I want to show that the branch of the complex square root function $\sqrt{}:\Omega \rightarrow \mathbb{C}$,
$x+iy\mapsto u(x,y)+iv(x,y)=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+i\frac{y}{|y|}\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$
on $\Omega=\{x+iy:x>0\}$ (the complement of the negative real axis $z\leq0$) is analytic using the theorem: If $u(x,y)$, $v(x,y)$ have continuous first-order partials and satisfy the Cauchy-Riemann equations, then $\sqrt{}$ is analytic.
This has been fine when $y\neq0$. But when $y=0$, by definition $\sqrt{x+iy}=\sqrt{x}$ since $x>0$, which does not satisfy the Cauchy-Riemann equations as I understand. I looked around a bit and one source just seemed to ignore this. Where is my confusion? Thank you.
The way you wrote $v$ makes it non-obvious that $v$ is differentiable (or even continuous) when $y = 0$. But you can express $v$ somewhat differently, e.g. $$ v = \frac{y}{2u} = \frac{y}{\sqrt{2x+2\sqrt{x^2+y^2}}}$$ and then you can see that it does satisfy the C-R equations.