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
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 $$ g(z)=r^{1/2}\,\mathrm{e}^{i\varphi/2}=-r^{1/2}\,\mathrm{e}^{i\vartheta/2}=\sqrt{z}. $$