$$I = \oint_C\frac{\sin(\pi z)}{(z-1)^5}\,dz,\quad\text{$C$ is the circle $|z|=2$.}$$
I have that there is a singularity at $z = 1$. So, taking $z_0 = 1$ we can write
$$I = \oint_C\frac{f(z)}{(z-z_0)^5}\,dz\quad\text{where $f(z)=\sin(\pi z)$}\Rightarrow f^{(4)}(z) = \pi^4\sin(\pi z)$$
Using this derivative theorem in my notes that looks like this:
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}\,dz$$
I can write
$$I = \frac{2\pi i}{4!}\cdot f^{(4)}(1) = \frac{2\pi i}{4!}\cdot(0) = 0$$
Is this final answer of zero correct?
The Laurent series of $f(z)=\frac{\sin(\pi z)}{(z-1)^5}$ at $z=1$ can be derived from the Laurent series of $g(z)=\frac{\sin(\pi(z+1))}{z^5}=-\frac{\sin(\pi z)}{z^5}$ at $z=0$. Since the last function is an even function, the residue at $z=0$ is zero, hence the original integral equals zero, too: no explicit computation is really needed.