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# 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

### NCERT solutions of class 11 maths

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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.

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.

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

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

The expression given to us is

= –(1 – i –3 +3i)

= –(–2  + 2i)

= 2  – 2i

Q2.For any two complex number z1 and z2, prove that

Re(z1 z2) = Re z1Re z2 –­ Im z1Im z2

Let z1  = a + ib and z2 = c + id

z1 z2 = (a + ib)(c + id)

= ac + iad +ibc +i2bd

Re(z1z2) = ac – bd

= Rez1Rez2  – Imz1Imz2

Hence proved

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

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 z1=2+i,   z= -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

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

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,

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

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

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

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### NCERT solutions for class 12 maths

 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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