Let $\gamma(z_0,R)$ denote the circular contour $z_0+Re^{it}$ for $0\leq t \leq 2\pi$. Evaluate $$\int_{\gamma(0,1)}\frac{\sin(z)}{z^4}dz.$$
I know that \begin{equation} \int_{\gamma(0,1)}\frac{\sin(z)}{z^4}dz = \frac{1}{z^4}\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\right) = \frac{1}{z^3}-\frac{1}{6z}+\cdots \end{equation} but I'm not sure if I should calculate the residues and poles or to use Cauchy's formula?
Using Cauchy's formula would give $$ \frac{2\pi i}{1!} \frac{d}{dz}\sin(z),$$ evaluated at $0$ gives $2\pi i$? I'm not sure though, any help will be greatly appreciated.