Isolated pole; ∮(z-1)/((z+1)²(z-2))

Isolated pole

A point z = a is an isolated singularity of f(z) if f is analytic in a punctured neighbourhood 0 < |z − a| < δ but not at a. It is an isolated pole (of order m) if the Laurent series about a has only finitely many negative powers:

f(z) = b_m/(z − a)ᵐ + … + b₁/(z − a) + a₀ + a₁(z − a) + …, with b_m ≠ 0.

Equivalently f(z) = g(z)/(z − a)ᵐ with g analytic and g(a) ≠ 0.

Evaluation

f(z) = (z − 1)/((z + 1)²(z − 2)); C: |z − i| = 2 (centre i, radius 2).

Poles: z = −1 (double) and z = 2 (simple).

• Distance from i to −1 = |−1 − i| = √2 ≈ 1.41 < 2, so z = −1 is inside C.
• Distance from i to 2 = |2 − i| = √5 ≈ 2.24 > 2, so z = 2 is outside C.

Residue at the double pole z = −1:

Res = d/dz [ (z − 1)/(z − 2) ] at z = −1 = [ (z − 2) − (z − 1) ] / (z − 2)² = −1/(z − 2)² = −1/(−3)² = −1/9

By Cauchy's residue theorem, ∮_C f dz = 2πi × (sum of residues inside C):

Answer: ∮_C f(z) dz = 2πi(−1/9) = −2πi/9