I'm trying to understand the link between $ \mathbb R^2 $ and $ \mathbb C $. I know how to compute potentials when I know the gradient of it :
$$ f(x,y) = \left( \frac {-x}{(x^2+y^2)^2 } , \frac {-y}{(x^2+y^2)^2 } \right) $$ gives $$ F(x,y) = \frac {1}{2(x^2+y^2) } $$
My question is,
How can you find the same result with complex integration ?
My attempt :
using this formula (and my calculator I admit) $$\int f(z) dz = \int u dx - v dy + i \int v dx + u dy $$
$$ \int \int \frac {-x}{(x^2+y^2)^2 } +i \frac {-y}{(x^2+y^2)^2 } dx dy = \frac { \arctan (\frac y x) } { 2x} - i \frac { 2y \ln(y) + \ln( \frac 1 { y^2 }) y - 2 \arctan (\frac x y ) x } { 4 x y }$$
which looks like something not that afar from my result, but I'm not convinced.
So is it possible and how do you do this?