Author Topic: Question 2 from 2.5  (Read 961 times)

Meerna Habeeb

  • Newbie
  • *
  • Posts: 4
  • Karma: 6
    • View Profile
Question 2 from 2.5
« on: November 12, 2018, 12:47:53 AM »
I am not so sure about my work, can someone help and correct me if I am wrong, or add if I have something missing I should be adding it :) Thanks in advance

Yexuan Zhang

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
Re: Question 2 from 2.5
« Reply #1 on: November 12, 2018, 01:57:47 AM »
Here is my answer.
Hope it can help. :D

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2550
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Question 2 from 2.5
« Reply #2 on: November 12, 2018, 03:15:47 AM »
Post scans, not crappy snapshots

Zhijian Zhu

  • Newbie
  • *
  • Posts: 3
  • Karma: 0
    • View Profile
Re: Question 2 from 2.5
« Reply #3 on: November 12, 2018, 10:37:05 AM »
Hi, I got the similar results as yours. I think we are good.
« Last Edit: November 12, 2018, 10:44:33 AM by Zhijian Zhu »

Huanglei Ln

  • Jr. Member
  • **
  • Posts: 8
  • Karma: 7
    • View Profile
Re: Question 2 from 2.5
« Reply #4 on: November 14, 2018, 03:12:53 PM »
Sin(z)=0
∴z=0  or  z=kπ
When z=0,
Numerator: f(z)=z^2  ,f(0)=0
            f'(z)=2z,f'(0)=0
          f^'' (z)=2z ,f^'' (0)≠0
∴order=2
Denominator: g(z)=sinz ,g(0)=0
            g^' (z)=cosz ,g^' (0)≠0
∴order=1
2-1=1
Order of zero=1


When  z=kπ   (k≠0)
Numerator: f(z)=z^2,f(kπ)=(kπ)^2≠0
∴order=0
Denominator: g(z)=sinz ,g(kπ)=0
            g^' (z)=cosz ,g^' (kπ)≠0
∴order=1
1-0=1
It is simple pole.

Huanglei Ln

  • Jr. Member
  • **
  • Posts: 8
  • Karma: 7
    • View Profile
Re: Question 2 from 2.5
« Reply #5 on: November 14, 2018, 03:18:38 PM »
the answer may like this