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)`