Proving identity $\frac{1}{x+iy}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$

Question

Below are some of the identities provided early in Needhams "Visual Complex analysis"

I want to verify the identity in my title. From the identities given before it, I could only write

$$\frac{1}{x+iy}=\frac{1}{x^2+y^2}\big(-\arctan\frac{y}{x}\big)$$

Not sure how to derive that $\big(-\arctan\frac{y}{x}\big)=x-iy$

thanks

Answer 1

$$\frac{1}{x+iy} \frac{x-iy}{x-iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2-i^2y^2} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}$$ as required

Answer 2

Given $z=x+iy$, this is the same as showing: $$\frac 1z =\frac{\bar z}{|z|^2}$$ or rather that: $$|z|^2=z\bar z$$ which is trivial

Answer 3

Or in polar co-ordinates:

$\frac{1}{re^{i\theta}}=\frac{1}{r}e^{-i\theta} = \frac{re^{-i\theta}}{r^2}$