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Question
If `("a" + 3"i")/(2+ "ib")` = 1 − i, show that (5a − 7b) = 0
Solution
`("a" + 3"i")/(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
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