Evaluate the contour integral $\int_{\gamma(0,1)}\frac{\sin(z)}{z^4}dz.$ 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.

Cauchy's integral formula is
$$f^{(n)}(z) = \frac{n!}{2\pi i} \int_\gamma \frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta,$$
where $\gamma$ is a closed path winding once around $z$, and enclosing no singularity of $f$.
Thus in your example, $n = 3$, and you need the third derivative,
$$\int_{\gamma(0,1)} \frac{\sin z}{z^4}\,dz = \frac{2\pi i}{3!} \sin^{(3)} 0 = \frac{2\pi i}{6} (-\cos 0) = - \frac{\pi i}{3}.$$
