On the Laurent series of $\frac{\sin z}{z-1}$ at $z=1$

Question

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 $$\frac{\sin z}{z-1}=\sum_{n=0}^{\infty}(-1)^n\frac{(z-1)^{2n}}{(2n+1)!}\ ? $$

Answer 1

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 $$f^{(k)}(1)=\begin{cases}\sin(1)&k\in 4\mathbb N\\ \cos(1)&k\in 4\mathbb N+1\\ -\sin(1)&k\in 4\mathbb N+2\\ -\cos(1)&k\in 4\mathbb N+3.\end{cases}$$

Answer 2

Similar to @Surb's answer, you can write $$\frac{\sin(z)}{z-1}=\sum_{n=-1}^\infty \frac{\cos \left(1+n\frac{\pi }{2}\right)}{(n+1)!}(z-1)^n$$