# NCERT solution class 11 chapter 5 Complex Numbers and Quadratic Equations exercise 5.4 mathematics

## EXERCISE 5.4

#### Question 1:

Evaluate:  #### Question 2:

For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re zRe z2 – Im z1 Im z2 #### Question 3:

Reduce to the standard form.  #### Question 4:

If x – iy = prove that .  #### Question 5:

Convert the following in the polar form:

(i) , (ii) (i) Here,  Let cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1 + 1

⇒ r2 (cos2 θ + sin2 θ) = 2 ⇒ r2 = 2                                     [cos2 θ + sin2 θ = 1] z = r cos θ + i r sin θ This is the required polar form.

(ii) Here,  Let cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r2 (cos2 θ + sin2 θ) = 1 + 1 ⇒r2 (cos2 θ + sin2 θ) = 2

⇒ r2 = 2                        [cos2 θ + sin2 θ = 1] z = r cos θ + i r sin θ This is the required polar form.

#### Question 6:

Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax2 + bx + c = 0, we obtain

a = 9, b = –12, and c = 20

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576

Therefore, the required solutions are #### Question 7:

Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax2 + bx + c = 0, we obtain

a = 2, b = –4, and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8

Therefore, the required solutions are #### Question 8:

Solve the equation 27x2 – 10+ 1 = 0

The given quadratic equation is 27x2 – 10x + 1 = 0

On comparing the given equation with ax2 + bx + c = 0, we obtain

a = 27, b = –10, and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8

Therefore, the required solutions are #### Question 9:

Solve the equation 21x2 – 28+ 10 = 0

The given quadratic equation is 21x2 – 28x + 10 = 0

On comparing the given equation with ax2 + bx = 0, we obtain

a = 21, b = –28, and c = 10

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56

Therefore, the required solutions are #### Question 10:

If find . #### Question 11:

If a + ib = , prove that a2 + b2 =  On comparing real and imaginary parts, we obtain Hence, proved.

#### Question 12:

Let . Find

(i) , (ii)  (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 .

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 + 5i) is the conjugate of –6 – 24i.

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 and y are 3 and –3 respectively.

#### Question 15:

Find the modulus of . #### Question 16:

If (x + iy)3 = u + iv, then show that . On equating real and imaginary parts, we obtain Hence, proved.

#### Question 17:

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

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 . 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 + iB, then show that

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.  On squaring both sides, we obtain

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Hence, proved.

#### Question 20:

If , then find the least positive integral value of m.  