Finding an expression for the complex number Z^-1 So I want to find out an expression to express:
$$z^{-1}$$
I know the answer is:
$$z^{-1} = \frac{x-iy}{x^2+y^2}$$
But how would I go about proving this/the steps to this?

Set $z = x + iy$; then
$\dfrac{1}{z} = \dfrac{1}{ x + iy}. \tag{1}$
Multiply by
$1 = \dfrac{x - iy}{x - iy}: \tag{2}$
$\dfrac{1}{z} = \dfrac{(x - iy) \times 1}{(x - iy)(x + iy)} = \dfrac{x - iy}{x^2 + y^2}, \tag{3}$
since $(x - iy)(x + iy) = x^2 + y^2$.  And that does it!
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
