Originally Posted by

**WannaBe** 1. How can I calculate the residue in z=1 of the next function:

$\displaystyle f(z)=(2z-1)e^{\frac{z}{z-1}} $

$\displaystyle (2z-1)e^{z/(z-1)}=(2z-1)e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}$

$\displaystyle =2e\sum_{n=0}^{\infty}\frac{(z-1)^{1-n}}{n!}+e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}$

and the coefficient on the 1/(z-1) term is:

$\displaystyle 2e(1/2)+e=2e$

which is consistent with the numerical result:

Code:

In[13]:=
NIntegrate[(2*z - 1)*Exp[z/(z - 1)]*(1/4)*
I*Exp[I*t] /.
z -> 1 + (1/4)*Exp[I*t], {t, 0, 2*Pi}]
N[4*Pi*I*E]
Out[13]=
7.105427357601002*^-15 +
34.15893689068571*I
Out[14]=
0. + 34.158936890694264*I