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
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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 z_{1 }and z_{2}, prove that
Re(z_{1} z_{2}) = Re z_{1}Re z_{2} – Im z_{1}Im z_{2}
Answer.
Let z_{1 } = a + ib and z_{2} = c + id
z_{1} z_{2} = (a + ib)(c + id)
= ac + iad +ibc +i^{2}bd
= (ac-bd) + i(ad +bc)
Re(z_{1}z_{2}) = ac – bd
= Rez_{1}Rez_{2 } – Imz_{1}Imz_{2}
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 z_{1}=2+i,_{ }z_{2 }= -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
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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 |
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 | Chapter 16- Probability |
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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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