# Toronto Math Forum

## MAT334--2020F => MAT334--Tests and Quizzes => Quiz 5 => Topic started by: RunboZhang on November 05, 2020, 07:27:03 PM

Title: LEC5101-Quiz5-E
Post by: RunboZhang on November 05, 2020, 07:27:03 PM
$\textbf{Problem: }\\\\ \text{Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity.} \\\\ \text{If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole}\\\\$

$\textbf{Solution:}$

\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}

\text{Then it shows two singularities: } \begin{aligned} e^{z} = -1 ,\ e^{z} = 1 \end{aligned}

\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}

\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}

$\text{Therefore } z_1 \text{ is a removable singularity with value } \frac{1}{2}$

\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}

$\text{Hence } z_2 \text{ is a pole.}$

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

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