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Find the first three nonzero terms of the Laurent series of $f(z)=\frac{1}{z^2(e^{z}-e^{-z})}$ in the annulus $0 \leq \lvert z \rvert \leq \pi$.

Now, instead of $f(z)$ I found the Taylor series of its reciprocal. Then I tried to multiply it by the Laurent series of $f(z)$ and put it equal to one and then comparing the coefficients. My problem is that since zero is the essential singularity of $f(z)$, we would not have first terms in this case. The reason I think zero is the essential singularity is that we cannot remove zero if we multiply $f(z)$ by $z^n$ for some $n$. Maybe my definition of pole is not correct.

P.S. Dester
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Tim
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1 Answers1

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It is sufficient to derive three non-zero terms from the expansion of the denominator of $f(z)$ at $z=0$ and then perform a geometric series expansion. Since \begin{align*} e^z-e^{-z}&=\sum_{n=0}^{\infty}\frac{z^n}{n!}-\sum_{n=0}^{\infty}(-1)^n\frac{z^n}{n!}\\ &=2\sum_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}\tag{1} \end{align*} we obtain from (1)

\begin{align*} \color{blue}{f(z)}&=\frac{1}{z^2\left(e^z-e^{-z}\right)}\\ &=\frac{1}{2z^3\left(1+\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots\right)}\tag{2}\\ &=\frac{1}{2z^3}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^2}{3!}+\frac{z^4}{5!}+\cdots\right)^n\tag{3}\\ &=\frac{1}{2z^3}\left(1-\left(\frac{z^2}{3!}+\frac{z^4}{5!}\right)+\left(\frac{z^4}{3!3!}\right)+\cdots\right)\tag{4}\\ &\,\,\color{blue}{=\frac{1}{2z^3}-\frac{1}{12z}+\frac{7z}{720}+\cdots}\tag{5} \end{align*} and the first three non-zero terms are given in (5).

Comment:

  • In (2) we take the first three non-zero terms and do not consider higher order terms.

  • In (3) we make a geometric series expansion.

  • In (4) we observe it is sufficient to develop terms up to $n=2$ and we put all terms of $x$ with higher order than $4$ to $\cdots$.

  • In (5) we simplify and see the first three non-zero terms.

Markus Scheuer
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