9.(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

(3)

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 n

i   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 2

cos 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 0

Similarly, 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 gh   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 xcos i sine     1 1 3 9 3 9 1 2 i . . . i . . . x . x . . . . e e                   i 2 e i    9.(B)

x1

3 8 0 

x1

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   zz   , 2

 

 

99 99 99 99 1 1 1 1 2 1 z z           

Complex Numbers

HWT - 3

(4)

1.(A) log1 2 z 2 log1 2 zz 2 z

Distance of z from (2, 0) is less than distance of z from (0, 0)Re z

 

1

2.(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 1

i 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 iyz2

x2y2

2xyi

 

2 2 2 0 0 Re z   xy  or y x 2 2 2 2 z   xy  or x2y24  x  2 and y  2

So, the solutions of the system are z  2 2i

5.(C) xn has roots1 1, a , a , . . ., a1 2 n1xn 1

x1



xa1



xa2

. . . x

an1

1



2

1

1 1 n n x x a x a . . . x a x    or

1 x x2. . .xn1

xa1

 

xa2

. . . x

an1

Put x = 1 in above equation to get :n 

1 a1



1a2

. . .

1an1

6.(C) Let z1  anda bi z2 c di

 

1 2 zza  c b d i

1 2

0 Im zz   adbc or0 addc0 or d a

c

0  ac or d = 0 In both cases we find that z1z2

7.(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  iii  iii 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

(5)

9.(B)

cosi sin

 

cos3i sin3

 

cos5i sin5

. . . . cos

2n1

i sin

2n1

ei . ei3 . ei5. . . ei2n1   2

 

 

1 3 5 2 1 2 2 1 i n i . . . . n e e cos n i sin n                     

 

2 1

cos 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

 

0

1.(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 . . .nan1 1a2a2 . . .nan1 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 x1 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 z1z2r `and arg z

 

1    arg z

 

2

z1r cos

i sin

and z2r cos

i sin

z1z2

8.(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 52

x3

2y2 or5 2 2 6 4 0 xyx 

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

(6)

1.(C) x  2 i 3 

x2

2x24x  4 3  x24x 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 xxxx  xx xx  x xx  xx   xx    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, y0

2 2 2 2 2 1 2 1 1 2 az   z xyxyi   x yi xy   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 axy   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)

12

 

n 1 4

n

 

n 

 

2 nn2n

3 is the least value of n.

6.(C)

2 2 2 2 a bca bc . a bc  

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

abc   or a bc2 1

7.(A) eizeir cosi sin  eir cosr 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 1

9.(B) Let akibkrk

coski sink

 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 16

cos 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 i

cos cos i sin

e

e e

e cos cos i sin

                             

Complex Numbers

HWT - 6

(7)

1.(B)

 

8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 i i

sin 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

 

n1 

 

n  and1 0

   

2 n1 2 n 1 0

n6k1

3.(B) x22x cos 1

x cos

2sin20

x cos 0 and sin 0

n and cos  x 10r1 2

2 1

n n

xx 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

1x



 yx

 

y xy

0 or

 yxxxyx2

yxyy20 or  

1 x2y2

0 or x2y21  z  and1 z1 6.(B) Let z  . Thenx iy x2y21

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

 

3

arg 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) z11, z2 1, z3i, z4 i

       

2 2 2 2 2 2 2 2 1 2 3 4 1 1 1 1 1 1 0 zzzz     i  i     

Complex Numbers

HWT - 7

(8)

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) z2m1ei2m1

2 1

1 3 29 k m m

Im zsin 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 zkiz2 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 ibx iy

aib

3a3b i3 3a bi2 3a

 

b2   

3 2

 

2 3

3 3 a ab i a b b      3 2 3 xaab and y3a b b2  3 or x y

a2 3b2

 

3a2 b2

4a2 4b2 a b      5.(C) f

 

3k  1 3k6k3

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 ii      

   

 

2 1 1 2 1 1 2 2 2 2 n n n n n i i i i ii. i i           8.(B) z 1 i 3 22

 

 

2 3 3 3 arg zarg z  

9.(D)

sini cos

4i4

cosi sin

41. ei4cos4i sin4

10.(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 1

Complex Numbers

HWT - 8

(9)

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)

   

1i 5 1 i 5

 

2 5

 

ei 4 5

 

2 5

ei4

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 z1cos A i sin A, z 2cos Bi sin B, z3cos Ci sin C

z1z2z30  3 3 3   1 2 3 31 2 3 3 3 i A B C iA iB iC zzzz z ze .e . ee    But z13z23z33ei 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 C3

5.(B)  ei811  2 3 4 5 i8 11 i16 11 i22 11 i32 11 i40 11 e e e e e         Also   2345 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   2345ei1411ei611ei2011ei1211ei411

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

1

2 Re    

(10)

6.(C) 6 7 1 6 7 4 4 i x     

4x6

2 49 or 16x236 48 x 49 or 16x248x850 Now, 16x316x211x173x

16x248x85

 

2 16x248x85

 3 x

   

0 2 0  3 3 7.(B) 3i32ai2 

1 a i

   5

1 a 3

 

i 5 2a

As the above number is real 1   ora 3 0 a 2

8.(B) s 1 232. . . nn1 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 z

arg 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 22ab2

tan i log r i tan

a b a b               

(11)

1.(C) Let z x iy 2 1 1 z m z   2 2 2 2z1 m z1 or

2x1

  

2 2y 2m2

x1

2y2   or 4x24y24x 1 m2x2y22x1  2 4

m  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 x2y26x8y0

5.(B) aei2, bei2, cei2, dei2abcd 1 ei  e i  2cos

abcd                        6.(D) inin1in2in3in1  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

(12)

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 zzzz 0 1, 0 z   z = 0 Now, z3 zz4   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              

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