this is what I got
$f(z)=(z-3)\sin\frac{1}{z+2}=(z-3)\left(\frac{1}{z+2}-\frac{1}{3!}\left(\frac{1}{z+2}\right)^3+\frac{1}{5!}\left(\frac{1}{z+2}\right)^5+...\right)$ $=(z+2)\left(\frac{1}{z+2}-\frac{1}{3!}\left(\frac{1}{z+2}\right)^3+\frac{1}{5!}\left(\frac{1}{z+2}\right)^5+...\right)-5\left(\frac{1}{z+2}-\frac{1}{3!}\left(\frac{1}{z+2}\right)^3+\frac{1}{5!}\left(\frac{1}{z+2}\right)^5+...\right)$ $=1-\frac{5}{z+2}-\frac{1}{3!}\left(\frac{1}{z+2}\right)^2+\frac{5}{3!}\left(\frac{1}{z+2}\right)^3+\frac{1}{5!}\left(\frac{1}{z+2}\right)^4-\frac{5}{5!}\left(\frac{1}{z+2}\right)^5+...$
My question is how can I write the above sum in terms of a series?
Hint. One may write, for $|z+2|\ne0$, $$ \begin{align} (z-3)\sin\frac{1}{z+2}&=((z+2)-5)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!\left(z+2\right)^{2n+1}} \\\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!\left(z+2\right)^{2n}}-\sum_{n=0}^\infty\frac{5(-1)^n}{(2n+1)!\left(z+2\right)^{2n+1}} \\\\&=\sum_{m=0}^\infty\frac{\cos \frac{m\pi}{2}+5(m+1)\sin \frac{m\pi}{2}}{(m+1)!\left(z+2\right)^{m}}. \end{align} $$