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Residue theorem: ∮z²/(3+4z+z²) on |z|=2
Short answer
Cauchy's residue theorem: integral equals 2πi times sum of residues of enclosed singularities. f(z)=z²/((3+4z+z²) has simple poles at z=−1 (inside |z|=2) and z=−3 (outside). Residue at −1 is 1/2, so integral = πi.
State Cauchy's Residue theorem and use it to evaluate ∮_C z²/(3 + 4z + z²) dz where C is |z| = 2.
Cauchy's residue theorem
If f(z) is analytic inside and on a simple closed curve C except for a finite number of isolated singularities a₁, a₂, …, aₙ inside C, then
∮_C f(z) dz = 2πi Σ Res[f, aₖ].
Evaluation
f(z) = z²/(z² + 4z + 3) = z²/((z + 1)(z + 3)), C: |z| = 2.
Poles: z = −1 (|−1| = 1 < 2, inside) and z = −3 (|−3| = 3 > 2, outside). Both are simple.
Residue at z = −1:
Res = lim (z → −1) (z + 1) f(z) = z²/(z + 3) at z = −1 = 1/2
Therefore:
Answer: ∮_C z²/(3 + 4z + z²) dz = 2πi × (1/2) = πi
✓ Answer checked by Claude Opus 5.5 (high effort)
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