Find and classify all isolated singularities of $\displaystyle f(z)=\frac{1}{1+\sqrt{z}}$.
So if $1+\sqrt{z}=0$ then $\sqrt{z}=-1$. Therefore $z=1$. Hence, $1$ is an isolated singularity and it is a pole. I am not quite sure whether my argument is correct here. Is it correct? If not, how may I do it?
That depends on the branch of $\sqrt z$. Usually, $\sqrt z$ is taken to be the branch defined on $\mathbb C\setminus(-\infty,0]$ and $\sqrt 1=1$. In that case, $f$ does not have isolated singularities.