Class 11 Maths Ncert Solutions Chapter 5 Ex 5.3 Complex Numbers & Quadratic Equations PDF

Class 11 Maths Ncert Solutions Ex 5.3

Class 11 Maths Ncert Solutions Chapter 5 Ex 5.3

Q-1: Evaluate the following:

[i18+(1i)25]3

Sol:

[i18+(1i)25]3  =  [i4×4+2+1i4×6+1]3  =  [(i4)4i2+1(i6)4i]3  =  [i2+1i]3   [Since, i4 = 1]

=[1+1i×ii]3=[1+ii2]3=[1i]3

=(1)3[1+i]3=[1+i3+3i(i+1)]

=[1i3+3i+3i2]

= – [1 – i + 3i – 3] = – [-2 + 2i]

= 2 – 2i

 

 

Q-2: For the two given complex number z1 and z2 below prove that

Re(z1z2) = Re z1Re z2 – Im z1 Im z2

 

Sol:

Assuming z1 = a1 + ib1 and z2 = a2 + ib2

Therefore, z1z2 = (a1 + ib1) ( a2 + ib2)

= a1(a2 + ib2) + ib2(a2 + ib2)

= a1a2 + ia1b2 + ib1a2 + i2b1b2

= a1a2 + ia1b2 + ib1a2 – b1b2

= (a1a2 – b1b2) + i(a1b2 + b1a2)

Re(z1z2) = a1a2 – b1b2

Re(z1z2) = Rez1 Rez2  – Imz1 Imz2

Hence, Proved

 

 

Q-3: Express (114i21+i)(34i5+i) in the standard form i.e. a + ib.

 

Sol:

[114i21+i][34i5+i]=[(1+i)2(14i)(14i)(1+i)][34i5+i] =[1+i2+8i1+i4i4i2][34i5+i]=[1+9i53i][34i5+i] =[3+4i+27i36i225+5i15i3i2]=33+31i2810i=33+31i2(145i) =33+31i2(145i)×14+5i14+5i

Now, on multiplying and dividing by (14 + 5i) we will get:

=462+165i+434i+155i22[(14)2(5i)2]=307+599i2(19625i2) =307+599i2(221)=307+599i442 =307442+i599442

 

 

Q-4: If aib=xiyuiv then prove that (a2+b2)2=x2+y2u2+v2

 

Sol:

aib=xiyuiv =xiyuiv×x+iyu+iv

Now, on multiplying and dividing by (u + iv) we will get:

=(xu+yv)+i(xvyu)u2+v2

Therefore, (aib)2=(xu+yv)+i(xvyu)u2+v2

a22iabb2=(xu+yv)+i(xvyu)u2+v2

On comparing both sides, we will get:

a2b2=xu+yvu2+v2and2ab=xvyuu2+v2……………..(a)

(a2 + b2)2 = (a2 – b2)2 + 4a2b2

=(xu+yvu2+v2)2+(xvuu2+v2)2

Now, by using equation (a) we will get:

=x2u2+y2v2+2xyuv+x2v2+y2u22xyuv(u2+v2)2 =x2u2+y2v2+x2v2+y2u2(u2+v2)2 =x2(u2+v2)+y2(u2+v2)2(u2+v2)2 =(x2+y2)(u2+v2)(u2+v2)2=x2+y2u2+v2

Hence, proved

 

 

Q-5) Convert the following complex number in polar form:

1: 1+7i(2i)2

2:1+3i12i

 

Sol:

1: Here, z = 1+7i(2i)2

1+7i(2i)2=1+7i4+i24i=1+7i414i

=1+7i34i×3+4i3+4i=3+4i+21i+28i232+42

=3+4i+21i2832+42=25+25i25 = -i + 1

Assuming rcosΘ=1andrsinΘ=1

On squaring on both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2

Therefore, Modulus = 2        [ Since, r > 0]

Now, 2cosΘ=1and2sinΘ=1cosΘ=12andsinΘ=12

Here, Θ lies in 2nd quadrant.

Therefore, Θ = (ππ4)=3π4

Therefore, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)

=2[cos(3π4)+isin(3π4)]

 

2: Here, z = 1+3i12i

1+3i12i=1+3i12i×1+2i1+2i

=1+2i+3i61+4=5+5i5 = -i + 1

Assuming rcosΘ=1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2

Therefore, Modulus = 2        [ Since, r > 0]

Now,2cosΘ=1and2sinΘ=1

cosΘ=12andsinΘ=12

Here, Θ lies in 2nd quadrant.

Now, Θ = (ππ4)=3π4

Therefore, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)

=2[cos(3π4)+isin(3π4)]

 

 

Q-6: Solve the equation given below:

3y2 – 4y + 203 = 0

 

Sol:

Here, 3y2 – 4y + 203 = 0

The given equation can be also be written as,

9y2 – 12y + 20 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 9, b = -12 and c = 20

Now, the discriminant is:

D=b24ac=(12)24×9×20=144720=576

Now, the solution is:

y=b±D2a =(12)±5762(9) =12±576i218

=12±24i18=2±4i3

=23±43i

Therefore, y = =23±43i

 

 

Q-7: Solve the equation given below:

y2 – 2y + 32 = 0

 

Sol:

Here, y2 – 2y + 32= 0

The given equation can be also be written as:

2y2 – 4y + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 2, b = -4 and c = 3

Now, the discriminant is:

D=b24ac=(4)24×(2)×(3)=1624=8

Now, the solution is:

y=b±D2a =(4)±82(2) =4±8i24

=4±22i4=2±2i2

=1±22i

Therefore, y =1±22i

 

 

Q-8: Solve the equation given below:

27y2 – 10y + 1 = 0

 

Sol:

Here, 27y2 – 10y + 1 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 27, b = -10 and c = 1

Now, the discriminant is:

D=b24ac=(10)24×(27)×(1)=100108=8

Now, the solution is:

y=b±D2a =(10)±82(27) =10±8i254

=10±22i54=5±2i27

=57±227i

Therefore, y =57±227i

 

 

Q-9: Solve the equation given below:

21y2 – 28y + 10 = 0

 

Sol:

Here, 21y2 – 28y + 10 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 21, b = -28 and c = 10

Now, the discriminant is:

D=b24ac=(28)24×(21)×(10)=784840=56

Now, the solution is:

Since, y=b±D2a

=(28)±562(21) =28±56i242

=28±214i42=14±14i21=32±1421i

Therefore, y =32±1421i

 

 

Q-10:  z1 = 2 – i, z2 = 1 + i, find z1+z2+1z1z2+i

 

Sol:

Here, z1 = 2 – i, z2 = 1 + i

z1+z2+1z1z2+i=(2i)+(1+i)+1(2i)(1+i)+1

=422i=21i=21i×1+i1+i=2(1+i)1i2

=2(1+i)1+1=2(1+i)2=|1+i|=12+12=2

 

 

Q-11:  If x + iy = (a+i)22a2+1 then prove that

x2 + y2(a2+1)2(2a+1)2

 

Sol:

Here, x + iy = (a+i)22a2+1

=a2+i2+2ai2a2+1=a21+2ai2a2+1=a212a2+1+i2a2a2+1

On comparing both sides, we will get:

x=a212a2+1andy=2a2a2+1

Therefore, x2+y2=(a212a2+1)2+(2a2a2+1)2

=a4+12a2+4a2(2a+1)2=a4+1+2a2(2a2+1)2 =(a2+1)2(2a2+1)2

Hence, proved

 

 

Q-12: If z1 = 2 – i and z2 = -2 + i is given find out,

1: Im(1z1z1¯¯¯¯¯)   

2:Re(z1z2z1)

 

Sol:

Here, z1 = 2 – i and z2 = -2 + i

1. Im(1z1z1¯¯¯¯¯)

=1(2+i)(2+i)=122i2=14+1=15

On comparing both sides, we will get:

Im(1z1z1¯¯¯¯¯)=0

 

2. Re(z1z2z1)

z1z2 = (2 – i)(- 2 + i) = -4 + 2i + 2i – i2 = -4 + 4i – (-1) = -3 + 4i

z1¯¯¯¯¯=2+i z1z2z1¯¯¯¯¯=3+4i2+i

Now multiplying and dividing it by (2 – i), we will get:

=3+4i2+i×2i2i

=6+3i+8i4i222i2

=6+11i4(1)4+1

=2+11i5=25+115i

On comparing both the sides, we will get:

Re(z1z2z1¯¯¯¯¯)=25

 

 

Q-13: Find out the modulus and argument of the given complex number

1+2i13i

 

Sol:

Here, z = 1+2i13i

1+2i13i×1+3i1+3i=1+3i+2i+6i2129i2

=1+3i+2i+6i212+9=5+5i10

=510+510i=12+12i

Assuming rcosΘ=1/2andrsinΘ=1/2

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1/2)2+(1/2)2r2(cos2Θ+sin2Θ)=1/4+1/4r2(1)=1/2r=12

Therefore, Modulus = 12          [ Since, r > 0]

Now, 12cosΘ=12and12sinΘ=12

cosΘ=12andsinΘ=12

Here, Θ lies in 2nd  quadrant.

Therefore, Θ=(ππ4)=3π4

 

 

Q-14: Find the real number a and b if (3 + 5i) (a – ib) is the conjugate if -6 – 24i.

 

Sol:

Assuming z = (3 + 5i) (a – ib)

z = 3a + 5ai – 3bi – 5bi2 = 3a + 5ai – 3bi + 5b = (3a + 5b) + i(5a – 3b)

Therefore, z¯¯¯=(3a+5b)i(5a3b)

Here, given that z¯¯¯=624i

Therefore, (3a + 5b) – i(5a – 3b) = -6 – 24i

On comparing both sides, we will get:

3a + 5b = -6 . . . . . . . . . . . . (a)

5a – 3b = 24 . . . . . . . . . . . . (b)

On solving these two equations, we will get:

9a + 15b = -18

25a – 15b = 120

34a = 102

Therefore, a = 3

Now, 3(3) + 5b = -6       [From equation (a)]

5b = -15

Therefore, b = -3

Hence, a = 3 and b = -3

 

 

Q-15: Find the modulus of 1+i1i1i1+i

 

Sol:

1+i1i1i1+i=(1+i)2(1i)2(1i)(1+i)

=1+i2+2i1i2+2i1+1=4i2=2i

   1+i1i1i1+i=|2i|=22=2

 

 

Q-16: If (a + ib)3 = u + iv, then prove that:

ua+vb=4(a2b2)

 

Sol:

Here, (a + ib)3 = u + iv

a3 + (ib)3 + 3a(ib)(a + ib) = u + iv

a3 + i3b3 + 3a2bi + 3ab2i2 = u + iv

a3 – ib3 + 3a2bi – 3ab2 = u + iv

(a3 – 3ab2) + i(3a2b – b3) = u + iv

On comparing both the sides, we will get:

u = a3 – 3ab2     and      v = 3a2b – b3

Hence, Proved

 

 

Q-17: If A and B are two different complex numbers with |B| = 1, then find BA1A¯¯¯¯B

 

Sol:

Assuming, A = x + iy and B = u + iv

Here, |B| = 1,

Therefore, u2+v2=1

u2 + v2 = 1 . . . . . . . . . . . . (a)

BA1A¯¯¯¯B=(x+iy)(u+iv)1(uiv)(x+iy)

=(xu)+i(yv)1(xu+uiy+ivx+yv)

=(xu)+i(yv)(1xuyv)+i(xvyu)=|(xv)+i(yv)||(1xuyv)+i(xvyu)|

Since, z1z2=|z1||z2|

Therefore, =(xu)2+(yv)2(1xuyv)+i(xvuy)

=