# NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

**NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations** unraveled by Expert Teachers according to CBSE norms. **NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations** are created here to assist you with revising the total Syllabus and Score Maximum marks. Subscribe for our free online course class with best mathematics teacher.

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**NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations**

Get 100 percent exact **NCERT Solutions** for **Class 11 Maths Chapter 5** **Complex Numbers and Quadratic Equations**ย fathomed by the master of **Maths**. We give well-ordered answers for all unsolved questions given in** the Class 11 maths** course according to **CBSE Board** rules from the most recent **NCERT** book for Class 11 maths.ย

**NCERT Solutions for Class 11 Mathsย Chapter 5 Complex number and Quadratic equations**

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**NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations**

##### ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย

### ย ย ย ย ย ย ย ย ย Exercise 5.3

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**Q1.Solve the given equation xยฒ + 3 =0.**

Answer.

Using the quadratic equation formula,we have

xยฒ +0.x + 3 =0

Comparing it with axยฒ +bx +c =0

a = 1, b = 0, c = 3

x = ยฑiโ3

**Q2. Solve the equation 2xยฒ + x + 1 =0.**

Answer.

Comparing the equation 2xยฒ + x + 1 =0 with axยฒ + bx + cย = 0,we have

a =2, b =1, c = 1

**Q3.Solve the equation xยฒ + 3x + 9 = 0.**

Answer.Comparing the equation xยฒ + 3x + 9 = 0 with axยฒ +bx +c =0

a = 1, b=3, c=9

**Q4. Solve the equation โxยฒ +x โ 2 =0.**

Answer. Comparing the equation โxยฒ +x โ 2 =0 with axยฒ + bx + c =0

a = โ1, b = 1, c = โ2

**Q5.Solve the equation xยฒ + 3x + 5 = 0.**

Answer.Comparing the equation xยฒ + 3x + 5 = 0 with axยฒ + bx + c =0

a =1, b = 3, c = 5

**Q6. Solve the equation xยฒ โ x + 2 = 0.**

Answer.Comparing the equation xยฒ โ x + 2 = 0 with axยฒ + bx + c =0

a =1, b = โ1, c =2

**Q7. Solve the equation โ2xยฒ +x + โ2 =0.**

Answer.Comparing the equation โ2xยฒ +x + โ2 =0 with axยฒ + bx + c =0

a =โ2, b = 1, c =โยฒ

**Q8.Solve the equation โ3xยฒ โโ2x + 3โ3 =0.**

Answer.Comparing the equation โ3xยฒ โโ2x + 3โ3 =0 with axยฒ + bx + c =0

a =โ3, b= โโ2 , c = 3โ3

Answer.Comparing the equationย with axยฒ + bx + c =0

a =1, b = 1, c = 1/โ2

Answer.

Writing the given equation in the standard form by multiplying with โ2

โ2xยฒ +x + โ2ย = 0

Comparing the equation โ2xยฒ +x + โ2ย = 0 with axยฒ + bx + c =0

a =โ2, b = 1, c = โ2

### ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย Miscellanousย Exercise

Q1. Evaluate

Answer.

The expression given to us is

= โ(1 โ i โ3 +3i)

= โ(โ2ย + 2i)

= 2ย โ 2i

**Q2.For any two complex number z _{1 }and z_{2}, prove that**

**Re(z _{1} z_{2}) = Re z_{1}Re z_{2} โยญ Im z_{1}Im z_{2}**

Answer.

Let z_{1 }ย = a + ib and z_{2} = c + id

z_{1} z_{2} = (a + ib)(c + id)

= ac + iad +ibc +i^{2}bd

= (ac-bd) + i(ad +bc)

Re(z_{1}z_{2}) = ac โ bd

= Rez_{1}Rez_{2 }ย – Imz_{1}Imz_{2}

Hence proved

Answer.

Multiplying denominator and numerator by the conjugate of the denominator:28 +10i

Answer. We are given that

As we know (x +iy)(x โiy) = xยฒ โ (iy)ยฒ = xยฒ + yยฒ…..(i)

………(ii)

…………..(iii)

Multiplying(ii) and (iii),we have

From equation number (i)

Squaring both sides

Hence proved

**Q5.Convert the following into the polar form**

Multiplying denominator and numerator by the conjugate of the denominator;

(3 + 4i)

So, let z = โ1 + i

Comparing the terms of this complex number with the polar form of the complex number:

z=rcosฮธ + irsinฮธ

rcosฮธ = โ1……..(i)

rsinฮธ = 1………..(ii)

Squaring and adding both the equations

rยฒ(cosยฒฮธ + sinยฒฮธ) = (โ1)ยฒ + 1ยฒ =2

r = ยฑโ2ย [cosยฒฮธ + sinยฒฮธ = 1]

r is the modulus, so it is always +

โดr = โ2

Placing the value of r in eq.(i) and (ii)

โ2cosฮธ = โ1, โ2sinฮธ = 1

cosฮธ is โ and sinฮธ is + in second quadrant but sinฮธ and cosฮธ are + in l quadrant, therefore taking the value of cosฮธ, in second quadrant the value of argument

=(ฯ โ ฯ/4)= 3ฯ/4

Therefore required polar form of the given complex number will be a follows

(ii)ย The given complex number is

Multiplying denominator and numerator by the conjugate of the denominator:

(1 + 2i)

Let z = โ1 + i,

Comparing the terms of this complex number with the polar form of the complex number:

z=rcosฮธ + irsinฮธ

rcosฮธย =โ1……(i) rsinฮธ = 1…….(ii)

Squaring and adding both the equations

rยฒ(cosยฒฮธ + sinยฒฮธ) = (โ1)ยฒ + 1ยฒ =2

r = ยฑโ2ย [cosยฒฮธ + sinยฒฮธ = 1]

r is the modulus, so it is always +

โดr = โ2

Placing the value of r in eq.(i) and (ii)

โ2cosฮธ = โ1, โ2sinฮธ = 1

cosฮธ is โ and sinฮธ is + in second quadrant but sinฮธ and cosฮธ are + in l quadrant, therefore taking the value of cosฮธ, in second quadrant the value of argument

=(ฯ โ ฯ/4)= 3ฯ/4

Therefore required polar form of the given complex number will be a follows

Answer. The given eq. can be written as 9xยฒ โ12x +20= 0

Comparing the equation 9xยฒ โ12x +20= 0 with axยฒ + bx + c =0

a =9, b= โ12, c = 20

a = 1, b= โ2

**Q8.Solve the equation 27xยฒ โ 10x + 1 = 0.**

Answer.Comparing the equation 27xยฒ โ10 x + 1 = 0 with axยฒ + bx + c =0

a =27,b = โ10, c = 1

**Q9.Solve the equation 21xยฒ โ 28x + 10 = 0.**

Answer.Comparing the equation 21xยฒ โ 28x + 10 = 0 with axยฒ + bx + c =0

a = 21, b = โ28, c = 10

Answer. The given expression is as follows.

We are given that

z1 = 2โi, z2 = 1+i

iยฒ =โ1

As we know the modulus of complex number z=a + ib = โ(aยฒ +bยฒ)

Answer. Evaluating the right hand part

Comparing L.H.S and R.H.S, we have

So,it has been proved that

Q12. Let z_{1}=2+i,_{ย ย }z_{2ย }= -2 + i, Find

Answer. (i) We have to find out

z1.z2 =(2 โi)(โ2 +i)

= โ4 + 2i + 2i โiยฒ

= โ4ย +4i โ(โ1)

=โ4 +4i + 1

= โ3 +4i

Therefore

Multiplying dinominator and numerator by the conjugate of denominator in R.H.S

So, writing it in the standard form of the complex number a + ib

Comparing L.H.S and R.H.S, we have

**Q13.Find the modulus and argument of the complex number**

Answer. The given complex number is

Multiplying its denominator and numerator by the conjugate of the denominator:

1 +3i

Comparing it withย the polar form of the complex number,we have

Squaring adding both the terms

rยฒ(cosยฒฮธ + sinยฒฮธ) =(โ1/2)ยฒ +(1/2)ยฒ

rยฒ = 1/4 + 1/4

r is the modulus,so neglecting minus sign

We have

cosฮธ is โ and sinฮธ is + in ll quadrant but sinฮธ is + in l quadrant also,therefore taking the value of cosฮธ

Value of argument will lie in ll quadrant and its value will be = ฯ โ ฮธ

Therefore the modulus and argument of the given complex number is 1/โ2 and 3ฯ/4.

Q14.ย Find the real numbers x and y if (x โiy)(3 + 5i) is the conjugate of (โ6 โ24i).

Answer.We are given that (x โiy)(3 + 5i) is the conjugate of

(โ6 โ24i)

(x โiy)(3 + 5i) = 3x + 5xi โi3y โ5iยฒy

= 3x + (5x โ3y)i +5yย ย ย ย ย ย ย ย [iยฒ = โ1]

=(3x + 5y) + (5x โ3y)i…..(i)

conjugate of (โ6 โ24i) = โ6 +24i…..(ii)

(i) is conjugate of (ii)

3x + 5y = โ6……(iii)

5x โ3y = 24……..(iv)

Multiplying eq.(iii) by 5 and eq.(iv) by 3 and adding both of the equation.

We get y = -3

Putting value of y=โ3, in equation (iii)

3x + 5รโ3 = โ6

3x = โ6 + 15

x = 3

Therefore the value of real number x and y will be x =3 and y =-3

Answer.

The complex number we are given that

Multiplying denominators and numerators of both the fractions by the conjugates of the denominatorsย of both the fractions (1 +i) and (1 โi) respectively.

= i + i = 2i

Therefore

2i = 0 + 2i

So,

Answer. We are given that

So,

Comparing terms of both sides,we have

We have to prove

L.H.S

Substituting the values of x and y in

= 4(xยฒย โย yยฒ)

Hence it has been proved that

Answer.

Let ฮฑ = a + ib and ฮฒ = c + id

โ cยฒ + dยฒ = 1….(i)

………(ii)

=1

**Q18. Find the number of non-zero integral solutions of the equation**

Answer.The given eq. isย

x =0

Number of non-zero integral solutions of the equation are=0

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Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | ย Chapter 16- Probability |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

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Chapter 8-Application of Integrals |

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