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.

The function $f(z)=\frac{\sin{z}}{z^4}$ has a third order pole $z=0$ inside $\gamma$. By Cauchy's residue theorem, the integral
$$ \int_{\gamma}f(z)\,dz=2{\pi}i\cdot\underset{z=0}{\rm res}\, f(z),$$
where the residue at the third order pole $z=0$ can be calculated using formula
$$\underset{z=0}{\rm res}\, f(z)=\frac{1}{(3-1)!}\cdot\lim_{z\to 0}[z^3\cdot f(z)]^{(3-1)}=\frac{1}{2}\cdot\biggl(\frac{\sin{z}}{z}\biggr)''\biggr|_{z=0}=-\frac{1}{6}. $$
Alternatively, though in fact by defintition, the residiue at $z=0$ can be found as the $c_{-1}$ coefficient at the term $z^{-1}$ of the Laurent series in the ring $0<|z|<\infty$ for $f(z)$ having two isolated singularities on the complex plane: a pole at $z=0$ and an essential singularity at $z=\infty$. More precisely,
$$f(z)=\frac{1}{z^4}\cdot (z-\frac{z^3}{3!}+\frac{z^5}{5!}-\dots)=\frac{1}{z^3}-\frac{1}{6z}+\frac{z}{5!}-\dots\,,\quad 0<|z|<\infty, $$
whence readily follows 
$$\underset{z=0}{\rm res}\, f(z)\overset{def}{=}c_{-1}=-\frac{1}{6}\,.$$
Hence, by Cauchy's residue theorem, the integral
$$ \int_{\gamma}f(z)\,dz=-\frac{2{\pi}i}{6}=-\frac{{\pi}i}{3}\,.$$
