Free download NCERT Solutions for Class 11 Maths Chapter 5** Complex Numbers and Quadratic Equations Ex 5.1, Ex 5.2, Ex 5.3 and Miscellaneous Exercise** PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2020-2021.

Topics and Sub Topics in Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations:

Section Name | Topic Name |

5 | Complex Numbers and Quadratic Equations |

5.1 | Introduction |

5.2 | Complex Numbers |

5.3 | Algebra of Complex Numbers |

5.4 | The Modulus and the Conjugate of a Complex Number |

5.5 | Argand Plane and Polar Representation |

5.6 | Quadratic Equations |

**Contents**show

**NCERT Solutions for Class 11 Maths Chapter 5**

**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).

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