NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations - Future Study Point

# NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations are solved by expert teachers according to  CBSE Board norms. Every Complex Number and Quadratic Equations exercise question with Solutions will assist you with revising the total syllabus and score excellent marks since NCERT solutions are the best input material for the preparation of the exams. You can subscribe us  for our free online study material which is avaiable in the form of NCERT solutions of science and maths,previous years’ question papers with solutions,latest sample papers of science and maths,our examination tips for achieving excellent marks in maths and science.

## NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

With Future Study Point Learning site you will get the chance of reading free e-books of science, maths and competitive exams.   FREE Online Tests and assignments given by us will improve your reasoning skills in maths and science.

Get 100 percent exact NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations created  by the master of Maths. We give well-ordered answers for all unsolved questions given in the Class 11 maths textbook according to CBSE Board norms from the latest NCERT text book for Class 11 maths.

## NCERT solutions of maths 11 th class Chapter 5 Complex number and Quadratic Equation.

Exercises 5.1 & 5.2

Exercise 5.3

### CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

## NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

### Exercise 5.1

Q1. Express the given complex number in the form of a + ib.

Solution:

= –3i²

= –3 × –1

= 3 = 3 + 0.i ( a =3, b= 0).

Q2.Express the given complex number in the form a  + ib, : I9 + i19.

Solution: I9 + i19

= (I2)4i  + (i2)9i

= 1 × i +(-1)i

= i – i =0 = 0 +i0 (a=0, b= 0)

Q3.Express the given complex number in the form of a + ib:i-39

Solution:

= i = 0 + i

Q4. Express the given complex number in the form a+ib:3(7 + i7) + i(7 +i7).

Solution:

3(7 + i7) + i(7 +i7)

=21 + 21i + 7i + 7i²

=21 + 28i  –7

=14 + 28i

=14 + i28

Q5.Express the given complex number in the form a + ib:(1 – i) – (–1 +i6).

Solution:

(1 – i) – (–1 +i6)

=1 – i + 1 – 6i

=2 – 7i

=2 –i7

Q6. Express the given complex number in the form a + ib:

Solution:

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q7.Express the given complex number in the form a + ib:

Solution:

Q8.Express the given complex number in the form a + ib:

Solution:

=[(1 – i)²]²

=[1 + i² – 2i]²

=(1 – 1 – 2i)²

=(–2i)² =4i²= 4 × –1 = –4[ i² = –1]

= –4 + 0i

Q9.Express the given complex number in the form a + ib:

Solution:

Q10. Express the given complex number in the form a + ib:

(i² = –1)

Q11.Find the multiplicative inverse of the complex number 4 – 3i.

The multiplicative inverse of  4 – 3i, will be

Multiplying its denominator and numerator by the conjugate of the denominator: 4 + 3i.

(i² = –1)

(i² = –1)

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q12. Find the multiplicative inverse of the complex number √5 + 3i.

Answer.  The multiplicative inverse of  √5 + 3i will be

Multiplying its denominator and numerator by the conjugate of the denominator: √5 – 3i.

Q13. Find the multiplicative inverse of the complex number: -i.

then mulplicative inverse of z  will be

= i

Q14. Express the following expression in the form of a + ib:

Study class 11 th CBSE maths NCERT solutions Trigonometric functions

### NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Exercise 5.2

Q1.Find the modulus and argument of the complex number z = –1 – i√3.

Modulus of the complex number :z=a  + ib is given by

Comparing the given complex number (–1 – i√3 ) with the standard form (a +ib) of complex number

a = –1, b  = –√3

The argument of the given complex number is given by θ as follows.

b = –√3 , a = –1, the complex number will lie in the third quadrant in the argand plane.

As θ lies in lll quadrant so the value of argument will be –(π – θ).

Therefore the argument of the given complex number will be as follows.

And modulus = 2

Q2.Find the modulus and the argument of the complex number z= –√3+i.

We can calculate the argument and modulus by converting the complex number into the polar form.

z = –√3+i = rcosθ + irsinθ

rcosθ = –√3….(i)  rsinθ = 1….(ii)

r²(cos²θ +sin²θ) = (–√3)² +1²

Since cos²θ +sin²θ =1,so

r²  = 4

r =±2

r = 2[ r is the modulus so avoiding the value, –2

Placing this value of r in eq.(i) and (ii),we get

2cosθ = –√3, 2sinθ = 1

As the value of sinθ is (+) and of cosθ (–), so the argument of the complex number will lie in the second quadrant.

Therefore taking 2cosθ = –√3 [ value of sinθ is + in l and ll quadrant]

Therefore the modulus and the argument of the complex number will be 2 and 5π/6.

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q3.Convert the given complex number into the polar form:1 –i.

Answer.In the given complex number 1 – i

a  = 1, b = –1, in the argand plane the position of the complex number is (a, b) ,b(imaginary part)   is shown along y-axis and a(real part) is shown along x-axis. The distance from origin(0,0) of the complex number r is called modulus  and the angle made by r with the x-axis θ is known as argument.

Therefore

a = rcosθ, b = rsinθ

If z = a + ib

Substituting value of a and b by rcosθ and rsinθ

z = rcosθ + irsinθ

r = √(a² + b²) = √[1² +(–1)²] =√2

Value of cosθ is + and of sinθ is – in iv quadrants, therefore

Taking sinθ = –1/√2, as the value of cosθ is + in l and lV quadrant

So,

It is clear that that argument will be

The polar form of the given complex number will be

Q4.Convert the given complex number into the polar form:–1+i.

Polar form of the complex number is = rcosθ +irsinθ

rcosθ = –1….(i)

rsinθ = 1……(ii)

Squaring and adding both equation (i) and (ii)

r²(cos²θ +sin²θ) = 2

r² = 2  (cos²θ +sin²θ =1)

r = ±√2

r is the modulus of the complex number ,so

r = √2

rcosθ = –1, rsinθ = 1

Value of cosθ is – and of sinθ is + in l and ll quadrant,so argument will lie in ll.

Since the value of sinθ is + in l and ll quadrant, therefore taking value of cosθ.

cos(π –θ)= –cosθ

Placing the value of θ and r in z = rcosθ + irsinθ.

The polar form of the complex number z will be as follows

z = rcosθ + irsinθ

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q5.Convert the given complex number into the polar form:–1–i.

z = rcosθ + irsinθ

rcosθ = –1….(i)

rsinθ = –1…..(ii)

Squaring and adding both eq.(i) and (ii)

r²(cos²θ + sinθ) = 2

r = √2                [cos²θ + sin²θ]

√2cosθ = –1, √2sinθ = –1

The value of cosθ and sinθ is – in lll  quadrant, taking the value of sinθ or cosθ

Q6.Convert the given complex number into the polar form:-3.

Answer. The given complex number is =–3

z= –3 + 0.i

The polar form of the complex number is written as follows

z = rcosθ +irsinθ

rcosθ = –3….(i) rsinθ =0…..(ii)

r²(cos²θ + sin²θ) =9

r =±3       [cos²θ + sin²θ]

r is the modulus of the complex number, so avoiding – sign.

r = 3.

3cosθ = –3, 3sinθ = 0

cosθ = –1, sinθ = 0

Value of cosθ is – and of sinθ is+ in ll quadrant

Value of sinθ is + in l and ll quadrant ,so taking the value of cosθ

cosθ = –cos0 =cos(π – 0)

cosθ = cosπ

θ = π

Therefore the polar form of the given complex number will be

z = 3cosπ + i3sinπ

Q7.Convert the given complex number into the polar form:√3 +i.

Answer. The given complex number is z  = √3 +i

The polar form of the complex number is written as follows

z = rcosθ + irsinθ

rcosθ = √3….(i) rsinθ = 1…..(ii)

r²(cos²θ + sin²θ)  =  4     [cos²θ + sin²θ]

r = ±2

r is the modulus of the complex number, so avoiding – sign.

r = 2

2cosθ = √3, 2sinθ = 1

The value of sinθ and cosθ is + in l quadrant,so taking either of the equation

The polar form of the complex number will be as follows

Q8.Convert the given complex number into the polar form:i

The given complex number is =i

z = 0 + i

The polar form of the complex number is written as follows

z = rcosθ + irsinθ

rcosθ = 0….(i) rsinθ = 1…..(ii)

r²(cos²θ + sin²θ)  =       [cos²θ + sin²θ]

r ±1

r is the modulus of the complex number, so avoiding – sign.

r = 1

cosθ = 0, sinθ =1

cosθ and sinθ both are + in l quadrant , so taking either of the equations

cosθ =0

The polar form of the complex number will be as follows

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

### NCERT Solutions for class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions for class 9 science

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

### NCERT Solutions for class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for Class 10 Science

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

### 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

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### 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

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

Scroll to Top