Developing Laurent series on three types of regions: centered at the origin, and non-centered at the origin
Solution 1:
That's not correct in what concerns the development into Lauren series in the annulus $2<|z-1|<3$. For such a $z$ we have\begin{align}\frac1{2-z}&=\frac1{1-(z-1)}\\&=-\frac1{z-1}\frac1{1-\frac1{z-1}}\\&=-\frac1{z-1}\sum_{n=0}^\infty\left(\frac1{z-1}\right)^n\\&=-\frac1{z-1}\sum_{n=0}^\infty(z-1)^{-n}\\&=-\frac1{z-1}\sum_{n=-\infty}^0(z-1)^n\\&=-\sum_{n=-\infty}^0(z-1)_{n-1}\\&=-\sum_{n=-\infty}^{-1}(z-1)^n.\end{align}