- 2069 Poush
Laurent series of (1-cos z)/z³
Short answer
Using the Maclaurin series of cos z, f(z) = 1/(2z) − z/24 + z³/720 − …, valid for 0<|z|<∞. The principal part is 1/(2z), so z=0 is a simple pole and the residue is 1/2.
Find the Laurent's series expansion of f(z) = (1 - cos z)/z³, 0 < |z| < R.
Use the Maclaurin series of cos z, which converges for all z:
cos z = 1 − z²/2! + z⁴/4! − z⁶/6! + …
so
1 − cos z = z²/2! − z⁴/4! + z⁶/6! − z⁸/8! + …
Divide by z³:
f(z) = (1 − cos z)/z³ = 1/(2z) − z/24 + z³/720 − z⁵/40320 + …
Answer:
f(z) = Σ (n = 1 to ∞) (−1)ⁿ⁺¹ z^(2n − 3) / (2n)!
valid for 0 < |z| < ∞.
- Principal part: 1/(2z) only, so z = 0 is a simple pole.
- Residue at z = 0 is the coefficient of 1/z, which is 1/2.
✓ Answer checked by Claude Opus 5.5 (high effort)
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