Evaluate ∮(4-3z)/(z(z-1)(z-2)) on |z|=3/2

Cauchy's integral formula: if f(z) is analytic inside and on a simple closed contour C, and a is a point inside C, then

f(a) = (1/2πi) ∮_C f(z)/(z − a) dz, that is, ∮_C f(z)/(z − a) dz = 2πi f(a).

Solution

C: |z| = 3/2. The poles are z = 0, 1, 2. Since |0| < 3/2 and |1| < 3/2, but |2| > 3/2, the poles z = 0 and z = 1 are inside C, and z = 2 is outside.

Partial fractions:

(4 − 3z)/(z(z − 1)(z − 2)) = A/z + B/(z − 1) + C/(z − 2)

A = (4 − 0)/((0 − 1)(0 − 2))   = 4/2  = 2
B = (4 − 3)/(1·(1 − 2))        = 1/(−1) = −1
C = (4 − 6)/(2·(2 − 1))        = −2/2  = −1

So the integrand is 2/z − 1/(z − 1) − 1/(z − 2).

Apply CIF with f(z) = 1 for the first two terms. The last term, −1/(z − 2), is analytic inside C, so its integral is 0 by Cauchy's theorem.

∮ 2/z dz        = 2 · 2πi · 1 = 4πi
∮ −1/(z − 1) dz = −2πi
∮ −1/(z − 2) dz = 0

I = 4πi − 2πi + 0 = 2πi

Answer: 2πi