Question 1: Evaluate:
Answer:
Question 2: For any two complex numbers 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:
Question 3: Reduce to the standard form.
Answer:
[ On Multiplying numerator and denomunator by (14 + 5i)
Question 4: If x – iy = prove that .
Answer:
[ On Multiplying numerator and denomunator by (c + id)]
Question 5: 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
r^{2} (cos^{2} θ + sin^{2} θ) = 1+ 1
⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 2
⇒ r^{2} = 2 [cos^{2} θ + sin^{2} 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
r^{2} (cos^{2} θ + sin^{2} θ) = 1+ 1
⇒r^{2} (cos^{2} θ +sin^{2} θ) = 2
⇒ r^{2} = 2 [cos^{2} θ + sin^{2} θ = 1]
∴z = r cos θ + i r sin θ
This is the required polar form.
Question 6: Solve the equation
Answer:
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax^{2} + bx + c = 0, we obtain
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–12)^{2} – 4 × 9 × 20 = 144 – 720 = –576
Therefore, the required solutions are
Question 7: Solve the equation
Answer:
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax^{2} + bx + c = 0, we obtain
a = 2, b = –4, and c = 3
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–4)^{2} – 4 × 2 × 3 = 16 – 24 = –8
Therefore, the required solutions are
Question 8: Solve the equation 27x^{2} – 10x + 1 = 0
Answer:
The given quadratic equation is 27x^{2} – 10x + 1 = 0
On comparing the given equation with ax^{2} + bx + c = 0, we obtain
a = 27, b = –10, and c = 1
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–10)^{2} – 4 × 27 × 1 = 100 – 108 = –8
Therefore, the required solutions are
Question 9: Solve the equation 21x^{2} – 28x + 10 = 0
Answer:
The given quadratic equation is 21x^{2} – 28x + 10 = 0
On comparing the given equation with ax^{2} + bx + c = 0, we obtain
a = 21, b = –28, and c = 10
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–28)^{2} – 4 × 21 × 10 = 784 – 840 = –56
Therefore, the required solutions are
Question 9: If find .
Answer:
Question 10: If a + ib = , prove that a^{2} + b^{2} =
Answer:
On comparing real and imaginary parts, we obtain
Hence, proved.
Question 10: 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
Question 11: 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.
Question 12: 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
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.
Question 13: Find the modulus of .
Answer:
Question 14: If (x + iy)^{3} = u + iv, then show that .
Answer:
On equating real and imaginary parts, we obtain
Hence, proved.
Question 15: If α and β are different complex numbers with = 1, then find .
Answer:
Let α = a + ib and β = x + iy
It is given that,
Question 16: Find the number of nonzero integral solutions of the equation .
Answer:
Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0.
Question 17: If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a^{2} + b^{2}) (c^{2}+ d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} +B^{2}.
Answer:
On squaring both sides, we obtain
(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}
Hence, proved.
Question 18: 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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