I have a function $f(z)$ with imaginary part $v(x,y) = \frac x{x^2 + y^2}$. How do I find the real part of this function?
I am trying to solve this using the Cauchy-Riemann equations. I have found that $v_x = \frac{y^2-x^2}{(x^2+y^2)^2}$ and $v_y = -\frac{2xy}{(x^2+y^2)^2}$.
I have substituted them into the CR equations and integrated with respects to $x$ and $y$ respectively. However, I am finding it hard to get a final solution for $u(x,y)$, please help.
Provided that $f(z)$ is analytic, so the Cauchy-Riemann equations hold in the first place, we have
$$v(x,y) = \frac x{x^2+y^2} \implies v_y = u_x = -\frac{2xy}{(x^2+y^2)^2} \\ \implies u(x,y) = \int -\frac{2xy}{(x^2+y^2)} \, dx = \frac y{x^2+y^2} + g(y)$$
Use the other CR condition to determine another form of $u(x,y)$, which you should be able to reconcile with the first solution.