Complex Numbers and Quadratic Equations
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NCERT Solutions for Class 11 Maths Chapter 5
Complex Numbers and Quadratic Equations Class 11
Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.1, 5.2, 5.3, miscellaneous Solutions
Exercise 5.1 : Solutions of Questions on Page Number : 103
Q1 :
Express the given complex number in the form a + ib:
Answer :
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Q2 :
Express the given complex number in the form a + ib: i9 + i19
Answer :
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Q3 :
Express the given complex number in the form a + ib: i-39
Answer :
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Q4 :
Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)
Answer :
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Q5 :
Express the given complex number in the form a + ib: (1 - i) - (-1 + i6)
Answer :
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Q6 :
Express the given complex number in the form a + ib:
Answer :
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Q7 :
Express the given complex number in the form a + ib:
Answer :
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Q8 :
Express the given complex number in the form a + ib: (1 - i)4
Answer :
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Q9 :
Express the given complex number in the form a + ib:
Answer :
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Q10 :
Express the given complex number in the form a + ib:
Answer :
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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
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Q12 :
Find the multiplicative inverse
of the complex number
Answer :
Let z =
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Therefore, the
multiplicative inverse of is given by
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Q13 :
Find the multiplicative inverse of the complex number -i
Answer :
Let z = –i
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Therefore, the multiplicative inverse of –i is given by
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Q14 :
Express the following expression in the form of a + ib.
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Answer :
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Exercise 5.2 : Solutions of Questions on Page Number : 108
Q1 :
Find the modulus and the argument
of the complex number
Answer :
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On squaring and adding, we obtain
Since both the values of sin θand cos θ are negative and sinθand cosθare negative in III quadrant,
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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 :
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On squaring and adding, we obtain
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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
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This is the
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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,
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|
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
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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
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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
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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
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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
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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, andc = 1
Therefore, the discriminant of the given equation is D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7 Therefore, the required solutions are
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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
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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
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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
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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
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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
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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
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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
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Therefore, the required solutions are
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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 =
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Therefore, the required solutions are
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Exercise Miscellaneous : Solutions of Questions on Page Number : 112
Q1 :
Evaluate:
Answer :
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Q2 :
For any two complex numbers z1 and z2, prove that Re (z1z2) = Re z1 Re z2 - Im z1 Im z2
Answer :
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Q3 :
Reduce to the standard form.
Answer :
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Q4 :
If x –
iy = prove that .
Answer :
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Q5 :
Convert the following in the polar form:
(i) , (ii)
Answer :
(i) Here,
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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]
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∴z = r cos θ + i r sin θ
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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]
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∴z = r cos θ + i r sin θ
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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
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
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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
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
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Q10 :
If find .
Answer :
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Q11 :
If find 
.
Answer :
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Q12 :
If a + ib = , prove that a2 + b2 =
Answer :
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On comparing real and imaginary parts, we obtain
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Hence, proved.
Q13 :
Let 
. Find
(i) 
, (ii) 
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Answer :
(i) 
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On multiplying numerator and denominator by (2 – i), we obtain
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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
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On squaring and adding, we obtain
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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
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It is given
that,
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Equating real and imaginary parts, we obtain
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Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
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Putting the value of x in equation (i), we obtain
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Thus, the values of x and y are 3 and –3 respectively.
Q16 :
Find the modulus of .
Answer :
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Q17 :
If (x + iy)3 = u + iv, then show that .
Answer :
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On equating real and imaginary parts, we obtain
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Hence, proved.
Q18 :
If
α and β are
different complex numbers with = 1, then find 
.
Answer :
Let α = a + ib and β = x + iy
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It is given that,
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Q19 :
Find the number of non-zero integral
solutions of the equation .
Answer :
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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 :
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On squaring both sides, we obtain
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
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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).