TOPICS

Miscellaneous

Complex Numbers And Quadratic Equations

**Question-1 :-**

**Question-2 :-** For any two complex numbers z₁ and z₂, prove that Re (z₁.z₂) = Re z₁.Re z₂ – Imz₁.Imz₂.

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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1)^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 1 r^{2}(cos^{2}θ + sin^{2}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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1)^{2}+ 1^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1 + 1 r^{2}(cos^{2}θ + sin^{2}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 3x^{2} - 4x + 20/3 = 0.

We have, 3x^{2}- 4x + 20/3 = 0 Hence, 9x^{2}- 12x + 20 = 0 Now, a = 9; b = -12; c = 20 b^{2}- 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 x^{2} - 2x + 3/2 = 0.

We have, x^{2}- 2x + 3/2 = 0 Hence, 2x^{2}- 4x + 3 = 0 Now, a = 2; b = -4; c = 3 b^{2}- 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 27x^{2} - 10x + 1 = 0.

We have, 27x^{2}- 10x + 1 = 0. Now, a = 27; b = -10; c = 1 b^{2}- 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 21x^{2} - 28x + 10 = 0

We have, 21x^{2}- 28x + 10 = 0 Now, a = 21; b = -28; c = 10 b^{2}- 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 r^{2}cos^{2}θ + r^{2}sin^{2}θ = (-1/2)^{2}+ (1/2)^{2}r^{2}(cos^{2}θ + sin^{2}θ) = 1/4 + 1/4 r^{2}(cos^{2}θ + sin^{2}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.

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} = 2^{x}.

|1 - i|^{x}= 2^{x}(√1² + (-1)²)^{x}= 2^{x}(√2)^{x}= 2^{x}2^{x/2}= 2^{x}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 (a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}.

(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 (a^{2}+ b^{2}) (c^{2}+ d^{2}) (e^{2}+ f^{2}) (g^{2}+ h^{2}) = A^{2}+ B^{2}.

**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).

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