Vtyjkyuk

I saw the solution for this in Palka's book and one of the branches was defined as follows.

$$
g(z) =
begincases
sqrtz, & zin D ,Im(z)geq0 \
-sqrtz, & zin D ,Im(z)<0
endcases
$$

I am aware that $Argz$ is not continous on the non positive real axis and as the book says $g$ has been constructed to prevent discontinuities. Is $g$ continuous on $D=mathbbC$ $[0,infty)$? I don't understand how this function was constructed any help will be much appreciated. Thanks

Apparently, $sqrtz$ is the unique holomorphic function defined in $mathbb Csmallsetminus(-infty,0]$, with $sqrt1=1$.

This statement, you are referring to says that the other square root of $z$, which is defined also as holomorphic function, now in $mathbb Csmallsetminus[0,infty)$, and is denoted by $g$, AGREES with $sqrtz$, for $Im z>0$, while for $Im z<0$, it agrees with $-sqrtz$.

To understand this better, in the first case (of $sqrtz$),
in the domain $mathbb Csmallsetminus(-infty,0]$ complex numbers are expressed as
$z=r,mathrme^ivartheta$, with $varthetain(-pi,pi)$, and
$$
sqrtz=r^1/2,mathrme^ivartheta/2.
$$
In the case of $g$, in $mathbb Csmallsetminus[0,infty)$ complex numbers are expressed as
$z=r,mathrme^ivarphi$, with $varphiin(0,2pi)$, and
$$
g(z)=r^1/2,mathrme^ivarphi/2.
$$
When $Im z>0$, i.e., $vartheta, varphiin(0,pi)$ the functions $sqrtz$ agree. When $Im z<0$, which happens if $varthetain (-pi,0)$ and $varphiin(pi,2pi)$, then
$$
fracivarphi2-fracivartheta2=ipi,
$$
and hence
$$
g(z)=r^1/2,mathrme^ivarphi/2=-r^1/2,mathrme^ivartheta/2=sqrtz.
$$