#### 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* + *ib*: *i*^{9} + *i*^{19}

#### Answer:

#### 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 + *i*7) + *i*(7 + *i*7)

#### Answer:

#### Question 5:

Express the given complex number in the form *a* + *ib*: (1 – *i*) – (–1 + *i*6)

#### 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 – 3*i*

#### Answer:

Let *z* = 4 – 3*i*

Then,

=
4 + 3*i *and

Therefore,
the multiplicative inverse of 4 – 3*i* 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 sin*θ* and cos*θ* 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* cos*θ* = 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 *x*^{2} + 3 = 0

#### Answer:

The
given quadratic equation is *x*^{2} + 3 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 1, *b* = 0, and *c* = 3

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = 0^{2} – 4 × 1 × 3 = –12

Therefore, the required solutions are

#### Question 2:

Solve the equation 2*x*^{2} + *x* + 1 = 0

#### Answer:

The
given quadratic equation is 2*x*^{2} +* x *+
1 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 2, *b* = 1, and* c *=
1

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = 1^{2} – 4 × 2 × 1 = 1 – 8 = –7

Therefore, the required solutions are

#### Question 3:

Solve the equation *x*^{2} + 3*x* + 9 = 0

#### Answer:

The
given quadratic equation is *x*^{2} + 3*x* + 9 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 1, *b* = 3, and *c* = 9

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = 3^{2} – 4 × 1 × 9 = 9 – 36 = –27

Therefore, the required solutions are

#### Question 4:

Solve the equation –*x*^{2} + *x* – 2 = 0

#### Answer:

The
given quadratic equation is –*x*^{2} + *x *–
2 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = –1, *b* = 1, and *c* = –2

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = 1^{2} – 4 × (–1) × (–2) = 1 – 8 = –7

Therefore, the required solutions are

#### Question 5:

Solve the equation *x*^{2} + 3*x* + 5 = 0

#### Answer:

The
given quadratic equation is *x*^{2} + 3*x* + 5 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 1, *b* = 3, and *c* = 5

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = 3^{2} – 4 × 1 × 5 =9
– 20 = –11

Therefore, the required solutions are

#### Question 6:

Solve the equation *x*^{2} – *x* + 2 = 0

#### Answer:

The
given quadratic equation is *x*^{2} – *x* + 2 = 0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* = 1, *b* = –1, and *c* = 2

Therefore, the discriminant of the given equation is

D
= *b*^{2} – 4*ac* = (–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 *ax*^{2} + *bx* + *c* = 0, we obtain

*a *=, *b* = 1, and *c* =

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac *=
1^{2} – =
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 *ax*^{2} + *bx* + *c* = 0, we obtain

*a* =, *b* =,
and *c* =

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac* =

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 *ax*^{2} + *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 *ax*^{2} + *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 z_{1} and z_{2}, prove that

Re (z_{1}z_{2}) = Re z_{1 }Re z_{2} – Im z_{1} Im z_{2 }

#### 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 *r *cos *θ* = –1 and *r* sin *θ* = 1

On squaring and adding, we obtain

*r*^{2} (cos^{2} *θ* + sin^{2} *θ*) = 1 + 1

⇒ *r*^{2} (cos^{2} *θ* + sin^{2} *θ*) = 2
⇒ *r*^{2} = 2 [cos^{2} *θ* + sin^{2} *θ* = 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

*r*^{2} (cos^{2} *θ* + sin^{2} *θ*) = 1 + 1
⇒*r*^{2} (cos^{2} *θ* + sin^{2} *θ*) = 2

⇒ *r*^{2} = 2 [cos^{2} *θ* + sin^{2} *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 *ax*^{2} + *bx* + *c* = 0, we obtain

*a* =
9, *b* = –12, and *c* = 20

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (–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 *ax*^{2} + *bx* + *c* = 0, we obtain

*a* =
2, *b* = –4, and *c* = 3

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (–4)^{2} – 4 × 2 ×
3 = 16 – 24 = –8

Therefore, the required solutions are

#### Question 8:

Solve the equation 27*x*^{2} – 10*x *+ 1 = 0

#### Answer:

The given
quadratic equation is 27*x*^{2} – 10*x* + 1 =
0

On
comparing the given equation with *ax*^{2} + *bx* + *c* = 0, we obtain

*a* =
27, *b* = –10, and *c* = 1

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (–10)^{2} – 4 × 27 ×
1 = 100 – 108 = –8

Therefore, the required solutions are

#### Page No 113:

#### Question 9:

Solve the equation 21*x*^{2} – 28*x *+ 10 = 0

#### Answer:

The given
quadratic equation is 21*x*^{2} – 28*x* + 10
= 0

On
comparing the given equation with *ax*^{2} + *bx *+ *c *= 0, we obtain

*a* =
21, *b* = –28, and *c* = 10

Therefore, the discriminant of the given equation is

D = *b*^{2} – 4*ac* = (–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 *a*^{2} + *b*^{2} =

#### 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
+ 5*i*) is the conjugate of –6 – 24*i*.

#### 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.

#### 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 + *i*B, then show that

(*a*^{2} + *b*^{2}) (*c*^{2} + *d*^{2})
(*e*^{2} + *f*^{2}) (*g*^{2} + *h*^{2}) = A^{2} + B^{2}.

#### Answer:

On squaring both sides, we obtain

(*a*^{2} + *b*^{2}) (*c*^{2} + *d*^{2}) (*e*^{2} + *f*^{2})
(*g*^{2} + *h*^{2}) = A^{2} + B^{2}

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).

**NCERT Solutions for Class 11 Chemistry Chapters**

- Chapter 1 – Sets
- Chapter 2 – Relations and Functions
- Chapter 3 – Trigonometric Functions
- Chapter 4 – Principle of Mathematical Induction
- Chapter 5 – Complex Numbers and Quadratic Equations
- Chapter 6 – Linear Inequalities
- Chapter 7 – Permutations and Combinations
- Chapter 8 – Binomial Theorem
- Chapter 9 – Sequences and Series
- Chapter 10 – Straight Lines
- Chapter 11 – Conic Sections
- Chapter 12 – Introduction to Three Dimensional Geometry
- Chapter 13 – Limits and Derivatives
- Chapter 14 – Mathematical Reasoning
- Chapter 15 – Statistics
- Chapter 16 – Probability

**NCERT Solutions for Class 11:**