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SchroedingersLion

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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

Now, you found the residuum of the original function to be 1, i.e. a

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

Now you just have to calculate a

To explain your error: The residuum of the substituted function is not exactly the same as the residuum of the original function.

- #3

Belgium 12

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Hi,

thank you very much for the answer

Belgium 12

thank you very much for the answer

Belgium 12

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