= Solution
{c}
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 $y^2=(r-x)(r+x)$,
$$
u^2=\frac{r+x}2,\qquad v^2=\frac{r-x}2,\qquad2uv=y.
$$
Consequently
$$
\boxed{(u+iv)^2=x+iy=z.}
$$
One can verify <holomorphy> directly:
$$
u_x=\frac u{2r}=v_y,\qquad u_y=\frac v{2r}=-v_x.
$$
Here $r>0$ throughout the domain, so all partial derivatives are continuous. Therefore \b[$f$ is holomorphic], specifically the <principal square root> on the cut plane, with <derivative> $f'(z)=1/[2f(z)]$.
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