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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?

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

1 Answer 1

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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}

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    $\begingroup$ (+1) but this can be simplified a small bit: $(k+1)k!=(k+1)!$ $\endgroup$
    – robjohn
    Nov 10, 2020 at 15:12
  • $\begingroup$ @robjohn Yes, I am an idiot :D thank you $\endgroup$ Nov 10, 2020 at 15:24
  • $\begingroup$ Thank you so much all. $\endgroup$ Nov 10, 2020 at 15:50
  • $\begingroup$ @ArianaTibor You're welcome. I recall to you that, if you think that my answer solves your problem, you can accept it. $\endgroup$ Nov 11, 2020 at 7:26
  • $\begingroup$ Hi, I'm still confuse and can't figure out how we arrive the expression in red from the previous one. Sorry $\endgroup$ Nov 11, 2020 at 15:49

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