Evaluate a contour integral with Taylor coefficients I am working on the following exercise:

Evaluate 
  $$\int_{\lvert z-1 \rvert = 2} \frac{\sin z}{z^4} dz $$
  with the formula for the Taylor coefficients of holomorphic functions
  $$c_n = \frac{1}{2\pi i} \int_{\lvert z-z_0 \rvert = r} \frac{f(z)}{(z-z_0)^{n+1}} dz$$

I should not use residue calculus. The only thing I can think of is using the formula for $c_n$ to write (I know the Taylor Series of $sin$)
$$c_3 = \frac{1}{2\pi i} \int_{\lvert z-1 \rvert = 2} \frac{\sin z}{(z-1)^{4}} dz,$$  but as you can see the denominator $(z-1)$ does not work...
Could you help me?

The only singularity of $\frac{\sin z}{z^4}$ inside $|z-1|\leq 2$ is at the origin, and it is a triple pole. By Cauchy's theorem and the residue theorem it follows that
$$ \oint_{|z-1|=2}\frac{\sin(z)}{z^4}\,dz = \oint_{|z|=1}\frac{\sin(z)}{z^4}\,dz = \oint_{|z|=1}\frac{z-\frac{z^3}{6}}{z^4}\,dz = 2\pi i\cdot\left(-\frac{1}{6}\right)=\color{red}{-\frac{\pi i}{3}}.$$
