editted in: it is a stipulation of the question to do this via expanding the sine function as a power series. deepest apologies for any confusion
Can i get a workings check please; appreciate it.
$$\int_{S_1(0)} \frac{\sin(z)}{z^8}= \int_{S_1(0)}\frac{1}{z^{8}}\left[ \sum_{n=0}^{\infty}\frac{(-1)^{n}z^{2n+1}}{(2n+1)!}\right]$$ $$=\sum_{n=0}^{\infty}\int_{S_1(0)}\frac{1}{z^{8}}\left[\frac{(-1)^{n}z^{2n+1}}{(2n+1)!}\right]$$ $$=\sum_{n=0}^{\infty}\int_{S_1(0)}\left[\frac{1}{z^{8}}\frac{(-1)^{n}z^{2n+1}}{(2n+1)!}\right]$$ $$=\sum_{n=0}^{\infty}\int_{S_1(0)}\left[\frac{(-1)^{n}z^{2n-7}}{(2n+1)!}\right]$$ $$=\left\{\begin{matrix} 0 ~ \forall n \in \mathbb{N}\setminus \{3\}\\ -\frac{2 \pi i }{7!} ~ \text{for } n =3 \end{matrix}\right.$$
Via the cauchy integral theorem and deformation theorem. Since $$\int_{S_{1}(0)}z^{n} = 0~\forall n \neq -1$$ and $2 \pi i$ otherwise
Since the co-effecient is also given by $$f^{(n)}(0) = \frac{n!}{2 \pi i}\int_{S_{1}(0)} \frac{f(w)}{w^{n+1}}dw \implies$$ $$\frac{2 \pi i}{n!}f^{(n)}(0) = \int_{S_{1}(0)} \frac{f(w)}{w^{n+1}}dw$$ to make it crystal clear. $$\frac{2 \pi i}{7!}\sin^{(7)}(0) = \oint_{S_{1}(0)} \frac{\sin(z)}{z^8}dz$$
$\sin$ repeats itself every fourth differentiation so we have $\sin(z) = \sin^{(4)}(z)$ and $-\cos(z) = \sin^{(3)}(z)$ which makes the final answer $$\int_{S_1(0)} \frac{\sin(z)}{z^8}= -\frac{2 \pi i}{7!}$$
and so the results agree
now specific things to check for me if you please are the signs, i havent dropped an i or a -1 somewhere have i? and just the logic, i havent made any glaring mistakes have i?
Cheers