Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz6 => Topic started by: Victor Ivrii on November 17, 2018, 04:14:52 PM

Locate each of the isolated singularities of the given function $f(z)$ and tell whether it is a removable singularity, a pole, or an essential singularity.
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:
$$
f(z) =\pi \cot(\pi z).
$$

My solution

To begin, $f(z) = \pi \text{cot}(\pi z) = \pi \frac{\text{cos}(\pi z)}{\text{sin}(\pi z)}$
Let the numerator be denoted by $h(z) = \pi \text{cos}(\pi z)$,
and the denominator be $g(z) = \text{sin}(\pi z)$
Then the denominator $g(z) = \text{sin}(\pi z) = 0 \implies z = k, k \in \mathbb{Z}$
Now, $h(k) = \pi \text{cos}(k\pi) = \pi \text{ or } \pi \neq 0 \implies \text{order = 0}$
$g(k) = \text{sin}(k\pi) = 0$
$g'(z)= \pi \text{cos}(\pi z) \implies g'(k) = \pi \text{cos}(k\pi) = \pi \text{ or } \pi \neq 0 \implies \text{order = 1}$
$\therefore$ this function has a pole singularity with order of pole = 1  0 = 1

Why 2k? It should just be k.
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \pi \ cot\ ( \pi z)\\
\\
=\frac{\pi cos( \pi z)}{sin( \pi z)}
\end{array}$
So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.

Why 2k? It should just be k.
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \pi \ cot\ ( \pi z)\\
\\
=\frac{\pi cos( \pi z)}{sin( \pi z)}
\end{array}$
So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.
You are right :( , it was a rudimentary mistake. I have corrected my answer :) thanks

No problem! :)

Nikita, your solution is unreadable