# NCERT Solutions for Class 11 Maths Exercise 5.3 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**

Get 100 percent exact **NCERT Solutions** for **Class 11 Maths Chapter 5** **Complex Numbers and Quadratic Equations** 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.

**NCERT Solutions for Class 11 Maths 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**

### Exercise 5.3

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**Q1.Solve the given equation x² + 3 =0.**

Answer.

Using the quadratic equation formula,we have

x² +0.x + 3 =0

Comparing it with ax² +bx +c =0

a = 1, b = 0, c = 3

x = ±i√3

**Q2. Solve the equation 2x² + x + 1 =0.**

Answer.

Comparing the equation 2x² + x + 1 =0 with ax² + bx + c = 0,we have

a =2, b =1, c = 1

**Q3.Solve the equation x² + 3x + 9 = 0.**

Answer.Comparing the equation x² + 3x + 9 = 0 with ax² +bx +c =0

a = 1, b=3, c=9

**Q4. Solve the equation –x² +x – 2 =0.**

Answer. Comparing the equation –x² +x – 2 =0 with ax² + bx + c =0

a = –1, b = 1, c = –2

**Q5.Solve the equation x² + 3x + 5 = 0.**

Answer.Comparing the equation x² + 3x + 5 = 0 with ax² + bx + c =0

a =1, b = 3, c = 5

**Q6. Solve the equation x² – x + 2 = 0.**

Answer.Comparing the equation x² – x + 2 = 0 with ax² + bx + c =0

a =1, b = –1, c =2

**Q7. Solve the equation √2x² +x + √2 =0.**

Answer.Comparing the equation √2x² +x + √2 =0 with ax² + bx + c =0

a =√2, b = 1, c =√²

**Q8.Solve the equation √3x² –√2x + 3√3 =0.**

Answer.Comparing the equation √3x² –√2x + 3√3 =0 with ax² + bx + c =0

a =√3, b= –√2 , c = 3√3

Answer.Comparing the equation with ax² + bx + c =0

a =1, b = 1, c = 1/√2

Answer.

Writing the given equation in the standard form by multiplying with √2

√2x² +x + √2 = 0

Comparing the equation √2x² +x + √2 = 0 with ax² + bx + c =0

a =√2, b = 1, c = √2

### Miscellanous Exercise

Q1. Evaluate

Answer.

The expression given to us is

= –(1 – i –3 +3i)

= –(–2 + 2i)

= 2 – 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} – Im z_{1}Im z_{2}**

Answer.

Let z_{1 } = 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 – bd

= Rez_{1}Rez_{2 } – 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 –iy) = x² – (iy)² = x² + y²…..(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 = –1 + i

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

z=rcosθ + irsinθ

rcosθ = –1……..(i)

rsinθ = 1………..(ii)

Squaring and adding both the equations

r²(cos²θ + sin²θ) = (–1)² + 1² =2

r = ±√2 [cos²θ + sin²θ = 1]

r is the modulus, so it is always +

∴r = √2

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

√2cosθ = –1, √2sinθ = 1

cosθ is – and sinθ is + in second quadrant but sinθ and cosθ are + in l quadrant, therefore taking the value of cosθ, in second quadrant the value of argument

=(π – π/4)= 3π/4

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

(ii) The given complex number is

Multiplying denominator and numerator by the conjugate of the denominator:

(1 + 2i)

Let z = –1 + i,

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

z=rcosθ + irsinθ

rcosθ =–1……(i) rsinθ = 1…….(ii)

Squaring and adding both the equations

r²(cos²θ + sin²θ) = (–1)² + 1² =2

r = ±√2 [cos²θ + sin²θ = 1]

r is the modulus, so it is always +

∴r = √2

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

√2cosθ = –1, √2sinθ = 1

cosθ is – and sinθ is + in second quadrant but sinθ and cosθ are + in l quadrant, therefore taking the value of cosθ, in second quadrant the value of argument

=(π – π/4)= 3π/4

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

Answer. The given eq. can be written as 9x² –12x +20= 0

Comparing the equation 9x² –12x +20= 0 with ax² + bx + c =0

a =9, b= –12, c = 20

a = 1, b= –2

**Q8.Solve the equation 27x² – 10x + 1 = 0.**

Answer.Comparing the equation 27x² –10 x + 1 = 0 with ax² + bx + c =0

a =27,b = –10, c = 1

**Q9.Solve the equation 21x² – 28x + 10 = 0.**

Answer.Comparing the equation 21x² – 28x + 10 = 0 with ax² + bx + c =0

a = 21, b = –28, c = 10

Answer. The given expression is as follows.

We are given that

z1 = 2–i, z2 = 1+i

i² =–1

As we know the modulus of complex number z=a + ib = √(a² +b²)

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,_{ }z_{2 }= -2 + i, Find

Answer. (i) We have to find out

z1.z2 =(2 –i)(–2 +i)

= –4 + 2i + 2i –i²

= –4 +4i –(–1)

=–4 +4i + 1

= –3 +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 with the polar form of the complex number,we have

Squaring adding both the terms

r²(cos²θ + sin²θ) =(–1/2)² +(1/2)²

r² = 1/4 + 1/4

r is the modulus,so neglecting minus sign

We have

cosθ is – and sinθ is + in ll quadrant but sinθ is + in l quadrant also,therefore taking the value of cosθ

Value of argument will lie in ll quadrant and its value will be = π – θ

Therefore the modulus and argument of the given complex number is 1/√2 and 3π/4.

Q14. Find the real numbers x and y if (x –iy)(3 + 5i) is the conjugate of (–6 –24i).

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

(–6 –24i)

(x –iy)(3 + 5i) = 3x + 5xi –i3y –5i²y

= 3x + (5x –3y)i +5y [i² = –1]

=(3x + 5y) + (5x –3y)i…..(i)

conjugate of (–6 –24i) = –6 +24i…..(ii)

(i) is conjugate of (ii)

3x + 5y = –6……(iii)

5x –3y = 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=–3, in equation (iii)

3x + 5×–3 = –6

3x = –6 + 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 denominators of both the fractions (1 +i) and (1 –i) 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(x² – y²)

Hence it has been proved that

Answer.

Let α = a + ib and β = c + id

⇒ c² + d² = 1….(i)

………(ii)

=1

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

Answer.The given eq. is

x =0

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

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**NCERT Solutions for class 11 maths**

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 | 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 |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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