Given the function $$f(z) = z^2 \cdot \frac{{e^{1/z}}}{{z-1}}$$, we want to find its Laurent series expansion around the point $z = 0$. This is my approach. I am not sure about the ending. First, let's consider the function $$g(z) = \frac{{e^{1/z}}}{{z-1}}.$$ We can expand $g(z)$ in a Taylor series as follows:

$$ g(z) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+k)!}}\right) \left(\frac{1}{z}\right)^n $$

Now, let's multiply $g(z)$ by $z^2$:

$$ z^2 \cdot g(z) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+k)!}}\right) z^{2-n} $$

To obtain the Laurent series, we need negative powers of $z$. Let's rewrite the series in a more convenient form:

$$ z^2 \cdot g(z) = \sum_{n=-\infty}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+1+k)!}}\right) z^n $$

Therefore, the Laurent series expansion of $$f(z) = z^2 \cdot \frac{{e^{1/z}}}{{z-1}}$$ around $z = 0$ is:

$$ f(z) = \sum_{n=-\infty}^{\infty} \left(\sum_{k=0}^{\infty} \frac{1}{{(n+1+k)!}}\right) z^n $$


1 Answer 1


Your very first Laurent expansion is incorrect. I don't know where: $$g(z)=\sum_{n\ge0}z^{-n}\cdot\sum_{k\ge0}\frac{1}{(k+n)!}$$Came from. Let's do this more carefully. We know that, for $|z|<1$: $$g(z)=-\exp(z^{-1})(1-z)^{-1}=-\exp(z^{-1})\sum_{m\ge0}z^m=-\left(\sum_{m\ge0}\frac{1}{m!}z^{-m}\right)\left(\sum_{m\ge0}z^m\right)$$Now, we expand the product in series: $$\begin{align}g(z)&=-\sum_{n\in\Bbb Z}z^n\cdot\sum_{k\ge\max(0,n)}\frac{1}{(k-n)!}\\&=-\sum_{n\ge0}z^ne-\sum_{n<0}z^n\sum_{k\ge-n}\frac{1}{k!}\\&=\frac{e}{z-1}-\sum_{n\ge1}z^{-n}\sum_{m\ge n}\frac{1}{m!}\end{align}$$This is correct because each term $z^n$ can be reached as $z^m\cdot z^{n-m}$ where both $m,m-n$ are nonnegative for exactly all $m\ge\max(0,n)$.

This series representation is correct for all $z,|z|<1$.

The final Laurent series for $f$, valid in the annulus: $$A=\{z\in\Bbb C:|z|<1,\,z\neq0\}$$

Is: $$f(z)=(2-e)+(1-e)z-e\sum_{n\ge2}z^n-\sum_{n\ge1}z^{-n}\left(\sum_{m\ge0}\frac{1}{(m+n+2)!}\right)$$

The residue of $f$ at zero is equal to $\frac{5}{2}-e$.


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