- 2074 Bhadra · 1+4 marks
Laurent series of sin z / z⁶
Short answer
Taylor series has only non-negative powers; Laurent series includes negative powers and holds in an annulus. For f(z) = sin z / z⁶, the Laurent series is Σ_{n=0}^∞ (−1)ⁿ z^{2n−5}/(2n+1)! for 0<|z|<∞. The pole at z=0 is of order 5, residue 1/120.
Point out difference between Taylor's series and Laurent's series. Find Laurent's series of function f(z) = sin z / z⁶, 0 < |z| < R.
Taylor's series vs Laurent's series
| Point | Taylor's series | Laurent's series |
|---|---|---|
| Form | Only non-negative powers of (z − a) | Positive and negative powers of (z − a) |
| Requirement | f(z) analytic everywhere inside a circle around a | f(z) analytic only in an annulus r < |z − a| < R; may be singular at a |
| Region | Disc |z − a| < R | Ring r < |z − a| < R |
| Coefficients | aₙ = f⁽ⁿ⁾(a)/n! | aₙ = (1/2πi) ∮ f(w)/(w − a)ⁿ⁺¹ dw, n = 0, ±1, ±2, … |
| Principal part | None | Present if a is a singularity |
| Use | Expansion at ordinary points | Expansion near singularities; gives residue (coefficient of 1/(z − a)) |
Laurent series of f(z) = sin z / z⁶
sin z is entire and has the Maclaurin series
sin z = z − z³/3! + z⁵/5! − z⁷/7! + …
Divide by z⁶:
f(z) = 1/z⁵ − 1/(3! z³) + 1/(5! z) − z/7! + z³/9! − …
Answer:
f(z) = 1/z⁵ − 1/(6z³) + 1/(120 z) − z/5040 + z³/362880 − … = Σ (n = 0 to ∞) (−1)ⁿ z^(2n − 5) / (2n + 1)!
This is valid for 0 < |z| < ∞ (R = ∞). The highest negative power in the principal part is z⁻⁵, so z = 0 is a pole of order 5, and the residue (coefficient of 1/z) is 1/120.
✓ Answer checked by Claude Opus 5.5 (high effort)
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