MATHEMATICS XI

Solutions Chapter 5 – Complex Numbers and Quadratic Equations

Page No 103:

Question 1:

Express the given complex number in the form a + ib:

Answer:

Question 2:

Express the given complex number in the form a + ibi9 + i19

Answer:

Question 3:

Express the given complex number in the form a + ibi–39

Answer:

Page No 104:

Question 4:

Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)

Answer:

Question 5:

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)

Answer:

Question 6:

Express the given complex number in the form a + ib:

Answer:

Question 7:

Express the given complex number in the form a + ib:

Answer:

Question 8:

Express the given complex number in the form a + ib:  (1 – i)4

Answer:

Question 9:

Express the given complex number in the form a + ib:

Answer:

Question 10:

Express the given complex number in the form a + ib:

Answer:

Question 11:

Find the multiplicative inverse of the complex number 4 – 3i

Answer:

Let z = 4 – 3i

Then,

= 4 + 3and 

Therefore, the multiplicative inverse of 4 – 3i is given by

Question 12:

Find the multiplicative inverse of the complex number 

Answer:

Let z = 

Therefore, the multiplicative inverse ofis given by

Question 13:

Find the multiplicative inverse of the complex number –i

Answer:

Let z = –i

Therefore, the multiplicative inverse of –i is given by

Question 14:

Express the following expression in the form of a + ib.

Answer:

Page No 108:

Question 1:

Find the modulus and the argument of the complex number

Answer:

On squaring and adding, we obtain

Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant,

Thus, the modulus and argument of the complex number are 2 and respectively.

Question 2:

Find the modulus and the argument of the complex number

Answer:

On squaring and adding, we obtain

Thus, the modulus and argument of the complex number are 2 and respectively.

Question 3:

Convert the given complex number in polar form: 1 – i

Answer:

1 – i

Let r cos θ = 1 and r sin θ = –1

On squaring and adding, we obtain

 This is the required polar form.

Question 4:

Convert the given complex number in polar form: – 1 + i

Answer:

– 1 + i

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

It can be written,

This is the required polar form.

Question 5:

Convert the given complex number in polar form: – 1 – i

Answer:

– 1 – i

Let r cos θ = –1 and r sin θ = –1

On squaring and adding, we obtain

 This is the required polar form.

Question 6:

Convert the given complex number in polar form: –3

Answer:

–3

Let r cos θ = –3 and r sin θ = 0

On squaring and adding, we obtain

This is the required polar form.

Question 7:

Convert the given complex number in polar form: 

Answer:

Let r cos θ = and r sin θ = 1

On squaring and adding, we obtain

This is the required polar form.

Question 8:

Convert the given complex number in polar form: i

Answer:

i

Let r cosθ = 0 and r sin θ = 1

On squaring and adding, we obtain

This is the required polar form.

Page No 109:

Question 1:

Solve the equation x2 + 3 = 0

Answer:

The given quadratic equation is x2 + 3 = 0

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

a = 1, b = 0, and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 02 – 4 × 1 × 3 = –12

Therefore, the required solutions are

Question 2:

Solve the equation 2x2 + x + 1 = 0

Answer:

The given quadratic equation is 2x2 + x + 1 = 0

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

a = 2, b = 1, and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7

Therefore, the required solutions are

Question 3:

Solve the equation x2 + 3x + 9 = 0

Answer:

The given quadratic equation is x2 + 3x + 9 = 0

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

a = 1, b = 3, and c = 9

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27

Therefore, the required solutions are

Question 4:

Solve the equation –x2 + x – 2 = 0

Answer:

The given quadratic equation is –x2 + – 2 = 0

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

a = –1, b = 1, and c = –2

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7

Therefore, the required solutions are

Question 5:

Solve the equation x2 + 3x + 5 = 0

Answer:

The given quadratic equation is x2 + 3x + 5 = 0

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

a = 1, b = 3, and c = 5

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11

Therefore, the required solutions are

Question 6:

Solve the equation x2 – x + 2 = 0

Answer:

The given quadratic equation is x2 – x + 2 = 0

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

a = 1, b = –1, and c = 2

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7

Therefore, the required solutions are

Question 7:

Solve the equation 

Answer:

The given quadratic equation is

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

=b = 1, and c =

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 12 – = 1 – 8 = –7

Therefore, the required solutions are

Question 8:

Solve the equation 

Answer:

The given quadratic equation is

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

a =b =, and c =

Therefore, the discriminant of the given equation is

D = b2 – 4ac =

Therefore, the required solutions are

Question 9:

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 =b =, and c = 1

Therefore, the required solutions are

Question 10:

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 =b = 1, and c =

Therefore, the required solutions are

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).

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