Branches of the square root function in the domain $D=\mathbb{C}$\ $[0,\infty)$ I saw the solution for this in Palka's book and one of the branches was defined as follows.
$$
g(z) =
\begin{cases}
\sqrt{z}, & z\in D ,Im(z)\geq0 \\
-\sqrt{z}, & z\in D ,Im(z)<0
\end{cases}
$$ 
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  $D=\mathbb{C}$\ $[0,\infty)$? I don't understand how this function was constructed any help will be much appreciated. Thanks
 A: Apparently, $\sqrt{z}$ is the unique holomorphic function defined in $\mathbb C\smallsetminus(-\infty,0]$, with $\sqrt{1}=1$.
This statement, you are referring to says that the other square root of $z$, which is defined also as holomorphic function, now in $\mathbb C\smallsetminus[0,\infty)$, and is denoted by $g$, AGREES with $\sqrt{z}$, for $\Im z>0$, while for $\Im z<0$, it agrees with $-\sqrt{z}$.
To understand this better, in the first case (of $\sqrt{z}$), 
in the domain $\mathbb C\smallsetminus(-\infty,0]$ complex numbers are expressed as
$z=r\,\mathrm{e}^{i\vartheta}$, with $\vartheta\in(-\pi,\pi)$, and 
$$
\sqrt{z}=r^{1/2}\,\mathrm{e}^{i\vartheta/2}.
$$
In the case of $g$, in $\mathbb C\smallsetminus[0,\infty)$ complex numbers are expressed as
$z=r\,\mathrm{e}^{i\varphi}$, with $\varphi\in(0,2\pi)$, and 
$$
g(z)=r^{1/2}\,\mathrm{e}^{i\varphi/2}.
$$
When $\Im z>0$, i.e., $\vartheta, \varphi\in(0,\pi)$ the functions $\sqrt{z}$ agree. When $\Im z<0$, which happens if $\vartheta\in (-\pi,0)$ and $\varphi\in(\pi,2\pi)$, then
$$
\frac{i\varphi}{2}-\frac{i\vartheta}{2}=i\pi,
$$
and hence
$$
g(z)=r^{1/2}\,\mathrm{e}^{i\varphi/2}=-r^{1/2}\,\mathrm{e}^{i\vartheta/2}=\sqrt{z}.
$$
