If $f(z)=\frac{\sin(\sqrt z)}{\sqrt z}$, then $z=0$ is a removal singularity of $f$. (T/F)
We know $\lim_{z \rightarrow 0}\frac{\sin(z)}{z}=1$. Is the sam thought can be applied here?, Here $z=0$ is also branch point of $\sqrt z$ but $0$ is nt branch point of $f$. Help
Note that$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$$So, if $w$ is any square root of $z$, then$$\sin w=w-w\frac z{3!}+w\frac{z^2}{5!}-\cdots$$and therefore$$\frac{\sin w}w=1-\frac z{3!}+\frac{z^2}{5!}-\cdots$$So, and although your question is not very precise, I would say that $0$ is a removable singularity of your function.
So, if you define$$f(z)=1-\frac z{3!}+\frac{z^2}{5!}-\cdots,$$then, no matter how you defined $\sqrt z$ (as long as it is a square root of $z$, of course), then $\frac{\sin\sqrt z}{\sqrt z}=f(z)$ and $f$ is a holomorphic function.