Evaluate ∮z/((z-1)(z-3)) on |z|=3/2

Cauchy's integral formula: f(a) = (1/2πi) ∮_C f(z)/(z − a) dz for a inside C, with f analytic inside and on C.

Solution

C: |z| = 3/2. The pole z = 1 lies inside (1 < 3/2) and the pole z = 3 lies outside (3 > 3/2).

Write the integrand as

z/((z − 1)(z − 3)) = [ z/(z − 3) ] · 1/(z − 1).

Let f(z) = z/(z − 3). It is analytic inside and on C, since its only singularity z = 3 is outside C.

By CIF with a = 1:

∮_C z dz/((z − 1)(z − 3)) = 2πi f(1) = 2πi · 1/(1 − 3) = 2πi · (−1/2) = −πi

Answer: −πi