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