Author Topic: Spring 2020 Test 2 Monday Sitting Problem 3  (Read 4358 times)

Xinxuan Lin

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
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

  • Jr. Member
  • **
  • Posts: 11
  • Karma: 0
    • View Profile
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

  • Full Member
  • ***
  • Posts: 18
  • Karma: 0
    • View Profile
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