If x and y are real, solve the equation: (jx)/(1+jy) = (3x+j4)/(x+3y)

Here are the solutions to questions 10, 11, and 12. Question 10: If x and y are real, solve the equation: (jx)/(1+jy) = (3x+j4)/(x+3y) Step 1: Cross-multiply the equation. jx(x+3y) = (3x+j4)(1+jy) Step 2: Expand both sides of the equation. Remember that j^2 = -1. jx^2 + j3xy = 3x + j3xy + j4 + j^24y jx^2 + j3xy = 3x + j3xy + j4 - 4y Step 3: Rearrange terms to group real and imaginary parts. jx^2 - j4 = 3x - 4y j(x^2 - 4) = (3x - 4y) Step 4: Equate the real and imaginary parts. Since x and y are real, the left side is purely imaginary and the right side is purely real. For them to be equal, both must be zero. Equating the real parts: 3x - 4y = 0 (*) Equating the imaginary parts: x^2 - 4 = 0 (**) Step 5: Solve equation (**) for x. x^2 = 4 x = ± 2 Step 6: Substitute the values of x into equation (*) to find y. Case 1: x=2 3(2) - 4y = 0 6 - 4y = 0 4y = 6 y = (6)/(4) = (3)/(2) Case 2: x=-2 3(-2) - 4y = 0 -6 - 4y = 0 4y = -6 y = (-6)/(4) = -(3)/(2) The solutions are x=2, y=(3)/(2) and x=-2, y=-(3)/(2). Question 11: If z = (a+jb)/(c+jd), where a, b, c and d are real quantities, show that (a) if z is real then (a)/(b) = (c)/(d) (b) if z is entirely imaginary then (a)/(b) = -(d)/(c) 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) z = ((ac+bd) + j(bc-ad))/(c^2+d^2) z = (ac+bd)/(c^2+d^2) + j(bc-ad)/(c^2+d^2) Let the real part be X = (ac+bd)/(c^2+d^2) and the imaginary part be Y = (bc-ad)/(c^2+d^2). a) 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) This shows that if z is real, then (a)/(b) = (c)/(d). b) 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) This shows that if z is entirely imaginary, then (a)/(b) = -(d)/(c). Question 12: Given that (a+b)+j(a-b) = (1+j)^2 + j(2+j), obtain the values of a and b. Step 1: Simplify the right-hand side (RHS) of the equation. (1+j)^2 + j(2+j) Expand (1+j)^2: (1+j)^2 = 1^2 + 2(1)(j) + j^2 = 1 + 2j - 1 = 2j Expand j(2+j): j(2+j) = 2j + j^2 = 2j - 1 Now, add the simplified parts: 2j + (2j - 1) = 4j - 1 So, the RHS is -1 + 4j. Step 2: Equate the real and imaginary parts of the given equation. The equation is now: (a+b)+j(a-b) = -1 + 4j Equating the real parts: a+b = -1 (1) Equating the imaginary parts: a-b = 4 (2) Step 3: Solve the system of linear equations for a and b. Add equation (1) and equation (2): (a+b) + (a-b) = -1 + 4 2a = 3 a = (3)/(2) Substitute the value of a into equation (1): (3)/(2) + b = -1 b = -1 - (3)/(2) b = -(2)/(2) - (3)/(2) b = -(5)/(2) The values are a=(3)/(2), b=-(5)/(2). That's 2 down. 3 left today — send the next one.