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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.

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  • $\begingroup$ You can't set $y=0$ in $f(x,y)$ and then check the partials. Doing so is ok to check the partial derivatives $u_x,v_x,$ because they depend only on the values of $f$ on the $x$-axis, but the derivatives $u_y,v_y$ depend on the values of $f$ for slightly larger and smaller values of $y$. You have to compute the partial derivative first and then set $y=0$. $\endgroup$ – juan arroyo Jan 5 '17 at 19:37
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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.

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  • $\begingroup$ In this case, can we say $2uv=y$ then $v_x=-\frac{v}{u}u_x$ that does not satisfy the Cauchy-Riemann equations? $\endgroup$ – Nosrati Jan 5 '17 at 19:06
  • $\begingroup$ But it does satisfy them. From $2uv=y$ you get $2 u_x v + 2 u v_x = 0$ and $2 u_y v + 2 u v_y = 1$, from $u^2 - v^2 = x$ you get $2 u u_x - 2 v v_x = 1$ and $2 u u_y + 2 v v_y = 0$, and then $u_x = v_y = \dfrac{u}{2(u^2 + v^2)}$, $u_y = - v_x = \dfrac{v}{2(u^2+v^2)}$. $\endgroup$ – Robert Israel Jan 6 '17 at 1:32

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