Perhaps it is an easy question, but what would be the Laurent series expansion of the function $\frac{\sin z}{z-1}$ in powers of $z-1$, valid in the domain $0<|z-1|<\infty$?
Isn't that
$$\frac{\sin z}{z-1}=\sum_{n=0}^{\infty}(-1)^n\frac{(z-1)^{2n}}{(2n+1)!}\ ?
$$
Best Answer
No it's not correct. The Laurent series of $\sin(z)$ around $z=1$ is given by $$\sin(z)=\sum_{k=0}^\infty \frac{f^{(k)}(1)}{k!}(z-1)^k,$$
where $$f^{(k)}(1)=\begin{cases}\sin(1)&k\in 4\mathbb N\\ \cos(1)&k\in 4\mathbb N+1\\ -\sin(1)&k\in 4\mathbb N+2\\ -\cos(1)&k\in 4\mathbb N+3.\end{cases}$$