Given the function $$f(z)=\frac{y^2-x^2}{2y}-i|x+y|$$ I need to find where this function is differentiable( where $f'(z)$ exists).
What I did:
$u(x,y)=\frac{y^2-x^2}{2y}$, $v(x,y)=-|x+y|$
First noted that $u(x,y)$ is not real differentiable at $y=0$ and $v(x,y)=-|x+y|$ is not differentiable when $y=-x$. Then, I can say that $f(z)$ is real-differentiable at $D=C-\{x+iy:y=0,y=-x\}$.
Finding the partial derivatives in $D$: $u_x=-\frac{x}{y}$,$u_y=\frac{1}{2}\left(1+\frac{x^2}{y^2}\right)$,$v_x=v_y=-\frac{x+y}{|x+y|}$.
Ok, here's the problem: I'm very confuse because of the absolute value there. I separate into cases for solving the Cauchy-Riemann equations:
$x+y\geq 0\Rightarrow$ $$u_x=-\frac{x}{y}=-1=v_y$$ $$u_y=\frac{1}{2}\left(1+\frac{x^2}{y^2}\right)=1=-v_x$$ With solution $x=y \wedge x^2=y^2\Rightarrow x=y\wedge y=|x|\Rightarrow x=|x|\Rightarrow x\geq 0$
$x+y<0\Rightarrow$ $$u_x=-\frac{x}{y}=1=v_y$$ $$u_y=\frac{1}{2}\left(1+\frac{x^2}{y^2}\right)=-1=-v_x$$ With solution $y=-x\wedge x^2=-3y^2\Rightarrow x=0$.
So, I think the domain where the derivative exists is the intersection of the solutions plus the separation into cases. I found the solution $E=C-\{x+iy:x>0,y>-x\}$. Can someone check if this is correct,please? I really appreciate any suggestions or correction.