Question-1 :-
Question-2 :- For any two complex numbers z₁ and z₂, prove that Re (z₁.z₂) = Re z₁.Re z₂ – Imz₁.Imz₂.
Solution :-Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ Therefore, z₁.z₂ = (x₁ + iy₁)(x₂ + iy₂) = x₁.x₂ + x₁.iy₂ + x₂.iy₁ + i²y₁.y₂ = x₁.x₂ + x₁.iy₂ + x₂.iy₁ - y₁.y₂ = (x₁.x₂ + y₁.y₂) + i(x₁.y₂ + x₂.y₁) Re(z₁.z₂) = x₁.x₂ + y₁.y₂ Re (z₁.z₂) = Re z₁.Re z₂ – Imz₁.Imz₂
Question-3 :-
Question-4 :-
Question-5 :- Convert the following in the polar form:
Here, z = - 1 + i Now, -1 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1)2 + 12 r2 (cos2 θ + sin2 θ) = 1 + 1 r2 (cos2 θ + sin2 θ) = 2 r x 1 = √2 r = √2 Therefore, -1 = r cos θ and 1 = r sin θ cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4 [lies on II Quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
Here, z = - 1 + i Now, -1 = r cos θ, 1 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1)2 + 12 r2 (cos2 θ + sin2 θ) = 1 + 1 r2 (cos2 θ + sin2 θ) = 2 r x 1 = √2 r = √2 Therefore, -1 = r cos θ and 1 = r sin θ cos θ = -1/√2 and sin θ = 1/√2, which gives θ = π - π/4 = 3π/4 [lies on II Quadrant] Therefore, required polar form is z = r[cosθ + i sinθ] = √2[cos(3π/4) + i sin(3π/4)]
Question-6 :- Solve 3x2 - 4x + 20/3 = 0.
Solution :-We have, 3x2 - 4x + 20/3 = 0 Hence, 9x2 - 12x + 20 = 0 Now, a = 9; b = -12; c = 20 b2 - 4ac = (-12)2 - 4 x 9 x 20 = 144 - 720 = -576 Therefore, the solutions are given by x = (12 ± √-576)/18 = (12 ± 24i)/18 = 2/3 ± 4i/3
Question-7 :- Solve x2 - 2x + 3/2 = 0.
Solution :-We have, x2 - 2x + 3/2 = 0 Hence, 2x2 - 4x + 3 = 0 Now, a = 2; b = -4; c = 3 b2 - 4ac = (-4)2 - 4 x 2 x 3 = 16 - 24 = -8 Therefore, the solutions are given by x = (4 ± √-8)/4 = (4 ± 2√2i)/4 = 1 ± √2i/2
Question-8 :- Solve 27x2 - 10x + 1 = 0.
Solution :-We have, 27x2 - 10x + 1 = 0. Now, a = 27; b = -10; c = 1 b2 - 4ac = (-10)2 - 4 x 27 x 1 = 100 - 108 = -8 Therefore, the solutions are given by x = (10 ± √-8)/54 = (10 ± 2√2i)/54 = 5/27 ± √2i/27
Question-9 :- Solve 21x2 - 28x + 10 = 0
Solution :-We have, 21x2 - 28x + 10 = 0 Now, a = 21; b = -28; c = 10 b2 - 4ac = (-28)2 - 4 x 21 x 10 = 784 - 840 = -56 Therefore, the solutions are given by x = (28 ± √-56)/42 = (28 ± √56i)/42 = 2/3 ± √14i/21
Question-10 :-
Question-11 :-
Question-12 :-
Question-13 :- Find the modulus and argument of the complex number
Now, -1/2 = r cos θ, 1/2 = r sin θ By squaring and adding, we get r2 cos2 θ + r2 sin2 θ = (-1/2)2 + (1/2)2 r2 (cos2 θ + sin2 θ) = 1/4 + 1/4 r2 (cos2 θ + sin2 θ) = 1/2 r x 1 = 1/√2 r = 1/√2 Modulus = 1/√2 Therefore, -1/2 = r cos θ and 1/2 = r sin θ cos θ = -√2/2 = -1/√2 and sin θ = √2/2 = 1/√2, Since cos θ are negative and cosθ are negative in II quadrant, So, which gives θ = (π - π/4) = 3π/4 Arguments = 3π/4
Question-14 :- Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Solution :-We have, z = (x – iy) (3 + 5i) = 3x + 5ix - 3iy - 5iy² = (3x + 5y) + i(5x - 3y) Conjugate, z = (3x + 5y) - i(5x - 3y) Given that : –6 – 24i Therefore, (3x + 5y) - i(5x - 3y) = –6 – 24i 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 9x + 15y + 25x - 15y = -18 + 120 34x = 102 x = 3 Putting the value of x in equation (i), we obtain 3 x 3 + 5y = -6 9 + 5y = -6 5y = -6 - 9 5y = -15 y = -3 Thus, the values of x and y are 3 and –3 respectively.
Question-15 :-
Question-16 :-
(x + iy)³ = u + iv x³ + i³y³ + 3.x².iy + 3.x.(iy)² = u + iv x³ - iy³ + 3.x².iy - 3.x.y² = u + iv (x³ - 3xy²) + i(3x²y - y³) = u + iv Therefore, u = x³ - 3xy², v = 3x²y - y³![]()
Question-17 :- If α and β are different complex numbers with |β| = 1, then find
Let α = a + ib and β = x + iy It is given that, |β| = 1 Therefore, (√x² + y²)x = 1 x² + y² = 1![]()
Question-18 :- Find the number of non-zero integral solutions of the equation |1 - i|x = 2x.
Solution :-
|1 - i|x = 2x
(√1² + (-1)²)x = 2x
(√2)x = 2x
2x/2 = 2x
Now, x/2 = x
x = 2x
2x - x = 0
x = 0
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.
Question-19 :- If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
Solution :-(a + ib) (c + id) (e + if) (g + ih) = A + iB |(a + ib) (c + id) (e + if) (g + ih)| = |A + iB| |(a + ib)| x |(c + id)| x |(e + if)| x |(g + ih)| = |A + iB| [|z₁.z₂| = |z₁|.|z₂|] √a² + b² x √c² + d² x √e² + f² x √g² + h² = √A² + B² On squaring both sides, we obtain (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
Question-20 :-
Therefore, m = 4k, where k is some integer. Therefore, the least positive integer is 1. Thus, the least positive integral value of m is 4 (= 4 × 1).