I believe the answer to the second part of your question is 1/3....
with for k=0,1,2.....,n-1
Basically just sum over the residues evaluated at the n roots of 1/3...
Hope this helps...
1. How can I calculate the residue in z=1 of the next function:
??
I've tried using Laurent expansion but couldn't figure out how to expand this function into a series (using the known expansion of the exponent don't help us here... ).
2. Calculate the next Integral:
.
My problem with this question is that we have n singularities and they are all inside our contour... I've tried to paramterize but without any success..
Hope you'll be able to help me
Thanks in advance!
and the coefficient on the 1/(z-1) term is:
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