Given $v = \frac{-y}{x^2+y^2}$, find the real component of the function $u$.
I'm not sure why I'm not getting this. Here's my work so far:
$$v_x = \frac{2xy}{(x^2+y^2)^2} = -u_y \Rightarrow u(x,y) = \frac{x}{x^2+y^2} + \phi(x) $$ where $\phi(x)$ is (as of yet) undetermined. Continuing on, I get $$ u_x = \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} + \phi'(x) = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} = v_y $$ simplifying $$ \phi'(x) = \frac{2x^2+2y^2}{(x^2+y^2)^2} - \frac{2}{x^2+y^2}$$ multiplying the rightmost fraction by $x^2+y^2$ (top and bottom) we find that $$\phi'(x) = 0 \Rightarrow \phi(x) = C \in \mathbb{R}$$ so $$ u(x,y) = \frac{x}{x^2+y^2} + C$$
This satisfies first first of the Cauchy-Reimann conditions, but it does not seem to satisfy the second. Explicitly the condition that $u_x = v_y$ is $$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} $$ I don't believe that this is generally true… The only idea that I have is that perhaps $\phi(x) = C$ is a constant with respect to $x$, but not $y$ but I think that would contradict my first line of reasoning above.
Any ideas?
Hint:
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \iff$$ $$\iff \frac{2}{x^2+y^2} = \frac{2(x^2+y^2)}{(x^2+y^2)^2} $$
So your work seems correct !