Author Topic: Spring 2020 Test 2 Monday Sitting Problem 3  (Read 481 times)

Xinxuan Lin

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Spring 2020 Test 2 Monday Sitting Problem 3
« on: December 01, 2020, 01:42:57 PM »
f(z) = (z$^4$ -$\pi^4$)tan$^2$($\frac{z}{2}$)

Part b of this question is asking to determine the types of the singular point.

In solution, it says z=2n$\pi$ with n$\neq$ $\pm$ 1 are double zeros; z=(2n+1)$\pi$ with n$\neq$ $\pm$ 1 are double poles.

Could anyone explain why n$\neq$ $\pm$ 1 here? Why is not n$\neq$ -1, 0?

Thanks in advanced!


Jiaqi Bi

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Re: Spring 2020 Test 2 Monday Sitting Problem 3
« Reply #1 on: December 02, 2020, 03:34:35 PM »
What I got on these poles were the same as you did $n \neq 0\ \& -1$. I believe that was a typo.

Xuefen luo

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Re: Spring 2020 Test 2 Monday Sitting Problem 3
« Reply #2 on: December 03, 2020, 03:28:54 AM »
I got that there is no restriction on the n when z=2n𝜋, and n ≠ 0 & -1 when  z=(2n+1)𝜋