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    Home » NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
    Class 11 Math

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

    AdminBy AdminUpdated:August 11, 202312 Mins Read
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    Exercise 5.1 : Solutions of Questions on Page Number : 103

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

    Q1 :Express the given complex number in the form a + ib: 
    Answer :

     


    Q2 :Express the given complex number in the form a + ib: i9 + i19
    Answer :

     


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

    Answer :


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


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



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


    Answer :


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


    Answer :


    Q8 :Express the given complex number in the form a + ib: (1 – i)4
    Answer :


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


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


    Q11 :Find the multiplicative inverse of the complex number 4 – 3i
    Answer :
    Let z = 4 – 3i
    Then, = 4 + 3i and Therefore, the multiplicative inverse of 4 – 3i is given by


    Q12 :Find the multiplicative inverse of the complex number 

    Answer :
    Let z = Therefore, the multiplicative inverse of is given by


    Q13 :Find the multiplicative inverse of the complex number -i
    Answer :
    Let z = -i
    Therefore, the multiplicative inverse of -i is given by


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


    Answer :

     


    Exercise 5.2 : Solutions of Questions on Page Number : 108

    Q1 :Find the modulus and the argument of the complex number 
    Answer :

    On squaring and adding, we obtain

    Since both the values of sin θand cos θ are negative and sinθand cosθare negative in III quadrant, 
    Thus, the modulus and argument of the complex number  are 2 and  respectively.


    Q2 :Find the modulus and the argument of the complex number 


    Answer :

    On squaring and adding, we obtain

    Thus, the modulus and argument of the complex number  are 2 and  respectively.


    Q3 :Convert the given complex number in polar form: 1 – i
    Answer :
    1 – i
    Let rcos θ = 1 and rsin θ = -1
    On squaring and adding, we obtain

    This is the required polar form.


    Q4 :Convert the given complex number in polar form: – 1 + i
    Answer :
    -1 + i
    Let rcos θ =-1 and rsin θ = 1
    On squaring and adding, we obtain

    It can be written,

    This is the required polar form.


    Q5 :Convert the given complex number in polar form: – 1 – i
    Answer :
    – 1 – i
    Let rcos θ = -1 and rsin θ = -1
    On squaring and adding, we obtain

     This is the required polar form.


    Q6 :Convert the given complex number in polar form: -3
    Answer :
    -3
    Let rcos θ = -3 and rsin θ = 0
    On squaring and adding, we obtain

     This is the required polar form.


    Q7 :Convert the given complex number in polar form: 
    Answer :

    Let rcos θ =and rsin θ = 1
    On squaring and adding, we obtain

     This is the required polar form.


     

    Q8 : Convert the given complex number in polar form: i
    Answer:
    i
    Let rcosθ = 0 and rsin θ = 1
    On squaring and adding, we obtain

     This is the required polar form.


     

    Exercise 5.3 : Solutions of Questions on Page Number : 109

    Q1 :Solve the equation x2 + 3 = 0
    Answer :
    The given quadratic equation is x2 + 3 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 1, b = 0, and c = 3
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 02 – 4 × 1 × 3 = -12
    Therefore, the required solutions are


     

    Q2 :Solve the equation 2x2 + x + 1 = 0
    Answer :
    The given quadratic equation is 2x2 + x + 1 = 0
    On comparing the given equation with ax2  + bx + c = 0, we obtain
    a = 2, b = 1, and c = 1
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = – 7
    Therefore, the required solutions are  


     

    Q3 :Solve the equation x2 + 3x + 9 = 0
    Answer :
    The given quadratic equation is x2 + 3x + 9 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 1, b = 3, and c = 9
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = -27
    Therefore, the required solutions are


     

    Q4 :Solve the equation -x2 + x – 2 = 0
    Answer :
    The given quadratic equation is -x2 + x – 2 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = -1, b = 1, and c = -2
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 12 – 4 × (-1) × (-2) = 1 – 8 = -7
    Therefore, the required solutions are


     

    Q5 :Solve the equation x2 + 3x + 5 = 0
    Answer :
    The given quadratic equation is x2 + 3x + 5 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 1, b = 3, and c = 5
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = -11
    Therefore, the required solutions are


     

    Q6 :Solve the equation x2 – x + 2 = 0
    Answer :
    The given quadratic equation is x2 – x + 2 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 1, b = -1, and c = 2
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = (-1)2 – 4 × 1 × 2 = 1 -“ 8 = -7
    Therefore, the required solutions are


    Q7 :Solve the equation 


    Answer :
    The given quadratic equation is 
    On comparing the given equation with ax2 + bx + c = 0, we obtain

    a = , b = 1, and c = 
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = 12 – – = 1 – 8 = -7
    Therefore, the required solutions are


    Q8 :Solve the equation 
    Answer :

    The given quadratic equation is 

    On comparing the given equation with ax2 + bx + c = 0, we obtain

    a = , b =, and c = 

    Therefore, the discriminant of the given equation is

    D = b2 – 4ac = 

    Therefore, the required solutions are


    Q9 :Solve the equation 
    Answer :
    The given quadratic equation is 
    This equation can also be written as

    On comparing this equation with ax2 + bx + c = 0, we obtain
    a = , b = , and c = 1
    Therefore, the required solutions are


     

    Q10 :Solve the equation 
    Answer :
    The given quadratic equation is 
    This equation can also be written as

    On comparing this equation with ax2 + bx + c = 0, we obtain
    a = , b = 1, and c =

    Therefore, the required solutions are


     

    Exercise Miscellaneous : Solutions of Questions on Page Number : 112

    Q1 :Evaluate: 
    Answer :


     

    Q2 :For any two complex numbers z1 and z2, prove that
    Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
    Answer :


     

    Q3 :Reduce to the standard form. 
    Answer :


    Q4 :If x – iy = prove that 
    Answer :


     

    Q5 :Convert the following in the polar form:
    (i)  , (ii) 
    Answer :
    (i)
    Here, 
    Let r cos θ = -1 and r sin θ = 1
    On squaring and adding, we obtain
    r2 (cos2 θ + sin2 θ) = 1 + 1
    ⇒ r2 (cos2 θ + sin2 θ) = 2
    ⇒ r2 = 2         [cos2 θ + sin2 θ = 1]

    ∴ z = r cos θ + i r sin θ


    This is the required polar form
    (ii) Here, 
    Let r cos θ = -1 and r sin θ = 1
    On squaring and adding, we obtain
    r2 (cos2 θ + sin2 θ) = 1 + 1
    ⇒r2 (cos2 θ + sin2 θ) = 2
    ⇒ r2 = 2                            [cos2 θ + sin2 θ = 1]

    ∴z = r cos θ + i r sin θ

    This is the required polar form.


     

    Q6 :Solve the equation 
    Answer :
    The given quadratic equation is 
    This equation can also be written as

    On comparing this equation with ax2 + bx + c = 0, we obtain
    a = 9, b = -12, and c = 20
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = (-12)2 – 4 × 9 × 20 = 144 – 720 = -576

    Therefore, the required solutions are


     

    Q7 :Solve the equation 
    Answer :
    The given quadratic equation is 
    This equation can also be written as

    On comparing this equation with ax2 + bx + c = 0, we obtain
    a = 2, b = -4, and c = 3
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = (-4)2 – 4 × 2 × 3 = 16 – 24 = -8
    Therefore, the required solutions are


     

    Q8 :Solve the equation 27x2 – 10x + 1 = 0
    Answer :
    The given quadratic equation is 27x2 – 10x + 1 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 27, b = -10, and c = 1
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = (-10)2 – 4 × 27 × 1 = 100 – 108 = -8
    Therefore, the required solutions are


    Q9 :Solve the equation 21x2 – 28x + 10 = 0
    Answer :
    The given quadratic equation is 21x2 – 28x + 10 = 0
    On comparing the given equation with ax2 + bx + c = 0, we obtain
    a = 21, b = -28, and c = 10
    Therefore, the discriminant of the given equation is
    D = b2 – 4ac = (-28)2 – 4 × 21 × 10 = 784 – 840 = -56
    Therefore, the required solutions are


    Q10 :If find  
    Answer :


     

    Q11 :If find .
    Answer :


     

    Q12 :If a + ib = , prove that a2 + b2 = 

    Answer :

    On comparing real and imaginary parts, we obtain

    Hence, proved.


     

    Q13 :Let  . Find
    (i)  , (ii) 
    Answer :

    (i)

    On multiplying numerator and denominator by (2 – i), we obtain

    On comparing real parts, we obtain

    (ii) 

    On comparing imaginary parts, we obtain


    Q14 :Find the modulus and argument of the complex number .
    Answer :
    Let , then

    On squaring and adding, we obtain

    Therefore, the modulus and argument of the given complex number are respectively.


     

    Q15 :Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of -6 – 24i.
    Answer :
    Let 
    It is given that,

    Equating real and imaginary parts, we obtain
    3x + 5y =-6   ……. (i)
    5x -3y = 24   ……..(ii)
    Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

    Putting the value of x in equation (i), we obtain

    Thus, the values of x and y are 3 and -3 respectively.


     

    Q16 :Find the modulus of 
    Answer :


     

    Q17 :If (x + iy)3 = u + iv, then show that 
    Answer :

    On equating real and imaginary parts, we obtain

    Hence, proved.

    =1


    Q19 :Find the number of non-zero integral solutions of the equation 
    Answer :

    Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.


     

    Q20 :If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
    (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
    Answer :

    On squaring both sides, we obtain
    (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
    Hence, proved.


     

    Q21 :If , then find the least positive integral value of m.
    Answer :

    Therefore, the least positive integer is 1.
    Thus, the least positive integral value of m is 4 (= 4 × 1).

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    • Chapter 1 – Sets
    • Chapter 2 – Relations and Functions
    • Chapter 3 – Trigonometric Functions
    • Chapter 4 – Principle of Mathematical Induction
    • Chapter 5 – Complex Numbers and Quadratic Equations
    • Chapter 6 – Linear Inequalities
    • Chapter 7 – Permutation and Combinations
    • Chapter 8 – Binomial Theorem
    • Chapter 9 – Sequences and Series
    • Chapter 10 – Straight Lines
    • Chapter 11 – Conic Sections
    • Chapter 12 – Introduction to three Dimensional Geometry
    • Chapter 13 – Limits and Derivatives
    • Chapter 14 – Mathematical Reasoning
    • Chapter 15 – Statistics
    • Chapter 16 – Probability
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