I have a function $f(z)$ holomorphic in $\mathbb{C}\setminus\mathbb{R}^-$. I have these information:
- $f(x+i\epsilon) = f(x-i\epsilon)$ on $\mathbb{R}^+$ (the $\epsilon$ is indented as a shorthand for a limit);
- $f(x+i\epsilon) = - f(x-i\epsilon)$ on $\mathbb{R}^-$;
- $f(z)=\sqrt{z} + O\left(\frac{1}{\sqrt{z}}\right)$ for $|z|\rightarrow\infty$, $z\in\mathbb{C}\setminus\mathbb{R}^-$.
I'm asked to show that $f(z)=\sqrt{z}$.
I tried to write $f(x)=\sqrt{z}+\frac{h(x)}{\sqrt{x}}$. In such a way $h(z)$ is continuos on the whole $\mathbb{C}$ and holomorphic on $\mathbb{C}\setminus\mathbb{R}^-$... If it were holomorphic on the whole $\mathbb{C}$, it would be constant and so $h(z)=\lim_{z\rightarrow 0}h(z)=0$... but I cannot see a way to prove that $f(z)$ is indeed holomorpic even on $\mathbb{R}^-$. Can you help me?
Let $\sqrt{z}$ denote the square root which branch cut on the negative real axis. Is there a reason $f(z)=\sqrt{z}(1+\frac{1}{z})$ does not satisfy your conditions?