 Solution

For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Holomorphy implies these relations. Conversely, if the first partial derivatives exist and are continuous near a point, satisfying the equations there gives complex differentiability; satisfying them on an open set gives a holomorphic function on that set.

Let $r=\sqrt{x^2+y^2}$. On the cut plane, $r+x>0$: equality could occur only at a nonpositive real point, which is excluded. Thus $u=\sqrt{(r+x)/2}$ is strictly positive and $v=y/(2u)$ is well defined and smooth. Since $y^2=(r-x)(r+x)$,ConsequentlyOne can verify holomorphy directly:Here $r>0$ throughout the domain, so all partial derivatives are continuous. Therefore $f$ is holomorphic, specifically the principal square root on the cut plane, with derivative $f^{\prime }(z)=1/[2f(z)]$.