This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hey Hølüwårh, good to see you again. Here are the solutions to the problems: Question 10: To solve the equation (jx)/(1+jy) = (3x+j4)/(x+3y), we will cross-multiply and equate the real and imaginary parts. Step 1: Cross-multiply the equation. jx(x+3y) = (3x+j4)(1+jy) Step 2: Expand both sides of the equation. jx^2 + j3xy = 3x + j3xy + j4 + j^24y Since j^2 = -1, substitute this into the equation: jx^2 + j3xy = 3x + j3xy + j4 - 4y Rearrange the right side to group real and imaginary terms: j(x^2 + 3xy) = (3x - 4y) + j(3xy + 4) Step 3: Equate the real parts and the imaginary parts. Equating the real parts: 0 = 3x - 4y (*1) Equating the imaginary parts: x^2 + 3xy = 3xy + 4 (*2) Step 4: Solve the system of equations. From (*2): x^2 = 4 x = ± 2 From (*1), we have 3x = 4y, so y = (3x)/(4). Case 1: If x=2 y = (3(2))/(4) = (6)/(4) = (3)/(2) Case 2: If x=-2 y = (3(-2))/(4) = (-6)/(4) = -(3)/(2) The solutions are: x=2, y=(3)/(2) and x=-2, y=-(3)/(2) Question 11: Given z = (a+jb)/(c+jd), where a, b, c, d are real quantities. Step 1: Express z in the form X+jY by multiplying the numerator and denominator by the conjugate of the denominator. z = (a+jb)/(c+jd) × (c-jd)/(c-jd) z = ((a+jb)(c-jd))/(c^2 - (jd)^2) z = (ac - ajd + jbc - j^2bd)/(c^2 - j^2d^2) Since j^2 = -1: z = (ac - ajd + jbc + bd)/(c^2 + d^2) Group the real and imaginary parts: z = ((ac+bd) + j(bc-ad))/(c^2+d^2) z = (ac+bd)/(c^2+d^2) + j(bc-ad)/(c^2+d^2) Here, the real part is X = (ac+bd)/(c^2+d^2) and the imaginary part is Y = (bc-ad)/(c^2+d^2). a) If z is real: Step 2: If z is real, its imaginary part must be zero. Y = 0 (bc-ad)/(c^2+d^2) = 0 Since c^2+d^2 ≠ 0, the numerator must be zero: bc-ad = 0 bc = ad Divide both sides by bd (assuming b ≠ 0 and d ≠ 0): (bc)/(bd) = (ad)/(bd) (c)/(d) = (a)/(b) b) If z is entirely imaginary: Step 3: If z is entirely imaginary, its real part must be zero. X = 0 (ac+bd)/(c^2+d^2) = 0 Since c^2+d^2 ≠ 0, the numerator must be zero: ac+bd = 0 ac = -bd Divide both sides by bc (assuming b ≠ 0 and c ≠ 0): (ac)/(bc) = (-bd)/(bc) (a)/(b) = -(d)/(c) Question 12: Given the equation (a+b)+j(a-b) = (1+j)^2 + j(2+j), we need to find the values of a and b. Step 1: Simplify the right-hand side (RHS) of the equation. (1+j)^2 + j(2+j) Step 2: Expand (1+j)^2. (1+j)^2 = 1^2 + 2(1)(j) + j^2 = 1 + 2j - 1 = 2j Step 3: Expand j(2+j). j(2+j) = 2j + j^2 = 2j - 1 Step 4: Combine the simplified terms for the RHS. RHS = (2j) + (2j - 1) RHS = -1 + 4j Step 5: Equate the left-hand side (LHS) with the simplified RHS. (a+b)+j(a-b) = -1 + 4j Step 6: Equate the real parts and the imaginary parts. Equating the real parts: a+b = -1 (*1) Equating the imaginary parts: a-b = 4 (*2) Step 7: Solve the system of linear equations for a and b. Add (1) and (2): (a+b) + (a-b) = -1 + 4 2a = 3 a = (3)/(2) Substitute a = (3)/(2) into (*1): (3)/(2) + b = -1 b = -1 - (3)/(2) b = -(2)/(2) - (3)/(2) b = -(5)/(2) The values of a and b are: a=(3)/(2), b=-(5)/(2) Drop the next question! 📸