# Question about computing residues after substitution

• A
Hi members,
See attacged PDF file for my question

Thank you

#### Attachments

• 18-09-2018 15;34;27.PDF
135.5 KB · Views: 112

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.

jim mcnamara and Belgium 12
Hi,
thank you very much for the answer
Belgium 12