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Question
If x + iy = `sqrt(("a" + "ib")/("c" + "id")`, prove that (x2 + y2)2 = `("a"^2 + "b"^2)/("c"^2 + "d"^2)`
Solution
x + iy = `sqrt(("a" + "ib")/("c" + "id")`
∴ (x + iy)2 = `("a" + "ib")/("c" + "id")`
∴ x2 + 2xyi + y2i2 = `("a" + "ib")/("c" + "id") xx ("c" - "id")/("c" - "id")`
∴ x2 + 2xyi – y2 = `("ac" - "adi" + "bci" - "bdi"^2)/("c"^2 - "d"^2"i"^2)` ...[∵ i2 = –1]
∴ (x2 – y2) + 2xyi = `("ac" - "adi" + "bci" + "bd")/("c"^2 + "d"^2)`
∴ (x2 – y2) + 2xyi =`(("ac" + "bd") + ("bc" - "ad")"i")/("c"^2 + "d"^2)`
∴ (x2 – y2) + 2xyi = `(("ac" + "bd")/("c"^2 + "d"^2)) + (("bc" - "ad")/("c"^2 + "d"^2))"i"`
Equating the real and imaginary parts separately, we get,
x2 – y2 = `("ac" + "bd")/("c"^2 + "d"^2)` and 2xy = `("bc" - "ad")/("c"^2 + "d"^2)`
∴ (x2 + y2)2 = (x2 – y2)2 + 4x2y2
= (x2 – y2)2 + (2xy)2
= `(("ac" + "bd")/("c"^2 + "d"^2))^2 + (("bc" - "ad")/("c"^2 + "d"^2))^2`
= `(("ac" + "bd")^2 + ("bc" - "ad")^2)/("c"^2 + "d"^2)^2`
= `("a"^2"c"^2 + 2"abcd" + "b"^2"d"^2 + "b"^2"c"^2 - 2"abcd" + "a"^2"d"^2)/("c"^2 + "d"^2)^2`
= `(("a"^2"c"^2 + "b"^2"c"^2) + ("a"^2"d"^2 + "b"^2"d"^2))/("c"^2 + "d"^2)^2`
= `(("a"^2 + "b"^2)"c"^2 + ("a"^2 + "b"^2)"d"^2)/("c"^2 + "d"^2)^2`
= `(("a"^2 + "b"^2)("c"^2 + "d"^2))/("c"^2 + "d"^2)^2`
∴ (x2 + y2)2 = `("a"^2 + "b"^2)/("c"^2 + "d"^2)`
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