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.
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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, : I^{9} + i^{19}.
Solution: I^{9} + i^{19 }
= (I^{2})^{4}i + (i^{2})^{9}i
= 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-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 |
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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 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
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Chapter 7- Integrals | |
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