Determine the nature of an isolated singularity

Question

Using the Laurent series, determine the nature of the isolated singularity $z = \pi$ for the function

$$f(z) = \frac{1-e^{-z}}{\sin{z}}$$

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$$e^{-z} = \sum_{n=0}^\infty f_1^{(n)}(\pi)\frac{(z-\pi)^n}{n!} = e^{-\pi} - e^{-\pi}(z-\pi) + e^{-\pi}\frac12(z-\pi)^2 - e^{-\pi}\frac16(z-\pi)^3 + \ldots$$

$$\sin{z} = \sum_{n=0}^\infty f_2^{(n)}(\pi)\frac{(z-\pi)^n}{n!}=-(z-\pi)+\frac16(z-\pi)^3 -\frac1{120}(z-\pi)^5+\ldots $$

So we may write

\begin{align} f(z) &= \frac{1-e^{-\pi} + e^{-\pi}(z-\pi) - e^{-\pi}\frac12(z-\pi)^2 + e^{-\pi}\frac16(z-\pi)^3 - \ldots}{-(z-\pi)+\frac16(z-\pi)^3 -\frac1{120}(z-\pi)^5+\ldots} \\ &= \frac{1}{z-\pi} \frac{1-e^{-\pi} + e^{-\pi}(z-\pi) - e^{-\pi}\frac12(z-\pi)^2 + e^{-\pi}\frac16(z-\pi)^3 - \ldots}{-1+\frac16(z-\pi)^2 -\frac1{120}(z-\pi)^4+\ldots} \\ &\to \frac{1}{z-\pi} (e^{-\pi}-1) \quad \text{as $z\to\pi$.} \end{align}

Hence $z=\pi$ is a simple pole of $f$.

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My question is: Does this look correct? Is there an easier way to do this (less computationally involved) still using the Laurent Series method?

Answer 1

$$\frac{1-e^{-z}}{\sin z}=\frac{1-e^{-\pi} e^{-(z-\pi)}}{(z-\pi)\left(-1+\frac{(z-\pi)^2}6+\ldots\right)}=$$

$$=\frac{1-e^{-\pi}(1-(z-\pi)+\frac{(z-\pi)^2}2-\ldots}{-(z-\pi)}\left(1+\frac{(z-\pi)^2}6+\ldots\right)=$$

$$=\frac{1-e^{-\pi}+e^{-\pi}(z-\pi)-\ldots}{-(z-\pi)}\left(1+\frac{(z-\pi)^2}6+\ldots\right)=$$

$$=\frac{1-e^{-\pi}}{-(z-\pi)}+e^{-\pi}+\ldots$$

Thus, $\;z=\pi\;$ is a simple pole with residue $\;-(1-e^{-\pi})=e^{-\pi}-1\;$.

Another way :

$$\lim_{z\to\pi}(z-\pi)\frac{1-e^{-z}}{\sin z}\stackrel{\text{l'Hospital}}=\lim_{z\to\pi}\frac{1-e^{-z}+(z-\pi)e^{-z}}{\cos z}=e^{-\pi}-1$$