Given z = (a+jb)/(c+jd), where a, b, c, d are real quantities. a) Show that if z is real, then (a)/(b) = (c)/(d)

Okay, Hølüwårh, let's go through Question 11 again. 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. The conjugate of the denominator c+jd is c-jd. z = (a+jb)/(c+jd) × (c-jd)/(c-jd) Multiply the numerators and denominators: z = ((a+jb)(c-jd))/(c^2 - (jd)^2) Expand the numerator: (a+jb)(c-jd) = ac - ajd + jbc - j^2bd. Simplify the denominator: c^2 - (jd)^2 = c^2 - j^2d^2. Since j^2 = -1, substitute this into both numerator and denominator: z = (ac - ajd + jbc - (-1)bd)/(c^2 - (-1)d^2) z = (ac - ajd + jbc + bd)/(c^2 + d^2) Group the real and imaginary parts in the numerator: z = ((ac+bd) + j(bc-ad))/(c^2+d^2) Separate into real and imaginary components: 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) Show that if z is real, then (a)/(b) = (c)/(d) 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 is a sum of squares of real numbers and cannot be zero (unless c=0 and d=0, which would make the original denominator zero, an undefined case), the numerator must be zero: bc-ad = 0 Rearrange the terms: bc = ad To get the desired form (a)/(b) = (c)/(d), divide both sides by bd (assuming b ≠ 0 and d ≠ 0): (bc)/(bd) = (ad)/(bd) (c)/(d) = (a)/(b) b) Show that if z is entirely imaginary, then (a)/(b) = -(d)/(c) Step 3: If z is entirely imaginary, its real part must be zero. X = 0 (ac+bd)/(c^2+d^2) = 0 Again, since c^2+d^2 ≠ 0, the numerator must be zero: ac+bd = 0 Rearrange the terms: ac = -bd To get the desired form (a)/(b) = -(d)/(c), divide both sides by bc (assuming b ≠ 0 and c ≠ 0): (ac)/(bc) = (-bd)/(bc) (a)/(b) = -(d)/(c) That's 3 down. 2 left today — send the next one.