NCERT Solutions for Class 11 Science Math Chapter 5 Complex Numbers And Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Complex Numbers And Quadratic Equations are extremely popular among Class 11 Science students for Math Complex Numbers And Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

Page No 103:

Question 1:

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

Answer:

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Question 2:

Express the given complex number in the form a + ib: i9 + i19

Answer:

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Question 3:

Express the given complex number in the form a + ib: i–39

Answer:



Page No 104:

Question 4:

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

Answer:

Page No 104:

Question 5:

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

Answer:

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Question 6:

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

Answer:

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Question 7:

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

Answer:

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Question 8:

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

Answer:

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Question 9:

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

Answer:

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Question 10:

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

Answer:

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Question 11:

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

Answer:

Let z = 4 – 3i

Then, = 4 + 3i and

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

Page No 104:

Question 12:

Find the multiplicative inverse of the complex number

Answer:

Let z =

Therefore, the multiplicative inverse ofis given by

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Question 13:

Find the multiplicative inverse of the complex number –i

Answer:

Let z = –i

Therefore, the multiplicative inverse of –i is given by

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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.

Page No 108:

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.

Page No 108:

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.

Page No 108:

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.

Page No 108:

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.

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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.

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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.

Page No 108:

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

Page No 109:

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

Page No 109:

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

Page No 109:

Question 4:

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 = 12 – 4 × (–1) × (–2) = 1 – 8 = –7

Therefore, the required solutions are

Page No 109:

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

Page No 109:

Question 6:

Solve the equation x2x + 2 = 0

Answer:

The given quadratic equation is x2x + 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

Page No 109:

Question 7:

Solve the equation

Answer:

The given quadratic equation is

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

a =, b = 1, and c =

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 12= 1 – 8 = –7

Therefore, the required solutions are

Page No 109:

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

Page No 109:

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

Page No 109:

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:

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Question 2:

For any two complex numbersz1 and z2, prove that

Re (z1z2)= Re z1 Re z2 – Im z1 Im z2

Answer:

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Question 3:

Reduce to the standard form.

Answer:

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Question 4:

If xiy =prove that.

Answer:

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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

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.

Page No 112:

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

Page No 112:

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

Page No 112:

Question 8:

Solve the equation 27x2 – 10x + 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 – 28x + 10 = 0

Answer:

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

Page No 113:

Question 10:

If find .

Answer:

Page No 113:

Question 11:

If a + ib =, prove that a2 + b2 =

Answer:

On comparing real and imaginary parts, we obtain

Hence, proved.

Page No 113:

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

Page No 113:

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.

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Question 14:

Find the real numbers x and y if (xiy) (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 x and y are 3 and –3 respectively.

Page No 113:

Question 15:

Find the modulus of .

Answer:

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Question 16:

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

Answer:

On equating real and imaginary parts, we obtain

Hence, proved.

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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,

Page No 113:

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.

Page No 113:

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.

Page No 113:

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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