## NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise

#### Page No 112:

#### Question 1:

Evaluate:

#### Answer:

#### Question 2:

For any two complex numbers z

_{1}and z_{2}, prove that
Re (z

_{1}z_{2}) = Re z_{1 }Re z_{2}– Im z_{1}Im z_{2}#### Answer:

#### Question 3:

Reduce to the standard form.

#### Answer:

#### Question 4:

If

*x*–*iy*=prove that.#### Answer:

#### Question 5:

Convert the following in the polar form:

(i) , (ii)

#### Answer:

(i) Here,

Let

*r*cos*θ*= –1 and*r*sin*θ*= 1
On squaring and adding, we obtain

*r*

^{2}(cos

^{2}

*θ*+ sin

^{2}

*θ*) = 1 + 1

⇒

*r*^{2}(cos^{2}*θ*+ sin^{2}*θ*) = 2 ⇒*r*^{2}= 2 [cos^{2}*θ*+ sin^{2}*θ*= 1]
∴

*z*=*r*cos*θ*+*i**r*sin*θ*
This is the required polar form.

(ii) Here,

Let

*r*cos*θ*= –1 and*r*sin*θ*= 1
On squaring and adding, we obtain

*r*

^{2}(cos

^{2}

*θ*+ sin

^{2}

*θ*) = 1 + 1 ⇒

*r*

^{2}(cos

^{2}

*θ*+ sin

^{2}

*θ*) = 2

⇒

*r*^{2}= 2 [cos^{2}*θ*+ sin^{2}*θ*= 1]
∴

*z*=*r*cos*θ*+*i**r*sin*θ*
This is the required polar form.

#### Question 6:

Solve the equation

#### Answer:

The given quadratic equation is

This equation can also be written as

On comparing this equation with

*ax*^{2}+*bx*+*c*= 0, we obtain*a*= 9,

*b*= –12, and

*c*= 20

Therefore, the discriminant of the given equation is

D =

*b*^{2}– 4*ac*= (–12)^{2}– 4 × 9 × 20 = 144 – 720 = –576
Therefore, the required solutions are

#### Question 7:

Solve the equation

#### Answer:

The given quadratic equation is

This equation can also be written as

On comparing this equation with

*ax*^{2}+*bx*+*c*= 0, we obtain*a*= 2,

*b*= –4, and

*c*= 3

Therefore, the discriminant of the given equation is

D =

*b*^{2}– 4*ac*= (–4)^{2}– 4 × 2 × 3 = 16 – 24 = –8
Therefore, the required solutions are

#### Question 8:

Solve the equation 27

*x*^{2}– 10*x*+ 1 = 0#### Answer:

The given quadratic equation is 27

*x*^{2}– 10*x*+ 1 = 0
On comparing the given equation with

*ax*^{2}+*bx*+*c*= 0, we obtain*a*= 27,

*b*= –10, and

*c*= 1

Therefore, the discriminant of the given equation is

D =

*b*^{2}– 4*ac*= (–10)^{2}– 4 × 27 × 1 = 100 – 108 = –8
Therefore, the required solutions are

#### Page No 113:

#### Question 9:

Solve the equation 21

*x*^{2}– 28*x*+ 10 = 0#### Answer:

The given quadratic equation is 21

*x*^{2}– 28*x*+ 10 = 0
On comparing the given equation with

*ax*^{2}+*bx*+*c*= 0, we obtain*a*= 21,

*b*= –28, and

*c*= 10

Therefore, the discriminant of the given equation is

D =

*b*^{2}– 4*ac*= (–28)^{2}– 4 × 21 × 10 = 784 – 840 = –56
Therefore, the required solutions are

#### Question 10:

If find .

#### Answer:

#### Question 11:

If

*a*+*ib*=, prove that*a*^{2}+*b*^{2}=#### Answer:

On comparing real and imaginary parts, we obtain

Hence, proved.

#### Question 12:

Let . Find

(i) , (ii)

#### Answer:

(i)

On multiplying numerator and denominator by (2 –

*i*), we obtain
On comparing real parts, we obtain

(ii)

On comparing imaginary parts, we obtain

#### Question 13:

Find the modulus and argument of the complex number.

#### Answer:

Let, then

On squaring and adding, we obtain

Therefore, the modulus and argument of the given complex number are respectively.

#### Question 14:

Find the real numbers

*x*and*y*if (*x*–*iy*) (3 + 5*i*) is the conjugate of –6 – 24*i*.#### Answer:

Let

It is given that,

Equating real and imaginary parts, we obtain

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

Putting the value of

*x*in equation (i), we obtain
Thus, the values of

*x*and*y*are 3 and –3 respectively.#### Question 15:

Find the modulus of .

#### Answer:

#### Question 16:

If (

*x*+*iy*)^{3}=*u*+*iv*, then show that.#### Answer:

On equating real and imaginary parts, we obtain

Hence, proved.

#### Question 17:

If α and β are different complex numbers with = 1, then find.

#### Answer:

Let α =

*a*+*ib*and β =*x*+*iy*
It is given that,

#### Question 18:

Find the number of non-zero integral solutions of the equation.

#### Answer:

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.

#### Question 19:

If (

*a*+*ib*) (*c*+*id*) (*e*+*if*) (*g*+*ih*) = A +*i*B, then show that
(

*a*^{2}+*b*^{2}) (*c*^{2}+*d*^{2}) (*e*^{2}+*f*^{2}) (*g*^{2}+*h*^{2}) = A^{2}+ B^{2}.#### Answer:

On squaring both sides, we obtain

(

*a*^{2}+*b*^{2}) (*c*^{2}+*d*^{2}) (*e*^{2}+*f*^{2}) (*g*^{2}+*h*^{2}) = A^{2}+ B^{2}
Hence, proved.

#### Question 20:

If, then find the least positive integral value of

*m*.#### Answer:

Therefore, the least positive integer is 1.

Thus, the least positive integral value of

*m*is 4 (= 4 × 1)._{}

^{}