I would like to evaluate $$\int_{\partial \mathbb{B}(-i,3)}\frac{\sin(z)}{(z-3)^3}\mathrm{\mathop{d}}z$$
using Cauchy's Integral Formula.
Notation: We denote $\partial \mathbb{B}(a, R)$ to be the boundary of the ball $\mathbb{B}(a, R)$. In this question, it will be the circle p radius $R$, centre $a$.
The Cauchy Integral Formula that I am working with says:
Suppose that $f:E \rightarrow \mathbb{C}$ is holomorphic, $E$ is an open subset of $\mathbb{C}$, and $z_0 \in E$. Pick $\rho > 0$ such that $\overline{\mathbb{B}(z_0, \rho)} \in E$. Let $C \subset E$ be a closed curve such that $\exists$ region $\Omega \subset E \setminus \{z_0\}$ such that $\partial \Omega = C \cup (-\partial \mathbb{B} (z_0, \rho))$. Then $f(z_0) = \frac{1}{2\pi i}\oint \frac{f(z)}{z-z_0}\,\mathop{\mathrm{d}}z$ (integral is oriented counterclockwise but I can't get the $\LaTeX$ code \ointctrclockwise to work).
The answer says that the integral evaluates to $-\pi i \sin (2i)$ and it uses $f(z) = \sin (z)$, $z_0 = 2i$, and differentiates $f(z_0)$ twice.
I understand that I cannot choose the curve $C$ to be $\mathbb{B}(-i,3)$ AND choose $z_0 = 3$ since then $z_0 \notin C$ and we canot satisfy the hypotheses of the theorem. Then how do we choose $z_0$?
What you say about what the answer says and etc. doesn't seem to make much sense, since
$$|3+i|=\sqrt{3^3+1^2}=\sqrt{10}>3\implies\;f(z):=\frac{\sin z}{(z-3)^n}$$
is analytic on $\;|z+i|=3\;$ and in its interior and thus by CIT we get the integral equals zero.
I can't see what $\;z_0=2i\;$ has to do with all this...