## EXERCISE 5.4

#### 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).