Analytic Function II: Complex Calculation

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Discussion Overview

The discussion revolves around the properties of analytic functions, specifically focusing on the derivative of a polynomial function expressed in terms of its roots. Participants are exploring the relationship between the function and its derivative, as well as seeking assistance with a homework problem related to this topic.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant presents the function p(z) and requests to show that P'(z)/P(z) equals the sum of 1/(z-zj) for j from 1 to n, under the condition that z is not equal to any of the roots z1...zn.
  • Another participant questions the approach taken, suggesting that the thread is not meant for simply posting homework questions without effort.
  • Multiple participants express difficulty with the problem, indicating they have tried hard but are still stuck, and request help in a manner that emphasizes the need to show their attempts for better assistance.

Areas of Agreement / Disagreement

There is no clear consensus among participants. Some express frustration with the lack of effort shown in the initial post, while others are seeking help and indicate they have made attempts but are struggling.

Contextual Notes

Participants have not provided specific details about their attempts to solve the problem, which may limit the effectiveness of the assistance offered.

asqw121
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Let p(z)=A(z-z1)...(z-zn) where A and z1...zn are complex numbers and A not equal to 0 . Show that
P'(z)/P(z)=∑ (1/(z-zj)) z not equal z z1...zn
 
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Re: Analytic function

What have you tried? This isn't the place to dump homework and collect answers :(
 
Re: Analytic function III

Already try hard but I really stuck on this question
Please help
 
Re: Analytic function III

asqw121 said:
Already try hard but I really stuck on this question
Please help

If you show what you have tried, then our helpers can help, otherwise they are left to do the problem for you, which is of little help to you as far as you being able to work this problem and others like it yourself.
 

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