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\textbf{Problem: }\\\\
\text{Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity.} \\\\ \text{If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole}\\\\
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\textbf{Solution:}
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\text{Let } \\\\
\begin{gather}
\begin{aligned}
f(z) &= \frac{e^{z}-1}{e^{2z}-1} \\\\
&= \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)}
\end{aligned}
\end{gather}
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\text{Then it shows two singularities: }
\begin{aligned}
e^{z} = -1 ,\ e^{z} = 1
\end{aligned}
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\text{Therefore we have} \\\\
\begin{gather}
\begin{aligned}
z_1&=log(1)\\\\
&=ln|1|+iarg(1)\\\\
&= 2k\pi i, k\in \mathbb{Z}\\\\
z_2&=log(-1)\\\\
&=ln|-1|+iarg(-1)\\\\
&=(2k+1)\pi i, k\in \mathbb{Z}
\end{aligned}
\end{gather}
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\text{When } z = z_1 \\\\
\begin{gather}
\begin{aligned}
\lim_{z \to z_1} \frac{e^{z}-1}{e^{2z}-1} &= \lim_{z \to 2k\pi i} \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)} \\\\
&= \lim_{z \to 2k\pi i} \frac{1}{e^{z}+1}\\\\
&= \frac{1}{2}
\end{aligned}
\end{gather}
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$\text{Therefore } z_1 \text{ is a removable singularity with value } \frac{1}{2}$
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\text{When } z = z_2 \\\\
\begin{gather}
\begin{aligned}
\lim_{z \to z_2} \frac{e^{z}-1}{e^{2z}-1} &= \lim_{z \to (2k+1)\pi i} \frac{e^{z}-1}{(e^{z}-1)(e^{z}+1)} \\\\
&= \lim_{z \to (2k+1)\pi i} \frac{1}{e^{z}+1}\\\\
&= \infty
\end{aligned}
\end{gather}
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$\text{Hence } z_2 \text{ is a pole.}$
$\text{Since the numerator has an order of 0 and the denominator has an order of 1}$
$\text{Therefore } z_2 \text{ is a pole of order 1.}$
