### Author Topic: Question 2 from 2.5  (Read 2869 times)

#### Meerna Habeeb

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• Posts: 4
• Karma: 6
##### 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

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

#### Victor Ivrii

• Administrator
• Elder Member
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##### Re: Question 2 from 2.5
« Reply #2 on: November 12, 2018, 03:15:47 AM »
Post scans, not crappy snapshots

#### Zhijian Zhu

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• Posts: 3
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##### 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
##### 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
##### Re: Question 2 from 2.5
« Reply #5 on: November 14, 2018, 03:18:38 PM »
the answer may like this