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Residue theorem: ∮sin z / z⁶ on |z|=1
Short answer
Cauchy’s residue theorem: ∮_C f(z)dz = 2πi Σ Res(f, z_k) inside C. For ∮ sin z/z⁶ dz over |z|=1, the residue at z=0 is 1/5! = 1/120. Hence the integral equals 2πi·(1/120) = πi/60.
State Cauchy's residue theorem. Using it evaluate ∮_C (sin z)/z⁶ dz where C: |z| = 1.
Cauchy's residue theorem
If f(z) is analytic inside and on a simple closed contour C except for a finite number of singular points z₁, z₂, …, zₖ inside C, then
∮_C f(z) dz = 2πi Σ Res[f(z), zₖ],
that is, 2πi times the sum of the residues at the singularities inside C.
Solution
f(z) = sin z / z⁶ has its only singularity at z = 0, which lies inside C: |z| = 1.
The residue is the coefficient of 1/z in the Laurent series about z = 0. Use
sin z = z − z³/3! + z⁵/5! − z⁷/7! + …
sin z / z⁶ = 1/z⁵ − 1/(3! z³) + 1/(5! z) − z/7! + …
The coefficient of 1/z is 1/5! = 1/120. So Res(f, 0) = 1/120.
(z = 0 is a pole of order 5 because sin z has a simple zero.)
By the residue theorem:
∮_C sin z / z⁶ dz = 2πi · (1/120) = πi/60
Answer: πi/60 ≈ 0.0524 i
✓ Answer checked by Claude Opus 5.5 (high effort)
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