I have to find $\text{Res}\left(\sin\left(\frac{z}{z-1}\right),1\right)$
We have $$\sin\left(\frac{z}{z-1}\right) = \sin\left(1+\frac{1}{z-1}\right)\\= \left(1+\frac{1}{z-1}\right)-\frac{1}{3!}\left(1+\frac{1}{z-1}\right)^3+\frac{1}{5!}\left(1+\frac{1}{z-1}\right)^5 - \frac{1}{7!}\left(1+\frac{1}{z-1}\right)^7 + \cdots$$
Therefore, we have $$\text{Res}\left(\sin\left(\frac{z}{z-1}\right),1\right)=1-\frac{1}{3!}\binom{3}{1}+\frac{1}{5!}\binom{5}{1}-\frac{1}{7!}\binom{7}{1}+ \cdots \\ = 1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \cdots = \cos1$$
Alternate Proof
$$\sin\left(1+\frac{1}{z-1}\right) = \sin1 \cos\left(\frac{1}{z-1}\right) + \cos1 \sin\left(\frac{1}{z-1}\right)$$
We will get a $1/(z-1)$ term only in the expansion of $\sin\left(\frac{1}{z-1}\right)$, therefore the residue is $\cos1$
Am I right?