Use geometric series to find the Laurent series for f (z) = z / (z - 1)(z - 2) in each annulus
(a) Ann(1,0,1)
(b) Ann(1,1,∞)
Ann(a,r,R)
a= center, r=smaller radius, R=larger radius
Ann(1,0,1)=D(1,1)\{0}
My attempt:
f(z)= -1/(z-1) + 2/(z-2)
geometric series: Σ[n=0 to inf] z^n - 1/2...