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If a+3i2+ib = 1 − i, show that (5a − 7b) = 0

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Question

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

Sum
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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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Chapter 1: Complex Numbers - Exercise 1.1 [Page 6]

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