NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations - Future Study Point

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

class 11 complex number and quadratic equations

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.

class 11 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. 

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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 z1 and z2, prove that

Re(z1 z2) = Re z1Re z2 –­ Im z1Im z2

Answer.

Let z1  = a + ib and z2 = c + id

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

= ac + iad +ibc +i2bd

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

Re(z1z2) = ac – bd

= Rez1Rez2  – Imz1Imz2

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 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

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

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

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NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
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Chapter 7- Integrals
Chapter 8-Application of Integrals

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