Rd Sharma Xi 2018 Solutions for Class 11 Humanities Math Chapter 13 Complex Numbers are provided here with simple step-by-step explanations. These solutions for Complex Numbers are extremely popular among Class 11 Humanities students for Math Complex Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Humanities Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 13.3:

Question 1:

Evaluate the following:
(i) i457
(ii) i528
(iii) 1i58
(iv) i37+1i67
(v) i41+1i2579
(vi) (i77+i70+i87+i414)3
(vii)  i30+i40+i60
(viii) i49+i68+i89+i110

Answer:


i i457   =i4×114+1              =i4114×i               =i      i4=1ii i528  =i4×132              =i4132              =1      i4=1iii 1i58=1i4×14+2              =1i414×i2              =1i2      i4=1              =-1      i2=-1iv i37+1i67=i4×9+1+1i4×16+3                      =i49×i+1i416×i3                           =i-1i            i3=-i                      =i-1i×ii                      =i-ii2                      =i--i          i2=-1                              =2i  v i41+1i2579 =i4×10+1+1i4×64+19                             =i410×i+1i464×i9                               =i+1i9                      i4=1

                      =i+ii29  =i-i9          i2=-1=0

 vi i77+i70+i87+i4143=i4×19+1+i4×17+2+i4×21+3+i4×103+23                                          =i419×i+i417×i2+i421×i3+i4103×i2                                          =i-1-i-13         i4=1,i3=-i and i2=-1                                          =-23                                          =-8vii i30+i40+i60=i4×7+2+i4×10+i4×15                             =i47×i2+i410+i415                             =-1+1+1           i4=1,i2=-1                             =1 viii i49+i68+i89+i110=i4×12+1+i4×17+i4×22+1+i4×27+2                                       =i412×i+i417+i422×i+i427×i2                                       =i+1+i-1           i4=1,i2=-1                                       =2i



Page No 13.31:

Question 1:

Express the following complex numbers in the standard form a + i b:
(i) (1+i)(1+2i)
(ii) 3+2i-2+i
(iii) 1(2+i)2
(iv) 1-i1+i
(v) (2+i)32+3i
(vi) (1+i)(1+3i)1-i
(vii) 2+3i4+5i
(viii) (1-i)31-i3
(ix) (1+2i)-3
(x) 3-4i(4-2i)(1+i)
(xi) 11-4i-21+i1-4i5+i
(xii) 5+2i1-2i

Answer:

i 1+i 1+2i=1+2i+i+2i2=1+3i-2              i2=-1 =-1+3iii 3+2i-2+i=3+2i-2+i×-2-i-2-i=-6-3i-4i-2i24-i2             i2=-1=-6-7i+24+1=-4-7i5=-45-75iiii12+i2=14+i2+4i               i2=-1=13+4i=13+4i×3-4i3-4i=3-4i9-16i2=3-4i9+16=325-425i

iv1-i1+i=1-i1+i×1-i1-i=1+i2-2i1-i2                  i2=-1=-2i2=-iv2+i32+3i=4+i2+4i2+i2+3i           i2=-1=8+2i2+8i+4i+i3+4i22+3i            =2+11i2+3i×2-3i2-3i=4-6i+22i-33i24-9i2=37+16i4+9=3713+1613ivi1+i1+3i1-i=1+i1+3i1-i×1+i1+i=1+3i1+i2+2i1-i2               i2=-1=1+3i2i2=i1+3i=i+3i2=-3+ivii 2+3i4+5i=2+3i4+5i×4-5i4-5i=8-10i+12i-15i216-25i2           i2=-1=23+2i16+25=2341+241i

(viii) 1-i31-i3=1+i2-2i1-i1-i3          i2=-1=-2i1-i1-i3×1+i31+i3=-2i1+i3-i-i41-i6=-2i1-i-i-11-i2=-2i-2i2=-2+0i


(ix) 1+2i-3=11+2i3=11+8i3+6i+12i2=11-8i+6i-12       i2=-1 & i3=-i=1-2i-11=1-2i-11×-2i+11-2i+11=-2i+114i2-121=-2i+11-4-121=-2i+11-125=-11125+2i125

(x) 3-4i4-2i1+i=3-4i4+2i-2i2         i2=-1=3-4i6+2i=3-4i6+2i×6-2i6-2i=18-6i-24i+8i236-4i2=18-30i-836+4     =10-30i40=14-34i     

(xi) 11-4i-21+i3-4i5+i=1+i-2+8i1-4i1+i3-4i5+i=-1+9i1-3i+4i23-4i5+i=-1+9i5-3i3-4i5+i            i2=-1=-3+4i+27i-36i225+5i-15i-3i2=33+31i28-10i=33+31i28-10i×28+10i28+10i=924+330i+868i+310i2784-100i2=614+1198i884=307442+599442i

(xii) 5+2i1-2i=5+2i1-2i×1+2i1+2i=5+52i+2i+2i21-2i2=5+62i-21+2         i2=-1=3+62i3=1+22i

Page No 13.31:

Question 2:

Find the real values of x and y, if
(i) (x+iy)(2-3i)=4+i
(ii) (3x-2iy)(2+i)2=10(1+i)
(iii) (1+i)x-2i3+i+(2-3i)y+i3-i
(iv) (1+i)(x+iy)=2-5i

Answer:

i x+iy2-3i=4+i2x-3ix+2iy-3i2y=4+i2x+3y+i-3x+2y=4+iComparing both the sides: 2x+3y=4            ....(1)  -3x+2y=1        ....(2)Multiplying equation (1) by 3 and equation (2) by 2:    6x+9y=12          ...(3)-6x+4y=2            ...(4)Adding equations (3) and (4):13y=14y=1413Substituting the value of y in equation (1): 2x+3×1413=42x=4-42132x=1013x=513 x=513 and y=1413 

(ii) 3x-2iy2+i2=10 1+i 3x-2iy4+i2+4i=101+i 3x-2iy3+4i=101+i9x+12xi-6iy-8i2y=10+10i9x+8y+i12x-6y=10+10iComparing both the sides:9x+8y=10        ....(1)12x-6y=10or, 6x-3y=5       ...(2)Multiplying equation (1) by 3 and equation (2) by 8,27x+24y=30      ....(3)      48x-24y=40     ....(4)Adding equations (3) and (4): 75x=70 x=1415Substituting the value of x in equation (1): 9×1415+8y=1012615+8y=108y=10-126158y=2415y=15

(iii) 1+ix-2i3+i+2-3iy+i3-i=i1+i3-ix-2i3-i+2-3i3+iy+i3+i3+i3-i=i3x-ix+3ix-i2x-6i+2i2+6y+2iy-9iy-3i2y+3i+i29-i2=i4x+2ix-3i+9y-7iy-310=i4x+9y-3+i2x-3-7y=10iComparing both the sides: 4x+9y-3=04x+9y=3       ....(1) 2x-3-7y=102x-7y=13            ...(2)Multiplying equation (2) by 2: 4x-14y=26        ...(3) Subtracting equation (3) from (1):     4x+9y=    3      4x-14y=26     -  +          -              23y=-23 y=-1Substituting the value of y in equation (1): 4x-9=34x=12x=3 x=3 and y=-1

(iv) 1+ix+iy=2-5ix+iy+ix+i2y=2-5ix+iy+ix-y=2-5ix-y+iy+x=2-5iComparing both the sides,x-y=2        ...(1) x+y=-5     ...(2)Adding equations (1) and (2), 2x=-3x=-32Substituting the value of x in equation (1),-32-y=2y=-32-2y=-72x=-32 and y=-72

Page No 13.31:

Question 3:

Find the conjugates of the following complex numbers:
(i) 4 − 5 i
(ii) 13+5i
(iii) 11+i
(iv) (3-i)22+i
(v) (1+i)(2+i)3+i
(vi) (3-2i)(2+3i)(1+2i)(2-i)

Answer:

i Let z=4-5i   z¯=4+5i      z=a+ib, so  z¯=a-ibii Let z=13+5i        =13+5i×3-5i3-5i        =3-5i9-25i2        =3-5i9+25        =3-5i34  z¯=3+5i34iii Let z=11+i         =11+i×1-i1-i         =1-i1-i2         =1-i2 z¯=1+i2iv Let z=3-i22+i         =9-6i-12+i         =8-6i2+i×2-i2-i         =16-8i-12i+6i24-i2         =10-20i5         =2-4i z  =2+4iv Let  z=1+i2+i3+i         =2+i+2i+i23+i         =1+3i3+i         =1+3i3+i×3-i3-i         =3-i+9i-3i29-i2         =6+8i10         =3+4i5 z¯=3-4i5vi Let z=3-2i2+3i1+2i2-i         =6+9i-4i-6i22-i+4i-2i2         =6+6+5i2+2+3i         =12+5i4+3i×4-3i4-3i         =48-36i+20i-15i216-9i2         =63-16i25  z=63+16i25



Page No 13.32:

Question 4:

Find the multiplicative inverse of the following complex numbers:
(i) 1 − i
(ii) (1+i3)2
(iii) 4 − 3i
(iv) 5+3i

Answer:

i Let z=1-i .Then ,1z=11-i    =11-i×1+i1+i   =1+i1-i2   =121+i   =12+12iii z=1+3i2       =1+3i2+23i       =-2+23iThen, 1z=1-2+23i×-2-23i-2-23i               =-2-23i4-12i2               =-2-23i16               =-18-38iiii z=4-3iThen, 1z=14-3i×4+3i4+3i                =4+3i16-9i2                =4+3i25               =425+325iiv z=5+3iThen, 1z=15+3i×5-3i5-3i               =5-3i5-9i2               =5-3i5+9               =5-3i14               =514-314i

Page No 13.32:

Question 5:

If z1=2-i,z2=1+i, find z1+z2+1z1-z2+i

Answer:

Given: z1=2-i, z2=1+i z1+z2+1z1-z2+i=2-i+1+i+12-i-1-i+i                            =41-i                            =41-i   Also, 1-i=12+i2             a+bi=a2+b2                   =2              z1+z2+1z1-z2+i=42                          

Page No 13.32:

Question 6:

If z1=2-i,z2=-2+i, find
(i) Re z1z2z1
(ii) Im 1z1z1

Answer:

i z1=2-i, z2=-2+i, z1=2+i z1z2z1=2-i-2+i2+i                   =-4+2i+2i-i22+i                  =-3+4i2+i                 =-3+4i2+i×2-i2-i                =-6+3i+8i-4i222-i2               =-2+11i4--1               =-2+11i5Rez1z2z1=-25ii1z1z1=12-i2+i                    =122-i2                    =15Im1z1z1=0          Since no term containing i is present

Page No 13.32:

Question 7:

Find the modulus of 1+i1-i-1-i1+i

Answer:

1+i1-i-1-i1+i=1+i1+i-1-i1-i1-i1+i=1+i2+2i-1-i2+2i12-i2=4i2   i2=-1=2i 2i=02+22          =2                  a+bi=a2+b21+i1-i-1-i1+i=2

Page No 13.32:

Question 8:

If x+iy=a+iba-ib, prove that x2 + y2 = 1

Answer:

x+iy=a+iba-ibTaking mod on both the sides:x+iy=a+iba-ibx2+y2=a2+b2a2+b2x2+y2=1x2+y2=1Hence proved.

Page No 13.32:

Question 9:

Find the least positive integral value of n for which 1+i1-in is real.

Answer:

1+i1-in=1+i1-i×1+i1+in=1+i2+2i1-i2n=1-1+2i1+1n=2i2n=inFor in to be real, the least positive value of n will be 2. As i2=-1

Page No 13.32:

Question 10:

Find the real values of θ for which the complex number 1+i cosθ1-2i cosθ is purely real.

Answer:

1+icosθ1-2icosθ=1+icosθ1-2icosθ×1+2icosθ1+2icosθ=1+2icosθ+icosθ-2cosθ1+4cos2θ=1-2cosθ+i3cosθ1+4cos2θFor it to be purely real, the imaginary part must be zero.3cosθ=0This is true for odd multiples of π2.θ=2n+1π2, nZ

Page No 13.32:

Question 11:

Find the smallest positive integer value of m for which (1+i)n(1-i)n-2 is a real number.

Answer:

1+im1-im-2=1+im1-im×1-i2=1+i1-i×1+i1+im×1+i2-2i=1+i2+2i1-i2m×1-1-2i=1-1+2i1+1m×-2i=-2iim=-2im+1For this to be real, the smallest positive value of m will be 1. Thus, i1+1=i2=-1, which is real.

Page No 13.32:

Question 12:

If 1+i1-i3-1-i1+i3=x+iy, find (x, y).

Answer:

1+i1-i=1+i1-i×1+i1+i            =1+i212-i2            =12+i2+2i1+1       [ i2=-1]            =1-1+2i2            =2i2            =i                   ....(1)


Also,
1-i1+i=1-i1+i×1-i1-i            =1-i212-i2            =12+i2-2i1+1       [ i2=-1]            =1-1-2i2            =-2i2            =-i                   ....(2)

It is given that,
 1+i1-i3-1-i1+i3=x+iy(i)3-(-i)3=x+iy             [From (1) and (2)]i3+i3=x+iy2i3=x+iy0-2i=x+iy                     [ i3=-i]x=0 and y=-2

Thus, (xy) = (0, −2).

Page No 13.32:

Question 13:

If 1+i22-i=x+iy, find x + y.

Answer:

1+i22-i=12+i2+2i2-i            =1-1+2i2-i                   [ i2=-1]            =2i2-1×2+i2+i                  =2i(2+i)22-i2            =4i+2i24+1                       [ i2=-1]            =4i-25            =-25+45i                   ....(1)


It is given that,
 1+i22-i=x+iy-25+45i=x+iy             [From (1)]x=-25 and y=45

x+y=-25+45           =25

Thus, x + y = 25.

Page No 13.32:

Question 14:

If 1-i1+i100=a+ib, find (ab).

Answer:

1-i1+i=1-i1+i×1-i1-i         =1-i212-i2         =12+i2-2i1+1                  [ i2=-1]               =1-1-2i2         =-2i2         =-i                            ....(1)


It is given that,
 1-i1+i100=a+ib(-i)100=a+ib             [From (1)]i4×25=a+ib1+0i=a+ib                [ i4=1]a=1 and b=0

Thus, (ab) = (1, 0).

Page No 13.32:

Question 15:

If a=cosθ+isinθ, find the value of 1+a1-a.

Answer:

1+a1-a=1+cosθ+isinθ1-cosθ-isinθ          =(1+cosθ)+isinθ(1-cosθ)-isinθ×(1-cosθ)+isinθ(1-cosθ)+isinθ          =1-cosθ+isinθ+cosθ-cos2θ+icosθsinθ+isinθ-isinθcosθ+i2sin2θ(1-cosθ)2-i2sin2θ          =1-cos2θ-sin2θ+2isinθ1+cos2θ-2icosθ+sin2θ                              [ i2=-1]          =sin2θ-sin2θ+2isinθ2-2icosθ                                     [ cos2θ+sin2θ=1]          =isinθ1-cosθ          =2isinθ2cosθ22sin2θ2          =2icotθ2

Thus, 1+a1-a=2icotθ2.

Page No 13.32:

Question 16:

Evaluate the following:
(i) 2x3+2x2-7x+72, when x=3-5i2
(ii) x4-4x3+4x2+8x+44, when x=3+2i
(iii) x4+4x3+6x2+4x+9, when x=-1+i2
(iv) x6+x4+x2+1, when x=1+i2
(v) 2x4+5x3+7x2-x+41, when x=-2-3i

Answer:

i x=3-5i2x2=3-5i22         =9+25i2-30i4        =-16-30i4x3=-16-30i4×3-5i2        =-48+80i-90i+150i28       =-198-10i8 2x3+2x2-7x+72=2-198-10i8+2-16-30i4-73-5i2+72                                      =-198-10i-32-60i-42+70i+2884                                      =16 4                                      =4ii x=3+2ix2=3+2i2         =9+4i2+12i         =5+12ix3=x2×x         =5+12i×3+2i         =15+10i+36i-24         =-9+46ix4=x22         =5+12i2         =25+144i2+120i         =-119+120ix4-4x3+4x2+8x+44=-119+120i-4-9+46i+45+12i +83+2i+44                                             =-119+120i+36-184i+20+48i+24+16i+44                                             = 5iii x=-1+2ix2=-1+2i2         =1+2i2-22i         =-1-22ix3=-1-22i×-1+2i         =1-2i+22i-4i2         =5+2ix4=-1-22i2         =1+8i2+42i         =-7+42ix4+4x3+6x2+4x+9=-7+42i+45+2i+6-1-22i+4-1+2i+9                                           =-7+42i+20+42i-6-122i-4+42i+9                                           =12iv x=1+i2x2=1+i22         =1+i2+2i2         =2i2        =ix6=x23         =i3          =-ix2=ix4=x22         =i2         =-1Now,  x6+x4+x2+1=-i-1+i+1                                     =0

(v) x=-2-3ix2=-2-3i2       =(-2)2+(-3i)2+2(-2)(-3i)       =4+3i2+43i       =4-3+43i                          [ i2=-1]       =1+43ix3=1+43i×-2-3i       =-2-3i-83i-12i2       =10-93i                             [ i2=-1]x4=1+43i2       =1+48i2+83i       =-47+83i                         [ i2=-1]2x4+5x3+7x2-x+41=2(-47+83i )+510-93i+71+43i--2-3i+41                                      =-94+163i+50-453i+7+283i+2+3i+41                                      =6

Page No 13.32:

Question 17:

For a positive integer n, find the value of (1-i)n1-1in.

Answer:

(1-i)n1-1in=1-in1-i4in               [ i4=1]                        =1-in1-i3n                        =1-in1+in                   [ i3=-i]                        =(1-i)(1+i)n                        =(1-i2)n                        =2n                                    [ i2=-1]

Thus, the value of (1-i)n1-1in is 2n.



Page No 13.33:

Question 18:

If 1+iz=1-iz¯, then show that z=-iz¯.

Answer:

1+iz=1-iz¯zz¯=1-i1+izz¯=1-i1+i×1-i1-izz¯=1+i2-2i1-i2zz¯=1-1-2i1+1              [ i2=-1]zz¯=-2i2zz¯=-iz=-iz¯

Hence,  z=-iz¯.

Page No 13.33:

Question 19:

Solve the system of equations Rez2=0, z=2

Answer:

Let z=x+iy.
Then ,
z2=x+iy2   =x2+i2y2+2ixy   =x2-y2+2ixy          [ i2=-1]

and z=x2+y2

According to the question,
Rez2=0 and z=2x2-y2=0 and x2+y2=2x2-y2=0 and x2+y2=4On Adding both the equations, we get2x2=4x2=2x=±2y2=2y=±2

Thus, x=±2 and y=±2.

Page No 13.33:

Question 20:

If z-1z+1 is purely imaginary number (z-1), find the value of z.

Answer:

Let z=x+iy.
Then,
z-1z+1=x+iy-1x+iy+1          =x-1+iyx+1+iy×x+1-iyx+1-iy          =x2+x-ixy-x-1+iy+ixy+iy-i2y2x+12-i2y2          =x2+y2-1+2iyx2+1+2x+y2              [ i2=-1]

If z-1z+1 is purely imaginary number, then
Rez-1z+1=0x2+y2-1=0x2+y2=1z2=1z=1

Thus, the value of z is 1.

Page No 13.33:

Question 21:

If z1 is a complex number other than −1 such that z1=1 and z2=z1-1z1+1, then show that the real parts of z2 is zero.

Answer:

Let z=x+iy.
Then,
z2=z1-1z1+1   =x+iy-1x+iy+1   =x-1+iyx+1+iy×x+1-iyx+1-iy   =x2+x-ixy-x-1+iy+ixy+iy-i2y2x+12-i2y2   =x2+y2-1+2iyx2+1+2x+y2              [ i2=-1]

Now,
Rez2=x2+y2-1x2+y2+1+2x          =0                       [ z1=1x2+y2=1]

Thus, the real parts of z2 is zero.

Page No 13.33:

Question 22:

If z+1=z+21+i, find z.

Answer:

Let z=x+iy.
Then,
z+1=x+1+iyz+1=x+12+y2


z+1=z+21+ix2+2x+1+y2=x+iy+2+2ix2+2x+1+y2=x+2+iy+2x2+2x+1+y2=x+2 and y+2=0x2+2x+1+y2=x+22 and y=-2x2+2x+1+y2=x2+4x+4 and y=-2y2=2x+3 and y=-24=2x+3 and y=-22x=1 and y=-2x=12 and y=-2

z=x+iy=12-2i

Thus, z=12-2i

Page No 13.33:

Question 23:

Solve the equation z=z+1+2i.

Answer:

Let z=x+iy.
Then,
z=x2+y2


z=z+1+2ix2+y2=x+iy+1+2ix2+y2=x+1+iy+2x2+y2=x+1 and y+2=0x2+y2=x+12 and y=-2x2+y2=x2+1+2x and y=-2y2=2x+1 and y=-24=2x+1 and y=-22x=3 and y=-2x=32 and y=-2

z=x+iy=32-2i

​Thus, z=32-2i

Page No 13.33:

Question 24:

What is the smallest positive integer n for which 1+i2n=1-i2n?

Answer:

1+i2n=1-i2n1+i2n=1-i2n12+i2+2in=12+i2-2in1-1+2in=1-1-2in                   [ i2=-1]2in=-2in2in=-1)n(2in(-1)n=1n is a multiple of 2

Thus, the smallest positive integer n for which 1+i2n=1-i2n is 2.

Page No 13.33:

Question 25:

If z1, z2, z3 are complex numbers such that z1=z2=z3=1z1+1z2+1z3=1, then find the value of z1+z2+z3.

Answer:

z1+z2+z3=z1z1¯z1¯+z2z2¯z2¯+z3z3¯z3¯                    =z12z1¯+z22z2¯+z32z3¯                    =1z1¯+1z2¯+1z3¯                               [ z1=z2=z3=1]                   =1z1+1z2+1z3 ¯                   =1                                             1z1+1z2+1z3 =1


Thus, the value of z1+z2+z3 is 1.

Page No 13.33:

Question 26:

Find the number of solutions of z2+z2=0

Answer:

Let z=x+iy.
Then,
z=x2+y2


z2+z2=0x+iy2+x2+y22=0x2+i2y2+2ixy+x2+y2=0x2-y2+2ixy+x2+y2=02x2+2ixy=02x(x+iy)=0x=0 or x+iy=0x=0 or z=0

For x=0, z=0+iy

​Thus, there are infinitely many solutions of the form z=0+iy, yR.



Page No 13.39:

Question 1:

Find the square root of the following complex numbers:
(i) −5 + 12i
(ii) −7 − 24i
(iii) 1 − i
(iv) −8 − 6i
(v) 8 −15i
(vi) -11-60-1
(vii)  1+4-3
(viii) 4i
(ix) −i

Answer:

z=±z+Rez2+iz-Rez2     , if Im(z) >0z=±z+Rez2-iz-Rez2     , if Im(z) <0i z=-5+12i , Rez=-5 and z=25+144=13 Here, Im(z) >0 z=±z+Rez2+iz-Rez2            =±13-52+i13+52           =±4+i9           =±2+3iii z=-7-24i,  Rez=-7, z=49+576=25 Here,  Im(z) <0 z=±z+Rez2-iz-Rez2        =±25-72-i25+72        =±3-4iiii z=1-i , Rez=1, z=1+1=2Here, Im(z) <0 z=±z+Rez2-iz-Rez2        =±2+12-i2-12iv z=-8-6i,  Rez=-8, z=64+36=10Here,  Im(z) <0 z=±z+Rez2-iz-Rez2            =±10-82-i10+82           =±1-3iv z=8-15i,  Rez=8, z=64+225=17Here, Im(z) <0 z=±z+Rez2-iz-Rez2             =±17+82-i17+82             =±125-3ivi -11-60-1=-11-60i,  Rez=-11, z=121+3600=61Here, Im(z) <0             z=±z+Rez2-iz-Rez2             =±61-112-i61+112             =±5-6ivii z=1+43-1=1+43i,  Rez=1, z=1+16×3=7 Here, Im(z) >0 z=±z+Rez2+iz-Rez2             =±7+12+i7-12             =±2+3iviii z=0+4i,  Rez=0, z=4 Here, Im(z) >0 z=±z+Rez2+iz-Rez2            =±4+02+i4-02            =±2+i2            =±21+iix z=-i,  Rez=0, z=1Here, Im(z) <0     z=±z+Rez2-iz-Rez2            =±12-i12            =±121-i



Page No 13.4:

Question 2:

Show that 1 + i10 + i20 + i30 is a real number.

Answer:

1+i10+i20+i30=1+i4×2+2+i4×5+i4×7+2=1+i42×i2+i45+i47×i2=1+i2+1+i2                              i4=1=1-1+1-1                                i2=-1=0This is a real number.Hence proved.

Page No 13.4:

Question 3:

Find the values of the following expressions:
(i) i49 + i68 + i89 + i110
(ii) i30 + i80 + i120
(iii) i + i2 + i3 + i4
(iv) i5 + i10 + i15
(v) i592+i590+i588+i586+i584i582+i580+i578+i576+i574
(vi) 1+ i2 + i4 + i6 + i8 + ... + i20
(vii) (1 + i)6 + (1 − i)3

Answer:

i i49+i68+i89+i110=i4×12+1+i4×17+i4×22+1+i4×27+2=i412×i+i417+i422×i+i427×i2=i+1+i+i2                    i4=1=2i+1-1                       i2=-1   =2iii i30+i80+i120=i4×7+2+i4×20+i4×30=i47×i2+i420+i430=i2+1+1             i4=1=-1+2                i2=-1      =1iii i+i2+i3+i4=i-1-i+1           i2=-1, i3=-i and i4=1=0    iv i5+i10+i15=i4×1+1+i4×2+2+i4×3+3=i41×i+i42×i2+i43×i3=i+i2+i3                  i4=1=i-1-i                     i2=-1, i3=-i    =-1v i592+i590+i588+i586+i584i582+i580+i578+i576+i574=i4×148+i4×147+2+i4×147+i4×146+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2=i4148+i4147×i2+i4146+i4146×i2+i4146i4145×i2+i4145+i4144×i2+i4144+i4143×i2=1+i2+1+i2+1i2+1+i2+1+i2        i4=1=1-1+1-1+1-1+1-1+1-1       i2=-1=-1  


(vi)  1+i2+i4+i6+i8+...+i20i2=-1,i4=1,i6=-1,i8=1,.....i20=1 1+i2+i4+i6+i8+...+i20=1+-1+1+-1+1+-1+...+1+-1+1=5×1+-1+1          As, there are 11 terms=5×0+1=1


(vii) (1 + i)6 + (1 − i)3
= [(1 + i)2]3 + (1 − i)3
= [12 + i2 + 2i]3 + (13 − i3 + 3i− 3i)
= [1 − 1 + 2i]3 + (1 + i − 3 − 3i)           [∵ i2 = −1, i= −i]
= (2i)3 + (−2 − 2i)
= 8i3 − 2 − 2i
= −8i − 2 − 2i                                        [∵ i= −i]
= −10i − 2



Page No 13.57:

Question 1:

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1 + i
(ii) 3+i
(iii) 1 − i
(iv) 1-i1+i
(v) 11+i
(vi) 1+2i1-3i
(vii) sin 120°-i cos 120°
(viii) -161+i3

Answer:

i z=1+i r=z  =1+1  =2Let tan α=ImzReztan α =11   α  =π4Since point (1,1) lies in the first quadrant, the argument of z is given by θ=  α  =π4Polar form =rcos θ+isin θ                     =2cosπ4+isinπ4ii z=3+ir=z  =3+1  =4  =2Let tan α=ImzReztan α =13   α  =π6Since point (3,1) lies in the first quadrant, the argument of z is given by θ=  α  =π6

Polar form =r cosθ +isinθ                      =2 cos π6+isin π6 

(iii) z=1-i r=z  =1+1  =2Let tan α=ImzRez tanα =-11                 =π4α =π4Since point (1,-1) lies in the fourth quadrant, the argument of z is given by θ=-α =-π4Polar form =rcos θ+isin θ                    =2cos-π4+isin-π4                   =2cosπ4-isinπ4     

(iv) 1-i1+iRationalising the denominator:1-i1+i×1-i1-i 1+i2-2i1-i2       -2 i2           i2=-1-ir=z =0+1 =1Since point (0,-1) lies on the negative direction of the imaginary axis, the argument of z is given by 3π2.Polar form =rcos θ+isin θ                   =cos3π2+isin3π2                   = cos2π-π2+isin2π-π2                    =cosπ2-isinπ2      

(v) 11+iRationalising the denominator: 11+i×1-i1-i 1-i1-i2       1-i2           i2=-112-i2r=z =14+14 =12Let tan α =Im(z)Re(z)  tan α=12-12                     =1    α =π4Since point 12,-12 lies in the fourth quadrant, the argument is given by θ=-α=-π4Polar form =rcos θ+isin θ                    =12cos-π4+isin-π4                   =12 cosπ4-isinπ4                 

(vi) 1+2i1-3iRationalising the denominator:1+2i1-3i×1+3i1+3i 1+3i+2i+6i21-9i2       -5+5i10           i2=-1-12+i2r=z =14+14 =12Let tan α =Im(z)Re(z) Then, tan α=12-12                     =1    α =π4Since point -12,12 lies in the second quadrant, the argument is given byθ=π-α  =π-π4   =3π4Polar form =rcos θ+isin θ                     =12cos3π4+isin3π4                                  

(vii) sin 120°-i cos 120° 32+i2r=z  =34+14   =1Let tan α=Im(z)Re(z)Then, tan α=1232                     =13α=π6Since point 32,12 lies in the first quadrant, the argument is given by θ=α=π6Polar form =rcosθ+i sinθ                    = cos π6+i sin π6


(viii) -161+i3Rationalising the denominator: -161+i3×1-i31-i3 -16+163i1-3i2       -16+163i4           i2=-1-4+43ir=z =16+48 =8Let tan α =Im(z)Re(z) Then, tan α=43-4                     =3    α =π3Since the point -4,43 lies in the third quadrant, the argument is given by θ=π-α   =π-π3    =2π3Polar form =rcos θ+isin θ                      =8cos2π3+isin2π3                                   

Page No 13.57:

Question 2:

Write (i25)3 in polar form.

Answer:

i253=i75       =i4×18+3       =i418.i3       =i3           [ i4=1]       =-i         [ i3=-i]

Let z=0-i.
Then, z=02+-12=1.

Let θ be the argument of z and α be the acute angle given by tanα=ImzRez.
Then, 
tanα=10=α=π2

Clearly, z lies in fourth quadrant. So, arg(z) = -α=-π2.

∴ the polar form of z is zcosθ+isinθ=cos-π2+isin-π2.

Thus, the polar form of (i25)is cosπ2-isinπ2.

Page No 13.57:

Question 3:

Express the following complex in the form r(cos θ + i sin θ):
(i) 1 + i tan α
(ii) tan α − i
(iii) 1 − sin α + i cos α
(iv) 1-icosπ3+isinπ3

Answer:

(i) Let z= 1+itan α  tan α is periodic with period π. So, let us take α [0,π2)( π2, π]Case I:When α[0,π2)z= 1+itan α z=1+tan2α         =sec α                      0<α<π2          =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α         =tan αβ=α  As z lies in the first quadrant . Therefore, arg(z)=β=αThus, z in the polar form is given by z= sec α cosα+isin α Case II:z=1+i tan α z=1+tan2α         =sec α                     π2<α<π          =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α        =-tan αtan β =tan π-αβ=π-αAs, z lies in the fourth quadrant .  arg(z)=-β= α-πThus, z in the polar form is given by z=  -sec α cosα-π+isin α-π      



(ii) Let z= tan α-i  tan α is periodic with period π. So, let us take α [0,π2)( π2, π]Case I:z= tan α-i z=tan2+1         =sec α                      0<α<π2          =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α        = cot α        =cot α        =tan π2-αβ=π2-α  We can see that Re(z) >0 and Im (z) <0.So, z lies in the fourth quadrant . arg(z)=-β=α- π2Thus, z in the polar form is given by z= sec α cosα-π2+isin α-π2 Case II:z= tan α-i z=tan2+1         =sec α                     π2<α<π          =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α        = cot α        =-cot α        =tan α-π2β=α- π2We can see that Re(z) <0 and Im (z) <0.So, z lies in the third quadrant . arg(z)=π+β= π2+αThus, z in the polar form is given by z=  -sec α cosπ2+α+isin π2+α      

(iii) Let z= 1-sinα+icosα.sine and cosine functions are periodic functions with period 2π. So, let us take α [0, 2π]Now, z=1-sinα+icosαz=1-sinα2+cos2α=2-sinα=21-sinαz=2cosα2-sinα22=2cosα2-sinα2Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=cosα1-sinα=cos2α2-sin2α2cosα2-sinα22=cosα2+sinα2cosα2-sinα2tanβ=1+tanα21-tanα2=tanπ4+α2Case I: When  0α<π2In this case, we have,cosα2>sinα2 and π4+α2[π4, π2)z=2cosα2-sinα2and tanβ=tanπ4+α2=tanπ4+α2β=π4+α2Clearly, z lies in the first quadrant.Therefore, argz=π4+α2Hence, the polar form of z is 2cosα2-sinα2cosπ4+α2+isinπ4+α2Case II: When π2<α<3π2In this case, we have,cosα2<sinα2 and π4+α2π2, πz=2cosα2-sinα2=-2cosα2-sinα2and tanβ=tanπ4+α2=-tanπ4+α2=tanπ-π4+α2=tan3π4-α2β=3π4-α2Clearly, z lies in the fourth quadrant.Therefore, argz=-β=-3π4-α2=α2-3π4Hence, the polar form of z is -2cosα2-sinα2cosα2-3π4+isinα2-3π4Case III: When 3π2<α<2πIn this case, we have,cosα2<sinα2 and π4+α2π,5π4z=2cosα2-sinα2=-2cosα2-sinα2and tanβ=tanπ4+α2=tanπ4+α2=-tanπ-π4+α2=tanα2-3π4β=α2-3π4Clearly, z lies in the first quadrant.Therefore, argz=β=α2-3π4Hence, the polar form of z is -2cosα2-sinα2cosα2-3π4+isinα2-3π4     


(iv) Let z=1-icosπ3+isinπ3             =1-i12+i32             =2-2i1+i3×1-i31-i3             =2-2i-23i+23i21+3             =2-23-2i(1+3)4             =1-3+i(-1-3)2             =1-32+i(-1-3)2Now, z=1-32+i(-1-3)2z=1-322+-1-322        =1+3-234+1+3+234        =84        =2Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=1+321-32=1+31-3=tanπ4+tanπ31-tanπ4tanπ3tanβ=tanπ4+π3=tan7π12β=7π12Clearly, z lies in the fourth quadrant.Therefore, argz=-7π12Hence, the polar form of z is 2cos7π12-sin7π12

Page No 13.57:

Question 4:

If z1 and z2 are two complex numbers such that z1=z2 and arg(z1) + arg(z2) = π, then show that z1=-z2¯.

Answer:

Let θbe the arg(z1) and θbe the arg(z2).

It is given that z1=z2 and arg(z1) + arg(z2) = π.

Since, z1 is a complex number.

z1=z1cosθ1+isinθ1   =z2cosπ-θ2+isinπ-θ2   =z2-cosθ2+isinθ2   =-z2cosθ2-isinθ2   =-z2¯

Hence,  z1=-z2¯.

Page No 13.57:

Question 5:

If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, prove that argz1z4+argz2z3=0.

Answer:

Given that z1, z2 and z3, z4 are two pairs of conjugate complex numbers.

z1=r1eiθ1, z2=r1e-iθ1, z3=r2eiθ2 and z4=r2e-iθ2

Then,
z1z4=r1eiθ1r2e-iθ2=r1r2eiθ1-θ2argz1z4=θ1-θ2             ...(1) 
and
z2z3=r1e-iθ1r2eiθ2=r1r2ei-θ1+θ2argz2z3=θ2-θ1           ...(2)

argz1z4+argz2z3=θ1-θ2-θ1+θ2                                 =0

Hence,  argz1z4+argz2z3=0.



Page No 13.58:

Question 6:

Express sinπ5+i1-cosπ5 in polar form.

Answer:

Let z=sinπ5+i1-cosπ5z=sinπ52+1-cosπ52        =sin2π5+1+cos2π5-2cosπ5        =2-2cosπ5        =21-cosπ5        =22sin2π10        =2sinπ10Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=1-cosπ5sinπ5=2sin2π102sinπ10cosπ10=tanπ10β=π10Clearly, z lies in the first quadrant.Therefore, argz=π10Hence, the polar form of z is 2sinπ10cosπ10+isinπ10



Page No 13.62:

Question 1:

Write the values of the square root of i.

Answer:

Let the square root of i be x+iy.i=x+iyi=x2+y2i2+2ixyi=x2-y2+2ixy        Squaring both the sidesComparing both the sides:x2-y2=0       ...(i) and 2xy=1    ...iiBy equation (ii), we find that x and y are of the same sign.From equation i,   x=±y xy=12, x2=12x=±12, y=±12i=±121+i

Page No 13.62:

Question 2:

Write the values of the square root of −i.

Answer:

Let -i=x+iySquaring both the sides-i=x2+y2i2+2ixy2xy=-1        ...iand x2-y2=0    ...iiEquation ii shows that x and y are of opposite sign. From  ii, x=±yFrom i, 2x-x=-12x2=12x=±12                       Since x and y have opposite signs, y=-12 when x=12and vice versa-i=±121-i

Page No 13.62:

Question 3:

If x + iy = a+ibc+id, then write the value of (x2 + y2)2.

Answer:

x+iy=a+ibc+idTaking modulus on both the sides, x+iy=a+ibc+idx+iy=a+ibc+idx2+y2=a2+b2c2+d2                              x+iy=x2+y2Squaring both the sides,x2+y2=a2+b2c2+d2Squaring again, we get,x2+y22=a2+b2c2+d2

Page No 13.62:

Question 4:

If π < θ < 2π and z = 1 + cos θ + i sin θ, then write the value of z.

Answer:

π<θ<2π π2<θ2<π   Dividing by 2z=1+cosθ+isinθz=1+cosθ2+sin2θz=1+cos2θ+2cosθ+sin2θz=1+1+2cosθz=21+cosθz=2×2cos2θ2z=2cos2θ2z=-2cosθ2              Since π2<θ2<π , cosθ2 is negative

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Question 5:

If n is any positive integer, write the value of i4n+1-i4n-12.

Answer:

i4n+1-i4n-12=i-1i2           i4n=1, i-1=1i=i2-1i2=i2-12i=-1-12i=-2-2i   =-1i=-ii2     i2=-1=-i-1=i

Page No 13.62:

Question 6:

Write the value of i592+i590+i588+i586+i584i582+i580+i578+i576+i574.

Answer:

i592+i590+i588+i586+i584i582+i580+i578+i576+i574=i4×148+i4×147+2+i4×147+i146×4+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2=1+i2+1+i2+1i2+1+i2+1+i2        i4=1=1-1+1-1+1-1+1-1+1-1        i2=1=1-1=-1          

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Question 7:

Write 1 − i in polar form.

Answer:

z=1-i r=z  =1+1  =2Let tan α=ImzRez tanα =-11                 =π4α =π4Since point (1,-1) lies in the fourth quadrant, the argument of z is given by θ=-α =-π4Polar form =rcos θ+isin θ                    =2cos-π4+isin-π4                   =2cosπ4-isinπ4     

Page No 13.62:

Question 8:

Write −1 + 3in polar form

Answer:

Let z=-1+3i. Then , r=z=-12+32 =2Let tan α=Im(z) Re (z)               = 3α =π3Since the point representing z lies in the second quadrant. Therefore, the argument of z is given by θ=π-α   =π-π3   =2π3So, the polar form is rcosθ+isinθ z=2cos2π3+isin2π3

Page No 13.62:

Question 9:

Write the argument of −i.

Answer:

Let z=-iThen , Rez=0, Imz=-1Since, the point (0,-1) representing z=0-i lies on negative direction of imaginary axis.Therefore, arg (z) =-π2 or 3π2 

Page No 13.62:

Question 10:

Write the least positive integral value of n for which 1+i1-in is real.

Answer:

1+i1-in=1+i1-i×1+i1+in=1+i2+2i1-i2n=1-1+2i1+1n2i2n=inFor in to be real, the smallest positive value of n will be 2. As, i2=-1, which is real.

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Question 11:

Find the principal argument of 1+i32.

Answer:

z=1+i32  =1+3i2+23i  =1-3+23i  =-2+23i

Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=232=3tanβ=tanπ3β=π3Clearly, z lies in the second quadrant.Therefore, argz=π-π3=2π3.Hence, the principal argument of z is 2π3.

Page No 13.62:

Question 12:

Find z, if z=4 and arg(z)=5π6.

Answer:

We know that,

z=zcosarg(z)+isinarg(z)z=4cos5π6+isin5π6      =4-cosπ6+isinπ6      =4-32+12i      =-23+2i

Thus, z=-23+2i.

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Question 13:

If z-5i=z+5i, then find the locus of z.

Answer:

z-5i=z+5iz-5i2=z+5i2z-5iz-5i¯=z+5iz+5i¯                zz¯=z2z-5iz¯+5i=z+5iz¯-5izz¯+5zi-5z¯i-25i2=zz¯-5zi+5z¯i-25i25zi+5zi=5z¯i+5z¯i10zi=10z¯iz=z¯z is purely real

Hence, the locus of z is real axis.



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Question 14:

If a2+122a-i=x+iy, find the value of x2+y2.

Answer:

a2+122a-i=x+iy                 ....(1)a2+122a-i¯=x+iy¯a2+122a+i=x-iy             ....(2)On multiplying (1) and (2), we geta2+122a-i×a2+122a+i=x+iyx-iya2+142a2-i2=x2-i2y2a2+142a2+1=x2+y2

Hence, x2+y2=a2+144a2+1.

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Question 15:

Write the value of -25×-9.

Answer:

 -25×-9=5-1×3-1                      =5i×3i                      =15i2                      =-15

Hence, -25×-9=-15.

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Question 16:

Write the sum of the series i+i2+i3+....upto 1000 terms.

Answer:

We know that,
i+i2+i3+i4=i-1-i+1=0

i+i2+i3+....+i1000=i+i2+i3+i4+i5+i6+i7+i8+...+i997+i998+i999+i1000=i+i2+i3+i4+i4i+i4i2+i4i3+i4i4+...+i4249i+i4249i2+i4249i3+i4249i4=i+i2+i3+i4+i+i2+i3+i4+...+i+i2+i3+i4=0

Thus, the sum of the series i+i2+i3+....upto 1000 terms is 0.

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Question 17:

Write the value of argz+argz¯.

Answer:

Let z be a complex number with argument θ.
Then,
z=reiθz¯=reiθ¯=re-iθ
⇒ argument of z¯ is −θ.

Thus, argz+argz¯=0.

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Question 18:

If z+43, then find the greatest and least values of z+1.

Answer:

z+1=z+4-3          z+4+-3          3+3          =6Also, z+10Thus, 0z+16.

Hence, the greatest and least values of z+1 is 6 and 0.

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Question 19:

For any two complex numbers z1 and z2 and any two real numbers a, b, find the value of az1-bz22+az2+bz12.

Answer:

az1-bz22+az2+bz12=az1-bz2az1-bz2¯+az2+bz1az2+bz1¯                                     =az1-bz2az1¯-bz2¯+az2+bz1az2¯+bz1¯                                     =a2z1z1¯-abz1z2¯-abz2z1¯+b2z2z2¯+a2z2z2¯+abz1z2¯+abz2z1¯+b2z1z1¯                                     =a2+b2z1z1¯+a2+b2z2z2¯                                     =a2+b2z1z1¯+z2z2¯                                     =a2+b2z12+z22

Hence, az1-bz22+az2+bz12=a2+b2z12+z22.

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Question 20:

Write the conjugate of 2-i1-2i2.

Answer:

2-i1-2i2=2-i1+4i2-4i              =2-i1-4-4i              =2-i-3-4i              =-2+i3+4i              =i-23+4i×3-4i3-4i              =3i-4i2-6+8i32-42i2              =11i+4-69+16              =-225+1125i

∴ Conjugate of 2-i1-2i2=-225+1125i¯=-225-1125i

Hence, Conjugate of 2-i1-2i2 is -225-1125i.

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Question 21:

If n ∈ , then find the value of in+in+1+in+2+in+3.

Answer:

in+in+1+in+2+in+3=in+in.i+in.i2+in.i3=in+in.i+in.(-1)+in.(-i)=in+in.i-in-in.i=0

Thus, the value of in+in+1+in+2+in+3 is 0.

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Question 22:

Find the real value of a for which 3i3-2ai2+(1-a)i+5 is real.

Answer:

 3i3-2ai2+(1-a)i+5=-3i+2a+(1-a)i+5=(2a+5)+i(1-a-3)=(2a+5)+i(-2-a)

Since, 3i3-2ai2+(1-a)i+5 is real.

 Im3i3-2ai2+(1-a)i+5=0-2-a=0a=-2

Hence, the real value of for which 3i3-2ai2+(1-a)i+5 is real is −2.

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Question 23:

If z=2 and argz=π4, find z.

Answer:

We know that,
z=zcosargz+isinargz  =2cosπ4+isinπ4  =212+i12  =21+i

Hence, z=21+i.

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Question 24:

Write the argument of 1+i31+icosθ+isinθ.

Disclaimer: There is a misprinting in the question. It should be 1+i3 instead of 1+3.

Answer:

Let the argument of 1+i3 be α. Then,
tanα=31=tanπ3α=π3

Let the argument of 1+i be β. Then,
tanβ=11=tanπ4β=π4

Let the argument of cosθ+isinθ be γ. Then,
tanγ=sinθcosθ=tanθγ=θ

∴ The argument of 1+i31+icosθ+isinθ=α+β+γ=π3+π4+θ=7π12+θ

Hence, the argument of 1+i31+icosθ+isinθ is 7π12+θ.

Page No 13.63:

Question 1:

The value of (1+i)(1+i2)(1+i3)(1+i4) is
(a) 2
(b) 0
(c) 1
(d) i

Answer:

 (b) 0
(1+ i) (1 + i2) (1 + i3) (1 + i4)
= (1+ i) (1 - 1) (1 -i) (1 + 1)      (i2 = -1,  i3 = -i and i4  = 1)
= (1 + i) (0) (1 -i) (2)
= 0

Page No 13.63:

Question 2:

If 3+2i sinθ1-2i sinθ is a real number and 0 < θ < 2π, then θ =
(a) π
(b) π2
(c) π3
(d) π6

Answer:

(a) π

 Given:

3 + 2isinθ1 - 2i sinθ is a real number

On rationalising, we get,

3 + 2i sin θ1 - 2i sin θ×1 + 2i sin θ1 + 2i sin θ = (3 + 2i sin θ) (1 + 2i sin θ)(1)2 - (2i sin θ)2=3 + 2i sin θ + 6i sin θ + 4i2 sin2 θ1 + 4 sin2 θ=3 - 4 sin2 θ + 8i sin θ1 + 4sin2 θ               i2=-1=3 - 4 sin2 θ 1 + 4sin2 θ+ i8 sin θ1 + 4sin2 θ      

For the above term to be real, the imaginary part has to be zero.

8sinθ1+4sin2θ=08sinθ=0

For this to be zero,
sin θ= 0
θ = 0, π, 2π, 3π...
But 0<θ<2π
Hence, θ=π

Page No 13.63:

Question 3:

If (1+i)(1+2i)(1+3i)...(1+n i)=a+i b, then 2×5×10×....×(1+n2) is equal to
(a) a2+b2
(b) a2-b2
(c) a2+b2
(d) a2-b2
(e) a+b

Answer:

(c) a2+ b2

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get:
1+i1+2i1+3i.......1+ni=a+ib1+i1+2i1+3i.......1+ni can be written as 1+i 1+2i 1+3i.......1+ni

12 + 12 × 12 + 22 × 12 + 32 × ... × 1 + n2  = a2 + b2

2 × 5 × 10  × ... × 1 + n2  = a2 + b2

Squaring on both the sides, we get:

2×5×10×...×(1 + n2)  = a2 + b2
2

Page No 13.63:

Question 4:

If a+ib=x+iy, then possible value of a-ib is
(a) x2+y2
(b) x2+y2
(c) x + iy
(d) xiy
(e) x2-y2

Answer:

(d) x -iy

a +ib = x + iySquaring on both the sides, we get,a + ib = x2 + (iy)2  + 2ixya +ib = (x2 - y2) + 2ixy a = (x2 - y2) and b = 2xy a -ib =  (x2 - y2) - 2ixya -ib = x2 + i2y2 - 2ixy         i2=-1                    Taking square root on both the sides, we get:a -ib = x - iy



Page No 13.64:

Question 5:

If z=cosπ4+i sinπ6, then
(a)  z =1,arg(z)=π4
(b)  z =1,arg(z)=π6
(c)  z =32,arg(z)=5π24
(d)  z =32,arg(z)=tan-112

Answer:

(d)  z =32, arg (z)=tan-112
z=cosπ4+isinπ6z=12+12iz=122+14z=12+14z=34z=32tan α=Im(z)Re(z)        =12α=tan-112Since, the point z lies in the first quadrant .Therefore, arg(z) =α=tan-112

Page No 13.64:

Question 6:

The polar form of (i25)3 is
(a) cosπ2+i sinπ2
(b) cos π + i sin π
(c) cos π − i sin π
(d) cosπ2-i sinπ2

Answer:

(d) cosπ2 -i sinπ2
(i25)3 = (i)75
         = (i)4×18+ 3 
         = (i)3 
        
= -i            (i4  = 1)

Let z=0-i Since, the point (0,-1) lies on the negative direction of imaginary axis.Therefore, arg (z) = -π2

Modulus, r = z = 1 = 1

Polar form = r (cos θ + i sin θ)
                      = cos-π2+i sin-π2
                      = cosπ2-i sin π2

Page No 13.64:

Question 7:

If i2 = −1, then the sum i + i2 + i3 +... upto 1000 terms is equal to
(a) 1
(b) −1
(c) i
(d) 0

Answer:

(d) 0

i+ i2+i3 +i4... i1000  i+ i2+i3 +i4        [  i2 =-1, i3=-i and i4=1]= i -1-i+1 = 0      Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4.Hence, the sum of all terms, till 1000, will be zero.i+ i2+i3 +i4... i1000  = 0

Page No 13.64:

Question 8:

If z=-21+i3, then the value of arg (z) is
(a) π
(b) π3
(c) 2π3
(d) π4

Answer:

(c) 2π3
z = -21 + i3

Rationalising z, we get,

z=-21 + i3×1 - i31 - i3z=-2 + i231 + 3z= -1 + i32 z= -12 + i32
tan α=Im(z)Re(z)         =3α=π3Since, z lies in the second quadrant .Therefore, arg (z) =π-π3                                 =2π3

Page No 13.64:

Question 9:

If a = cos θ + i sin θ, then 1+a1-a=
(a) cotθ2
(b) cot θ
(c) i cotθ2
(d) i tanθ2

Answer:

(c) i cotθ2
a = cosθ +isinθ      given 1+a1-a = 1+cosθ + isinθ 1-cosθ - isinθ1+a1-a = 1+cosθ +isinθ 1-cosθ - isinθ×1-cosθ + isinθ1-cosθ + isinθ

1+a1-a = 1+isinθ2-cos2θ1-cosθ2-isinθ2

1+a1-a = 1-sin2θ +2isinθ- cos2θ1+cos2θ -2cosθ + sin2θ

1+a1-a = 1-sin2θ+cos2θ + 2isinθ1+sin2θ+cos2θ-2cosθ

1+a1-a = 2isinθ2(1-cosθ)
 1+a1-a=2isinθ2cosθ22sin2θ2
1+a1-a = icosθ2sinθ2
1+a1-a=i cotθ2

Page No 13.64:

Question 10:

If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n2)=
(a) aib
(b) a2b2
(c) a2 + b2
(d) none of these

Answer:

(c) a2+ b2

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get,

1+i1+2i1+3i......1+ni=a+ib1+i1+2i1+3i......1+ni can be wriiten as  1+i 1+2i 1+3i....1+ni12 + 12×12 + 22 ×12+ 32....×12 + n2 = a2+b2

2×5 ×10....×1 + n2 = a2+b2

Squaring on both the sides, we get:

2×5 ×10....×(1 + n2) = a2+b2
2×5×10×.....(1 + n2)  = a2 + b2

Page No 13.64:

Question 11:

If (a2+1)22a-i=x+iy, then x2+y2 is equal to
(a) (a2+1)44a2+1
(b) (a+1)24a2+1
(c) (a2-1)2(4a2-1)2
(d) none of these

Answer:

(a)a2+144a2+1
x+iy = a2+122a-i

Taking modulus on both the sides, we get:
x2+y2 = a2+124a2+1
Squaring both sides, we get,x2+y2=a2+144a2+1

Page No 13.64:

Question 12:

The principal value of the amplitude of (1 + i) is
(a) π4
(b) π12
(c) 3π4
(d) π

Answer:

(a)π4

Let z = (1+i)
tan α=Im(z) Re(z)         =1α=π4Since, z lies in the first quadrant . 
Therefore, arg (z) = π4

Page No 13.64:

Question 13:

The least positive integer n such that 2i1+in is a positive integer, is
(a) 16
(b) 8
(c) 4
(d) 2

Answer:

(b)  8Let z=2i1+iz=2i1+i×1-i1-iz=2i1-i1-i2z=2i1-i1+1        i2=-1z=2i1-i2z=i-i2z=i+1Now, zn=1+inFor n =2,z2=1+i2    =1+i2+2i    =1-1+2i    =2i                          ...(1) Since this is not a positive integer,For n=4,z4=1+i4    =1+i22    =2i2               Using (1)         =4i2    =-4                     ...(2)This is a negative integer.For n=8,z8=1+i8    =1+i42    =-42                              Using (2)    =16This is a positive integer.Thus, z=2i1+in is positive for n=8.Therefore, 8 is the least positive integer such that 2i1+in is a positive integer.

Page No 13.64:

Question 14:

If z is a non-zero complex number, then  z 2zz is equal to
(a) zz
(b)  z 
(c)  z 
(d) none of these

Answer:

(a) zz
z2zz = z2z2                   ( zz = z2)Let z =a+ib z = a2+ b2Let z = a-ibz = a2+ b2 z2zz= z2z2                 = zz                

Page No 13.64:

Question 15:

If a = 1 + i, then a2 equals
(a) 1 − i
(b) 2i
(c) (1 + i) (1 − i)
(d) i − 1.

Answer:

(b) 2i

 a = 1 + i                   
On squaring both the sides, we get,
a2 = (1 + i)2
a2  = 1 + i2  + 2i
a2  = 1-1 + 2i          (i2 = -1)
a2  = 2i

Page No 13.64:

Question 16:

If (x + iy)1/3 = a + ib, then xa+yb=
(a) 0
(b) 1
(c) −1
(d) none of these

Answer:

(d) none of these

x + iy13 = a + ibCubing on both the sides, we get:x + iy = a + ib3x + iy = a3  + ib3 + 3a2bi  + 3aib2x+ iy = a3  + i3b3 + 3a2ib + 3i2ab2x + iy = a3  - ib3 + 3a2ib - 3ab2        ( i2 = -1, i3 = -i)x + iy = a3 - 3ab2 + i-b3 + 3a2b x = a3 - 3ab2  and y = 3a2b- b3or ,xa = a2- 3b2  and yb = 3a2 - b2xa+ yb = a2- 3b2 + 3a2 - b2 xa+ yb = 4a2- 4b2

Page No 13.64:

Question 17:

(-2)(-3) is equal to
(a) 6
(b) -6
(c) i6
(d) none of these.

Answer:

(b) -6

-2×-3 = 2×3×-1×-1= 6×i×i = 6×i2= -6            i2 = -1



Page No 13.65:

Question 18:

The argument of 1-i31+i3 is
(a) 60°
(b) 120°
(c) 210°
(d) 240°

Answer:

(d) 240°

1-i31+i3Rationalising the denominator, 1-i31+i3×1-i31-i3=1+3i2-23 i1-3i2=-2-23 i4       i2=-1=-12-i32tan α=Im (z)Re (z)Then, tan α =-32-12                    =3 α=60°Since the points -12,-32 lie in the third quadrant, the argument is given by:θ=180°+60°   =240°

Page No 13.65:

Question 19:

If z=1+i1-i, then z4 equals
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1
Let z = 1+i1-i
Rationalising the denominator:
z = 1+i1-i×1+i1+i 
z=1+i2+2i1-i2
z= 1-1+2i1+1
z=2i2z=i
z4=i4Since i2=-1, we have:z4=i2×i2z4=1

Page No 13.65:

Question 20:

If z=1+2i1-(1-i)2, then arg (z) equal
(a) 0
(b) π2
(c) π
(d) none of these.

Answer:

(a) 0

Let z = 1+2i1-1-i2
z= 1+2i1-1+i2 -2i
z= 1+2i1-1-1-2i
z =1+2i1+2i 
 z=1Since point (1,0) lies on the positive direction of real axis, we have:  arg (z)=0

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Question 21:

If z=1(2+3i)2, than  z =
(a) 113
(b) 15
(c) 112
(d) none of these

Answer:

(a) 113
Let z= 12+3i2z= 14+9i2+12i z = 14-9+12i z= 1-5+12i

z = 1-5+12i×-5-12i-5-12i

z = -5-12i25+144z= -5169-12i169

 z = 251692+1441692 
z= 1169
z=113

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Question 22:

If z=1(1-i)(2+3i), than  z =
(a) 1
(b) 1/26
(c) 5/26
(d) none of these

Answer:

(b) 126
Let z = 11-i2+3iz = 12+i -3i2 z=12+i+3

z= 15+i×5-i5-i 

z= 5-i25-i2

z= 5-i25+1 

z= 5-i26

 z=526-i26

z = 25676+1676
z=126

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Question 23:

If z=1-cosθ+i sinθ, then  z =
(a) 2 sinθ2
(b) 2 cosθ2
(c) 2sinθ2
(d) 2cosθ2

Answer:

(c) 2 sinθ2

z= 1 -cosθ + i sinθz = 1-cosθ2 + sin2θz = 1+cos2θ - 2cosθ+ sin2θz = 1 + 1 - 2cosθz = 21-cosθz = 4sin2θ2z = 2sinθ2 

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Question 24:

If x+iy=(1+i)(1+2i)(1+3i), then x2 + y2 =
(a) 0
(b) 1
(c) 100
(d) none of these

Answer:

(c) 100

x + iy = (1+i)(1+2i)(1+3i)Taking modulus on both the sides:x + iy=(1+i)(1+2i)(1+3i)x + iy= 1+i×1+2i×1+3ix2+y2=12+1212+2212+32x2+y2 = 2510 x2+y2= 100Squaring both the sides, x2+y2 =100

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Question 25:

If z=11-cosθ-i sinθ, then Re (z) =
(a) 0
(b) 12
(c) cotθ2
(d) 12cotθ2

Answer:

(b) 12
z = 11-cosθ-isinθz = 11-cosθ-isinθ×1-cosθ+isinθ1-cosθ+isinθz = 1-cosθ+isinθ1-cosθ2 -isinθ2z= 1-cosθ+isinθ1+cos2θ -2cosθ +sin2θz=1-cosθ+isinθ1+1-2cosθz= 1-cosθ+isinθ2(1-cosθ)Re(z) =1-cosθ21-cosθ=12

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Question 26:

If x+iy=3+5i7-6i, then y =
(a) 9/85
(b) −9/85
(c) 53/85
(d) none of these

Answer:

(c) 5385
x+iy=3+5i7-6ix+iy=3+5i7-6i×7+6i7+6ix+iy = 21+53i+30i249-36i2x+iy = 21-30+53i49+36x+iy =-985+i5385On comparing both the sides: y = 5385

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Question 27:

If 1-ix1+ix=a+ib, then a2+b2=
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1

1-ix1+ix = a+ibTaking modulus on both the sides, we get:1-ix1+ix = a+ib12+x212+x2= a2+b2a2+b2 = 1Squaring both the sides, we get: a2+b2 = 1

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Question 28:

If θ is the amplitude of a+iba-ib, than tan θ =
(a) 2aa2+b2
(b) 2aba2-b2
(c) a2-b2a2+b2
(d) none of these

Answer:

(b) 2aba2-b2

z=a+iba-ib×a+iba+ibz=a2+i2b2+2abia2-i2b2z=a2-b2+2abia2+b2z=a2-b2a2+b2+i2aba2+b2Rez=a2-b2a2+b2, Imz=2aba2+b2tan α=ImzRez         =2aba2-b2α= tan-12aba2-b2Since, z lies in the first quadrant . Therefore, arg (z) =α= tan-12aba2-b2tan θ=2aba2-b2

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Question 29:

If z=1+7i(2-i)2, then
(a)  z =2
(b)  z =12
(c) amp (z) = π4
(d) amp (z) = 3π4

Answer:

(d) amp (z) = 3π4

 z=1+7i2-i2z=1+7i4+i2-4iz=1+7i4-1-4i          i2=-1z=1+7i3-4iz=1+7i3-4i×3+4i3+4iz=3+4i+21i+28i29-16i2z=3-28+25i9+16z=-25+25i25z=-1+itan α=ImzRez        =1α=π4Since, z lies in the second quadrant . Therefore, amp (z) = π-α                                  =π-π4                                      = 3π4                                                                          

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Question 30:

The amplitude of 1i is equal to
(a) 0
(b) π2
(c) -π2
(d) π

Answer:

(c) -π2

Let z= 1iz = 1i×iiz = ii2z =-i

Since , z 0,-1 lies on the negative imaginary axis . Therefore, arg (z)  = -π2 



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Question 31:

The argument of 1-i1+i is
(a) -π2
(b) π2
(c) 3π2
(d) 5π2

Answer:

(a) -π2

Let z = 1-i1+iz = 1-i1+i×1-i1-iz = 1+i2-2i1-i2z = 1-1-2i1+1z = -2i2z  =-iSince, z lies on negative direction of imaginary axis . Therefore, arg (z) = -π2

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Question 32:

The amplitude of 1+i33+i is
(a) π3
(b) -π3
(c) π6
(d) -π6

Answer:

(c) π6

Let z = 1+i33+iz =1+i33+i×3-i3-iz = 3+2i-3i23-i2z =3 +3 +2i4z=23+2i4z=32+12itan α=Im(z)Re(z)          =13α=π6Since, z lies in the first quadrant . Therefore, arg(z) = tan-1 13=π6

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Question 33:

The value of (i5 + i6 + i7 + i8 + i9) / (1 + i) is
(a) 12(1+i)
(b) 12(1-i)
(c) 1
(d) 12

Answer:

(a) 12(1+i)

i5+i6+i7+i8+i91+i=i-1-i+1+i1+i            As, i5=i, i6=-1, i7=-i, i8=1, i9=i=ii+1=ii+1×i-1i-1=ii-1i2-1=i2-i-2=121+i

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Question 34:

1+2i+3i21-2i+3i2 equals
(a) i
(b) −1
(c) −i
(d) 4

Answer:

(c) -i

Let z= 1+2i+3i21-2i+3i2z = 1+2i-31-2i-3z = -2+2i-2-2i×-2+2i-2+2iz = -2+2i2-22-2i2z = 4+4i2-8i4+4z = 4-4-8i8z= -8i8z = -i

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Question 35:

The value of i592+i590+i588+i586+i584i582+i580+i578+i576+i574-1 is
(a) −1
(b) −2
(c) −3
(d) −4

Answer:

(b) -2

i592+i590+i588+i586+i584i582+i580+i578+i576+i574-1=i4×148+i4×147+2+i4×147+i4×146+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2-1        i4=1 and i2=-1=1+i2+1+i2+1i2+1+i2+1+i2-1=1-1-1 =-2

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Question 36:

The value of (1+i)4+(1-i)4 is
(a) 8
(b) 4
(c) −8
(d) −4

Answer:

(c) -8

Using a4+b4 =a2+b22-2a2b2(1+i)4+(1-i)4 =1+i2+1-i22-21+i21-i2       =1+i2+2i+1+i2-2i2-21+i2+2i1+i2-2i      =1-1+2i+1-1-2i2-21-1+2i1-1-2i=0-22i-2i                                           i2 =-1=8i2=-8

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Question 37:

If z=a+ib lies in third quadrant, then z¯z also lies in third quadrant if

(a) a>b>0
(b) a<b<0
(c) b<a<0
(d) b>a>0

Answer:

Since, z=a+ib lies in third quadrant.
a<0 and b<0        ....(1)

Now,
z¯z=a+ib¯a+ib    =a-iba+ib    =a-iba+ib×a-iba-ib    =a2+i2b2-2abia2-i2b2    =a2-b2-2abia2+b2

Since, z¯z also lies in third quadrant.
a2-b2<0(a-b)(a+b)<0a-b>0 and a+b<0a>b           ....(2)

From (1) and (2),
b<a<0

Hence, the correct option is (c).

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Question 38:

If fz=7-z1-z2, where z=1+2i, then fz is

(a) z2
(b) z
(c) 2z
(d) none of these

Answer:

fz=7-z1-z2      =7-1+2i1-1+2i2      =7-1-2i1-12+22i2+4i      =6-2i1-1+4-4i      =6-2i4-4i      =6-2i4-4i×4+4i4+4i      =24+24i-8i-8i242-42i2      =24+16i+816+16      =32+16i32      =1+12i

Since z=1+2i,
z=12+22        =1+4        =5

fz=12+122            =1+14            =52            =z2

Hence, the correct answer is option (a).

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Question 39:

A real value of x satisfies the equation 3-4ix3+4ix=a-ib (a, b ), if a2+b2=

(a) 1
(b) −1
(c) 2
(d) −2

Answer:

a-ib=3-4ix3+4ix        =3-4ix3+4ix×3-4ix3-4ix        =9+16x2i2-24xi9-16x2i2        =9-16x2-i24x9+16x2a-ib2=9-16x2-i24x9+16x22a2+b2=9-16x22+24x29+16x22              =81+256x4-288x2+576x29+16x22              =81+256x4+288x29+16x22              =9+16x229+16x22              =1

Hence, the correct option is (a).

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Question 40:

The complex number z which satisfies the condition i+zi-z=1 lies on

(a) circle x2 + y2 = 1
(b) the x−axis
(c) the y−axis
(d) the line x + y = 1

Answer:

i+zi-z=1i+zi-z2=12i+zi-zi+zi-z¯=1i+zi-z-i+z¯-i-z¯=1-i2-zi+z¯i+zz¯-i2+zi-z¯i+zz¯=1-i2-zi+z¯i+zz¯=-i2+zi-z¯i+zz¯-zi+z¯i=zi-z¯iz¯i+z¯i=zi+zi2z¯i=2ziz¯=zz is purely real

Hence, the correct option is (b).

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Question 41:

If z is a complex number, then

(a) z2>z2
(b) z2=z2
(c) z2<z2
(d) z2z2

Answer:

It is obvious that, for any complex number z,
z2=z2

Hence, the correct option is (b).

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Question 42:

Which of the following is correct for any two complex numbers z1 and z2?

(a) z1z2=z1z2
(b) argz1z2=argz1 argz2
(c) z1+z2=z1+z2
(d) z1+z2z1+z2

Answer:

Since, it is known that
z1z2=z1z2,
argz1z2=argz1+argz2 and
z1+z2z1+z2

Hence, the correct option is (a).

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Question 43:

If the complex number z=x+iy satisfies the condition z+1=1, then z lies on

(a) x−axis
(b) circle with centre (−1, 0) and radius 1
(c) y−axis
(d) none of these

Answer:

z+1=1z+12=12z+1z+1¯=1z+1z¯+1=1zz¯+z+z¯+1=1zz¯+z+z¯=0Since, z=x+iyzz¯+z+z¯=0x+iyx-iy+x+iy+x-iy=0x2+y2+2x=0x+12+y-02=12which is the equation of a circle with centre (-1, 0) and radius 1

Hence, the correct option is (b).



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