Question

suppose f(z) =1/z. write f in the form f(z) = u(x,y) + iv(x,y), where z = x+iy and u and v are real-valued functions

Answer 1

$z=x+iy$, where $x,y\in{\mathbb{R}}$. We receive: $\frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{(x+iy)(x-iy)}=\frac{x-iy}{x^2+y^2}.$ From the latter expression we receive that $u=\frac{x}{x^2+y^2}$ and $v=\frac{-y}{x^2+y^2}$.

Answer: $f(z)=u(x,y)+iv(x,y),$ where $z=x+iy$, $x,y\in{\mathbb{R}}$; $u(x,y)=\frac{x}{x^2+y^2}$, $v(x,y)=\frac{-y}{x^2+y^2}$.