A Question about computing residues after substitution

[Belgium 12]

Hi members,
See attacged PDF file for my question

Thank you

 

[SchroedingersLion]

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.

 

[Belgium 12]

Hi,
thank you very much for the answer
Belgium 12