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Question
Answer the following:
If x + iy = `("a" + "ib")/("a" - "ib")`, prove that x2 + y2 = 1
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Solution
x + iy = `("a" + "ib")/("a" - "ib") = (("a" + "ib")("a" + "ib"))/(("a" - "ib")("a" + "ib"))`
= `("a"^2 + "i"^2"b"^2 + 2"abi")/("a"^2 - "i"^2"b"^2)`
= `(("a"^2 - "b"^2) + 2"abi")/("a"^2 + "b"^2)` ...[∵ i2 = – 1]
∴ x + iy = `("a"^2 - "b"^2)/("a"^2 + "b"^2) + (2"ab")/("a"^2 + "b"^2)"i"`
Equating real and imaginary parts, we get
x = `("a"^2 - "b"^2)/("a"^2 + "b"^2)` and y = `(2"ab")/("a"^2 + "b"^2)`
∴ x2 + y2 = `(("a"^2 - "b"^2)^2)/("a"^2 + "b"^2)^2 + (4"a"^2"b"^2)/("a"^2 + "b"^2)^2`
= `("a"^4 + "b"^4 - 2"a"^2 "b"^2 + 4"a"^2"b"^2)/("a"^2 + "b"^2)^2`
= `(("a"^2 + "b"^2)^2)/("a"^2 + "b"^2)^2`
∴ x2 + y2 = 1
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