NCERT Solutions For Class 11 Maths Chapter 5

NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations

NCERT solutions class 11 maths chapter 5 complex numbers and quadratic equations is one of the best tools to practice mathematics for class 11. The NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations is provided here so that students can understand the concepts of the chapter in details. The NCERT solutions for class 11 maths chapter 5 is created by expert teachers according to the latest CBSE syllabus. Get the NCERT solutions for class 11 maths chapter 5 pdf given below.

NCERT Solutions Class 11 Maths Chapter 5 Exercises

Exercise: 5.1

Q.1: Express the following complex number in x + iy form; (4i)(74i)

Sol:

(4i)(74i)  = -4 × 74 × i × i = -7 i2 = -7(-1)   [ Since, i2 = -1 ]

= 7

Q.2: Express the following complex number in x + iy form; i19+i9 

Sol:

i19+i9=i4×2+1+i4×4+3 = (i4)2i+(i4)4i3

=1×i+1×(i)   [ Since,  i4 = 1, i3 = -1 ]

= i + (-i)

= 0

 

 

Q.3: Express the following complex number in x + iy form; i39

 

Sol:

i39=i4×93 =(i4)9i3

=(1)9i3   [ Since, i4 = 1 ]   

=1i3=1i   [ Since, i3 = -i ]

=1i×ii

=ii2=i1 = i   [ Since, i2 = -1 ]

 

 

Q.4: Express the following complex number in x + iy form; 5(6 + 6i) + i(6 + 6i)

 

Sol:

5(6 + 6i) + i(6 + 6i) = 30 + 30i + 6i + 6i2

= 30 + 36i + 6(-1)    [ Since, i2 = -1 ]

= 30 – 6 + 36i

= 24 + 36i

 

 

Q.5: Express the following complex number in x + iy form; (2 – 3i) – (-4 + 5i)

Sol:

(2 – 3i) – (-4 + 5i) = 2 – 3i + 4 – 5i

= 6 – 8i

 

 

Q.6: Express the following complex number in x + iy form;

(14+i34)(6+i43)

 

Sol:

(14+i34)(6+i43)=14+i346i43

=(146)+i(3443)

=234+i(712)

=234i712

 

 

Q.7: Express the following complex number in x + iy form:

[(13+i73)+(4+i13)](43+i)

 

Sol:

[(13+i73)+(4+i13)](43+i)

=13+i73+4+i13+43i

=(13+43+4)+i(73+131)

=173+i53

 

 

Q.8: Express the following complex number in x + iy form: (1 – i)4

Sol:

 

(1 – i)4 = [(1i)2]2

=[12i+i2]2=[112i]2=(2i)2=(2i)×(2i)=4i2          [ Since i2 = -1 ]

= -4

 

 

Q.9: Express the following complex number in ‘x + iy’ form; (13+3i)3

Sol:

 

=(13+3i)3:

=(13)3+(3i)3+3(13)(3i)(13+3i)

=127+27i3+3i(13+3i)

=127+27i3+i+9i2

=12727i+i9      [Since, i3 = -i and i2 = -1]

=(1279)+i(27i+1) = 2422726i

 

 

Q.10: Express the following complex number in ‘x + iy’ form: (2i13)3

SoL:

 

(2i13)3=(1)3(2+i13)3 = [23+(i3)3+3(2)(i3)(2+i3)] = [8+i327+2i(2+i3)]

=[8+i327+4i+2i23)] = [8i27+4i23]      [ Since, i3 = -i and i2 = -1 ]

=[223+i10727]=221i10727

 

 

Q.11: Find the multiplicative inverse of the given complex number, 7 – 6i.

Sol:

 

Assuming, z = 7 – 6i

Now, z¯¯¯=7+6i

And, |z|2 = (7)2 + (-6)2 = 49 + 36 = 85

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=7+6i85=785+685i

 

 

Q.12: Find the multiplicative inverse of the given complex number, 7+4i

Sol:

Assuming, z = 7+4i

Now, z¯¯¯=7+4i

And, |z|2 = (7)2+(4)2=23

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=74i23=723423i

 

 

Q.13: Find the multiplicative inverse of the given complex number, – i

Sol:

 

Assuming, z = – i

Now, z¯¯¯=i

And, |z|2 = (-i)2 = 1

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=i1 = i

 

 

Q.14: Express the following complex number in ‘x + iy’ form: (3+i5)(3i5)(3+i2)(3i2)

Sol:

 

(3+i5)(3i5)(3+i2)(3i2)=(3)2(i5)23+i23+i2

Now, by using (a + b) (a – b) = (a – b)2

=95i222i=95×(1)22i=9+522i×ii=14i22i2

=14i22=7i2×22=72i2

 

 

Exercise: 5.2

Q.1: Find out the modulus and argument of the given complex number:

z=1i3

 

Sol:

Assuming rcosΘ=1andrsinΘ=3

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

(rcosΘ)2+(rsinΘ)2=(1)2+(3)2r2(cos2Θ+sin2Θ)=1+3r2(1)=4r=4=2

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

Now, 2cosΘ=1and2sinΘ=3cosΘ=12andsinΘ=32

Since, the values of both cosΘandsinΘare negative and they both are negative in 3rd quadrant

Therefore, Argument = (ππ3)=2π3

 

 

Q.2: Find out the modulus and argument of the given complex number

z=3+i

 

Sol:

Assuming rcosΘ=3andrsinΘ=1

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

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2

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

Now, 2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12

Here, Θ lies in the 2nd quadrant.

Therefore, Argument = (ππ6)=5π6

 

 

Q.3: Convert the following complex number in polar form: 1 – i

Sol:

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 4th quadrant.

Therefore, Θ = (π4)

Therefore, the required polar form is:

1i=rcosΘ+irsinΘ=2cos(π4)+i2sin(π4)

Therefore, 1i=2[cos(π4)+isin(π4)]

 

 

Q.4: Convert the following complex number in polar form;   -1 + i

 

Sol:

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.

Therefore, Θ = (ππ4)=3π4

So, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)

Therefore, -1 + i = 2[cos(3π4)+isin(3π4)]

 

 

Q.5: Convert the following complex number in polar form;  -3

Sol:

Assuming rcosΘ=3andrsinΘ=0

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

(rcosΘ)2+(rsinΘ)2=(3)2+(0)2r2(cos2Θ+sin2Θ)=9r2(1)=9r=9=3

Therefore, Modulus = 3    [ Since, r > 0]

Now, 3cosΘ=3and3sinΘ=0cosΘ=1andsinΘ=0

Therefore, Θ=π

So, the required polar form is:

3=rcosΘ+irsinΘ=3cos(π)+i.3sin(π)

Therefore, 3=3[cos(π)+isin(π)]

 

 

Q.6: Convert the following complex number in polar form;   -1 – i

Sol:

 

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

 

 

Q.7: Convert the following complex number in polar form: z=3+i

 

Sol:

z=3+i

Assuming rcosΘ=3andrsinΘ=1

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

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2

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

Now,

2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12

Here, Θ lies in 1st quadrant.

Therefore, Θ = (π6)

So, the required polar form is:

3+i=rcosΘ+irsinΘ=2cos(π6)+i2sin(π6)

Therefore, 3+i=2[cos(π6)+isin(π6)]

 

 

Q.8: Convert the following complex number in polar form: i

 

Sol:

Assuming rcosΘ=0andrsinΘ=1

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

(rcosΘ)2+(rsinΘ)2=(0)2+(1)2r2(cos2Θ+sin2Θ)=1r2(1)=1r=1=1

Therefore, Modulus = 1     [ Since, as r > 0)]

Now, cosΘ=0and3sinΘ=1

Therefore, Θ=π2

So, the required polar form is:

i=cos(π2)+isin(π2)

 

 

Miscellaneous Exercise

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