Author Topic: LEC5101-Quiz5-E  (Read 394 times)

RunboZhang

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LEC5101-Quiz5-E
« on: November 05, 2020, 07:27:03 PM »
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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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$\text{Hence } z_2 \text{ is a pole.}$

$\text{Since the numerator has an order of 0 and the denominator has an order of 1}$

$\text{Therefore } z_2 \text{ is a pole of order 1.}$