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I wish to compute $$ \int_C \frac{\sin z}{(z+1)^7}\ \text{d}z $$ where $C$ is the circle of radius $6$, centre $0$, positively oriented.

Now I know that $f(z) = \frac{\sin z}{(z+1)^7}$ is analytic on $\mathbb{C}\setminus\{-1\}$ since $f$ is the composition and quotient of analytic functions.

I tried to use the Cauchy-Goursat Extension which says $$ \int_C f(z)\ \text{d}z = \int_{C'} f(z)\ \text{d}z $$ where $C'$ is the circle of radius $1$, center $-1$, positively oriented. However this only reformulates the problem.

My second thought was to integrate around $-1$ with a keyhole-like contour with an $\varepsilon$-wide cut and show it goes to $0$ but this feels unnecessarily complicated. Am I missing something that'd simplify things?

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The pole $z=-1$ is inside the circle $C$. Hence, by the Residue Theorem, since the pole $z=-1$ has order $n=7$, we have that (see HERE) $$\int_C \frac{\sin z}{(z+1)^7}\ \text{d}z=2\pi i\cdot \mbox{Res}\left(\frac{\sin z}{(z+1)^7},-1\right)=2\pi i \cdot \frac{ D^{(6)}(\sin(z))_{z=-1}}{6!}.$$

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