Author Topic: Q6 TUT 5301  (Read 610 times)

Victor Ivrii

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Q6 TUT 5301
« 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:
$$
f(z) =\pi \cot(\pi z).
$$

Nikita Dua

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Re: Q6 TUT 5301
« Reply #1 on: November 17, 2018, 04:31:56 PM »
My solution

Xinyan Jiang

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Re: Q6 TUT 5301
« Reply #2 on: November 17, 2018, 05:19:25 PM »
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
« Last Edit: November 17, 2018, 05:32:10 PM by Xinyan Jiang »

Jeffery Mcbride

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Re: Q6 TUT 5301
« Reply #3 on: November 17, 2018, 05:24:26 PM »
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.

Xinyan Jiang

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Re: Q6 TUT 5301
« Reply #4 on: November 17, 2018, 05:34:28 PM »
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

Jeffery Mcbride

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Re: Q6 TUT 5301
« Reply #5 on: November 17, 2018, 05:39:12 PM »
No problem!  :)

Victor Ivrii

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Re: Q6 TUT 5301
« Reply #6 on: November 28, 2018, 04:40:47 AM »
Nikita, your solution is unreadable