$f(z)=\frac{\sin(\sqrt{z-1})}{(z-1)^2}$
I can see that there is a singularity at $z=1$, but I don't see how to find the order of it or the residue. I understand that with $\cos$ we can say
$\frac{\cos(\sqrt{z-1})}{(z-1)^2} = \frac{1}{(z-1)^2} \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{n-1})^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} (n-1)^{n-2}$
which tells us we have a double pole and residue $(-1)^1/(2)!=-1/2$. When we try and do this with $\sin$, however, we get $(n-1)^{n-3/2}$. Does this tell us anything about the residue or the order of the pole?
There is no $r>0$ such that $\sqrt {z-1}$ is analytic in $\{z:0<|z-1| <r\}$ irrespective of which of the two branches of square root you use. So we cannot treat $1$ as an isolated singularity of this function and classification of singularity at $z=1$, finding residue etc don't make sense.