I need to find the Laurent series of $$\sin\left( \frac{z}{1-z} \right)$$ at the point $z_0 = \infty$.
How I can decompose this? At point $z_0 = 1$ everything turned out well. There I tried 2 things. Firstly, I tried $$\frac{z}{1-z} = \frac{1}{\frac{1}{z} -1} \Rightarrow \sin\left(\frac{1}{\frac{1}{z} - 1}\right) = \sum_{n=0}^{\infty} (-1)^n \frac{1}{\left(\frac{1}{\frac{1}{z} - 1}\right)^{2n+1} (2n+1)!} $$ but it looks ridiculous. Next, I tried $$ \sin\left(\frac{1}{z-1} -1\right) = \sin\left(\frac{1}{z}\left(\frac{1}{1-\frac{1}{z}}\right)\right).$$ But from there on, I don't know what I should do.
The Laurent series you want is the same as the Laurent series of $f(1/z)$ at $z=0$.
$$f(1/z)=\sin\left(\frac1{z-1}\right)$$ which is indeed analytic at $0$.
Therefore, just proceed as you are finding its Taylor series. I do not expect that the series has a nice form.