Can you help me and discuss me on the question?
Expand $\displaystyle{f(z) = ze^{1/(z-1)}}$ in a Laurent series valid for $\displaystyle{\left|z-1\right|> 0}$.
I have no idea anything about the exponential complex form to fnd the Laurent series for that. I know that the laurent series of $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ for every complex $z$. Can I just substitiute it?
Since $$e^{w}=\sum_{k\geq0}\frac{w^{k}}{k!},\,\left|w\right|<+\infty$$ we have, taking $w=\frac{1}{z-1}$, that for $0<\left|z-1\right|<+\infty$ (these bounds follow from $\left|w\right|=\left|\frac{1}{z-1}\right|<+\infty$) we obtain
\begin{align} ze^{\frac{1}{z-1}}=\sum_{k\geq0}\frac{z}{k!\left(z-1\right)^{k}} & =\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k-1}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}} \\ & =z-1+\sum_{k\geq1}\frac{1}{k!\left(z-1\right)^{k-1}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}} \\ & =z-1+\sum_{k\geq0}\frac{1}{\left(k+1\right)!\left(z-1\right)^{k}}+\sum_{k\geq0}\frac{1}{k!\left(z-1\right)^{k}} \\ & =z-1+\sum_{k\geq0}\left(\frac{1}{\left(k+1\right)!}+\frac{1}{k!}\right)\frac{1}{\left(z-1\right)^{k}} \\ &= \color{red}{z-1+\sum_{k\geq0}\frac{k+2}{\left(k+1\right)!}\frac{1}{\left(z-1\right)^{k}}}. \end{align}