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?
 A: Hint: $$\frac{1}{z}=\frac{\bar{z}}{z\bar{z}}.$$
A: Our intuition from the real numbers tells us that the inverse of a real number $c$ is $1/c$.  
Likewise, we conjecture that the inverse of a complex $z \in \mathbb{C}$ is $\frac{1}{z} = \frac{1}{x + iy}$.  To get this into a more recognizable form, multiply the numerator and denominator by $z^* = x - iy$.  From here, we arrive at $\frac{x-iy}{x^2+y^2} = \frac{x}{x^2 + y^2} + \frac{y}{x^2+y^2}i$.  Certainly, our purported $z^{-1}$ is of the form $a+bi$ for $a, b \in \mathbb{R}$.  Next, we can simply multiply $z$ by our purported $z^{-1}$ to confirm that we do indeed get $1 + 0i$, which is the multiplicative identity in $\mathbb{C}$.
A: 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!!!
