Integral using cauchy theorem

1. Feb 9, 2014

Dassinia

Hello,
I don't get why using the fact that ∫dz/z = 2*pi*i accros the circle
This integral gives:

1/3∫(1/(z-2)-1/(z-1/2))dz = 1/3(-2*pi*i) accross the circle

Thanks !

2. Feb 9, 2014

Dick

Why do you think 1/z has anything to do with (1/3)*(1/(z-2)-1/(z-1/2))??