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

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

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

Exercises 5.1 & 5.2

Exercise 5.3

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

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   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:

Answer.

            (i² = –1)

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

Answer.

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.

Answer. Let  z = –i

then mulplicative inverse of z  will be

= i

 

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

Answer.

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.

Answer.

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.

Answer.

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)

Squaring and adding both equation

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.

Answer.

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.

Answer. z = –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)

Squaring and adding both equation

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)

Squaring and adding both equations

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

Answer.

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)

Squaring and adding both equations

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

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Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
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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
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Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
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Chapter 7- Co-ordinate geometry	Chapter 15-Probability
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Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
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Chapter 7-Control and Coordination	Chapter 15-Environment
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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
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