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If a + Ib = ("X" + "Iy")/("X" - "Iy"), Prove that "A"^2 +"B"^2 = 1 and "B"/"A" = (2"Xy")/("X"^2 - "Y"^2) - Mathematics

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Sum

If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`

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Solution

Here, a + ib = `("x" +"iy")/("x" - "iy")`        ....(i)

∴ a - ib = `("x" -"iy")/("x" + "iy")`              .....(ii)

Multiplying (i) and (ii), we obtain

`"a"^2 +"b"^2 = ("x" + "iy")/("x" - "iy") xx ("x" - "iy")/("x" + "iy")`

`"a"^2 +"b"^2 = 1`

Again a + ib = `("x" + "iy")/("x" - "iy") xx ("x" + "iy")/("x" + "iy")`


`= ("x"^2 - "y"^2 + 2"xyi")/("x"^2 + "y"^2)`


a + ib = `("x"^2 - "y"^2)/("x"^2 + "y"^2) + (2"xy")/("x"^2 + "y"^2)  "i"`

Equating real and imaginary part, we obtain

a= `("x"^2 - "y"^2)/("x"^2 + "y"^2)` and  b = `(2"xy")/("x"^2 + "y"^2)`


Now `"b"/"a" = (2"xy")/("x"^2 + "y"^2) xx ("x"^2 + "y"^2)/("x"^2 - "y"^2)`


`=(2"xy")/("x"^2 - "y"^2)`

Concept: Basic Concepts of Differential Equation
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