You cannot copy content of this page

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

EXERCISE 5.4


Page No 112:

Question 1:

Evaluate: 

Answer:

Question 2:

For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re zRe z2 – Im z1 Im z2

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 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

Answer:

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

Answer:

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

Answer:

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

Page No 113:

Question 9:

Solve the equation 21x2 – 28+ 10 = 0

Answer:

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 .

Answer:

Question 11:

If a + ib =, prove that a2 + b2 = 

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 + 5i) is the conjugate of –6 – 24i.

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 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 + iB, then show that

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

Answer:

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.

Answer:

Therefore, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!
Free Web Hosting