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Question 1:

Evaluate:

Solution 1:

Comments

Swapna Shree

Question 2:

For any two complex numbersz₁ and z₂, prove that

Re (z₁z₂)= Re z₁Re z₂ – Im z₁ Im z₂

Solution 2:

Comments

Swapna Shree

Question 3:

Reduce to the standard form.

Solution 3:

Comments

Swapna Shree

Question 4:

If x – iy =prove that.

Solution 4:

Comments

Swapna Shree

Question 5:

Convert the following in the polar form:

(i) , (ii)

Solution 5:

(i) Here,

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r² (cos²θ + sin²θ) = 1 + 1

⇒ r² (cos²θ + sin²θ) = 2
⇒ r² = 2 [cos²θ + sin²θ = 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

r² (cos²θ + sin²θ) = 1 + 1
⇒r² (cos²θ + sin²θ) = 2

⇒ r² = 2 [cos²θ + sin²θ = 1]

∴z = r cos θ + i r sin θ

This is the required polar form.

Comments

Swapna Shree

Question 6:

Solve the equation

Solution 6:

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

Comments

Swapna Shree

Question 7:

Solve the equation 27x² – 10x + 1 = 0

Solution 7:

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

Comments

Swapna Shree

Question 8:

Solve the equation 21x² – 28x + 10 = 0

Solution 8:

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

On comparing the given equation with ax² + bx + c = 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

Comments

Swapna Shree

Question 9:

If find .

Solution 9:

Comments

Swapna Shree

Question 10:

If a + ib =, prove that a² + b² =

Solution 10:

a+ib= $\tfrac{(x+i)^{2}}{2x^{2}+1}$ .......................(1)

taking conjugate on both sides ,

a-ib= $\tfrac{(x-i)^{2}}{2x^{2}+1}$ ........................(2)

multiplying (1) and (2)

(a+ib)(a-ib)= $\tfrac{(x+i)^{2}}{2x^{2}+1}$ * $\tfrac{(x-i)^{2}}{2x^{2}+1}$

a²+b²=[(x+i)(x-i)]²/[2x²+1]²

Comments

Afeefa T.

Question 11:

Let . Find

(i) , (ii)

Solution 11:

(i)

On multiplying numerator and denominator by (2 – i), we obtain

On comparing real parts, we obtain

(ii)

On comparing imaginary parts, we obtain

Comments

Swapna Shree

Question 12:

Find the modulus and argument of the complex number.

Solution 12:

Let, then

On squaring and adding, we obtain

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

Comments

Swapna Shree

Question 13:

Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Solution 13:

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.

Comments

Swapna Shree

Question 14:

Find the modulus of .

Solution 14:

Comments

Swapna Shree

Question 15:

If (x + iy)³ = u + iv, then show that.

Solution 15:

On equating real and imaginary parts, we obtain

Hence, proved.

Comments

Swapna Shree

Question 16:

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

Solution 16:

Let α = a + ib and β = x + iy

It is given that,

Comments

Swapna Shree

Question 17:

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

Solution 17:

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.

Comments

Swapna Shree

Question 18:

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

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

Solution 18:

On squaring both sides, we obtain

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

Hence, proved.

Comments

Swapna Shree

Question 19:

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

Solution 19:

Comments

Charlie

Question 20:

If x – iy =prove that.

Solution 20:

x-iy=......................(1)

taking conjugate on both sides,

$x+iy=\sqrt{\tfrac{a+ib}{c+id}}$ ......................(2)

multiply equations (1) and (2)

(x+iy)(x-iy)=* $\sqrt{\tfrac{a+ib}{c+id}}$

squaring on both sides

(x²+y²)²= $\frac{(a+ib)(a-ib)}{(c+id)(c-id)}$ = $\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$

hence proved

Comments

Afeefa T.

Question 21:

Solve the equation.

Solution 21:

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

Comments

Tasneem

All Exercises - Chapter 5 - Complex Numbers and Quadratic Equations

Exercise 5.1

Exercise 5.2

Exercise 5.3

Miscellaneous Exercise 5

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Learn Miscellaneous Exercise 5 with Free Lessons & Tips

All Exercises - Chapter 5 - Complex Numbers and Quadratic Equations

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