Mathematics Asked by Ariana Tibor on March 1, 2021

Can you help me and discuss me on the question?

Expand $displaystyle{f(z) = ze^{1/(z-1)}}$ in a Laurent series valid for $displaystyle{left|z-1right|> 0}$.

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 $$e^{w}=sum_{kgeq0}frac{w^{k}}{k!},,left|wright|<+infty$$ we have, taking $w=frac{1}{z-1}$, that for $0<left|z-1right|<+infty$ (these bounds follow from $left|wright|=left|frac{1}{z-1}right|<+infty$) we obtain

begin{align} ze^{frac{1}{z-1}}=sum_{kgeq0}frac{z}{k!left(z-1right)^{k}} & =sum_{kgeq0}frac{1}{k!left(z-1right)^{k-1}}+sum_{kgeq0}frac{1}{k!left(z-1right)^{k}} \ & =z-1+sum_{kgeq1}frac{1}{k!left(z-1right)^{k-1}}+sum_{kgeq0}frac{1}{k!left(z-1right)^{k}} \ & =z-1+sum_{kgeq0}frac{1}{left(k+1right)!left(z-1right)^{k}}+sum_{kgeq0}frac{1}{k!left(z-1right)^{k}} \ & =z-1+sum_{kgeq0}left(frac{1}{left(k+1right)!}+frac{1}{k!}right)frac{1}{left(z-1right)^{k}} \ &= color{red}{z-1+sum_{kgeq0}frac{k+2}{left(k+1right)!}frac{1}{left(z-1right)^{k}}}. end{align}

Answered by Marco Cantarini on March 1, 2021

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