# Find the Laurent Series expansion for $\displaystyle{f(z) = ze^{1/(z-1)}}$ valid for $\left|z-1\right|> 0$

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?

• Check when $e^z$ converges ($|z|<\infty$) and substitute it, and see it really works. Nov 10, 2020 at 8:53
• e^z converges because complex exponentials are differentiable. So what is next? Nov 10, 2020 at 9:56

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}
• (+1) but this can be simplified a small bit: $(k+1)k!=(k+1)!$