Can anyone offer advice on how one might use the residue of gamma to show the integral gamma (z) dz=-4piei/3 where C=(z:z[Z+2-i]=2).......or am I on the right track at all?
Two questions...
a) by 'gamma function' do you mean...
$\displaystyle \Gamma (z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt$
... or some else?...
c) which is the path in the complex plane along which you intend to integrate the 'gamma function'?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
It says here that the residue of the gamma function at the simple pole z=–n is $\displaystyle (-1)^n/n!\,.$ If C is meant to be the circle of radius 2 centred at –2+i then the poles inside the circle are at –1, –2 and –3, and $\displaystyle 2\pi i$ times the sum of the residues given by that formula is indeed $\displaystyle -4\pi i/3$.