(a + 3i)/(2 + ib) = 1 - i
a + 3i = (1 – i)(2 + ib) = 2 + bi – 2i – bi2
2 + (b – 2)i – b(-1) ……[∵ i2 = -1]
a + 3i = (2 + b) + (b – 2)i
Equating real and imaginary parts, we get
a = 2 + b and 3 = b – 2
a = 2 + b and b = 5
a = 2 + 5 = 7
5a – 7b = 5(7) – 7(5)
= 35 – 35 = 0