Question about computing residues after substitution

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  • Thread starter Belgium 12
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Hi members,
See attacged PDF file for my question

Thank you
 

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  • #2
The first option is correct. Your substitution function gives a residuum of 1/2.
Now, you need to transform it to the residuum of the original function.

Your function (around the pol z=1) can be written as a Laurent series: ##f(z)=\sum_{n=-1}^\infty a_n (z-1)^n ##. Note that all terms with n<-1 vanish because z=1 is a pol of first order. The residuum is defined as the first coefficient of the Laurent-series with negative index, i.e. a-1.

Now, you found the residuum of the original function to be 1, i.e. a-1=1.
If you do your substitution in the Laurentseries, namely (z-1)=2u, you will be left with a new Laurent series of a different function: ##g(u)=\sum_{n=-1}^\infty b_n (u-0)^n##, where ##b_n=a_n*2^n##. If you calculate the residuum of g in the usual way, you will get b-1=1/2 (which you got).
Now you just have to calculate a-1 from that.

To explain your error: The residuum of the substituted function is not exactly the same as the residuum of the original function.
 
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Likes jim mcnamara and Belgium 12
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Hi,
thank you very much for the answer
Belgium 12
 

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