--- title: "Laurent series of sin z / z⁶" subject: "Applied Mathematics" chapter: 1 key: "semester_4/applied_mathematics:laurent-sinz-over-z6" source: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-sinz-over-z6 site: BCT Engineering --- # Laurent series of sin z / z⁶ Applied Mathematics · Chapter 1: Complex Analysis - Asked in: 2074 Bhadra, 1+4 marks - Video: [Laurent Series of A Function f(z)=(z-sinz)/z^3|| Example 2 || Complex Analysis MAT324](https://www.youtube.com/watch?v=UztyBygjgoU) — Right Resource Institute - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-sinz-over-z6 - Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1 ## Question 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. ## 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. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement correctly distinguishes Laurent from Taylor series? A. Laurent has only non-negative powers B. Taylor is valid in an annulus, Laurent in a disc C. Laurent has negative powers too and is valid in an annulus where f may be singular at the centre D. Taylor has a principal part, Laurent does not Answer: C. Taylor: only non-negative powers, disc; Laurent: both signs, ring. (Key point: Taylor vs Laurent) 2. How is the Laurent series of sin z / z⁶ obtained? A. Use partial fractions B. Apply the residue formula only C. Expand in powers of (z-1) D. Divide the Maclaurin series of sin z by z⁶ term by term Answer: D. sin z is entire; dividing its Maclaurin series by z⁶ gives the series. (Key point: Method) 3. What is z = 0 and what is the residue? A. Pole of order 5; residue 1/120 B. Pole of order 6; residue 1/120 C. Pole of order 5; residue -1/6 D. Removable singularity; residue 0 Answer: A. Lowest power is z^-5 (order 5), and the 1/z coefficient is 1/5! = 1/120. (Key point: Nature of z=0) 4. For which z is the series valid? A. |z| < 1 B. 0 < |z| < ∞ C. 1 < |z| < ∞ D. |z| = 0 only Answer: B. sin z is entire so the only singularity is z=0; R = ∞. 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