complex analysis contour integrals using Cauchy's Residue theorem

Question

I am trying to find the following integral: $$\oint \frac{\cos(z-1)}{(z-1)^{5}}dz$$ over the region $c$, where $c$ is $\left | z-1 \right | =1$.
This is my solution so far :

$$\oint \frac{\cos(z-1)}{(z-1)^{5}}dz = \frac{1-\frac{(z-1)^{2}}{2!}+\frac{(z-1)^{4}}{4!}-...}{(z-1)^{5}} = \frac{1}{(z-1)^{5}} - \frac{1}{2!(z-1)^{3}} + \frac{1}{4!(z-1)} + ...$$

Is Residue equal to $\frac{1}{4!}$ and therefore the result of integration $2i\pi\times \frac{1}{4!}$?

Answer 1

$$\oint \frac{\cos(z-1)}{(z-1)^{5}}dz=\frac{2\pi i}{4!} \Bigg\{\frac{d^4}{dz^4}(\cos(z-1)) \,\,\, \mathrm{at} \,\,\,\, z=1 \,\,\,\, \Bigg\}=\frac{2 \pi i }{4!} $$

Answer 2

Hint: From general Cauchy formula for derivatives one has $$f^{(4)}(z_0)=\frac{4!}{2\pi i} \oint_{|z-z_0|=1}\frac{f(z)}{(z-z_0)^5}dz$$. So you need only to find your function and derivate it 4 times.