--- title: "Taylor series of (2z³+1)/(z²+z) about z = 1" subject: "Applied Mathematics" chapter: 1 key: "semester_4/applied_mathematics:taylor-series-2z3-about-1" source: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-2z3-about-1 site: BCT Engineering --- # Taylor series of (2z³+1)/(z²+z) about z = 1 Applied Mathematics · Chapter 1: Complex Analysis - Asked in: 2068 Magh, 5 marks - Video: [Taylor's Expansion of (2z³+1)/(z²+z) about z=1 | Complex Analysis || Complex Integration](https://www.youtube.com/watch?v=YYZks7QpY_s) — Swati Theng Mathematics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/taylor-series-2z3-about-1 - Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1 ## Question Find the Taylor's series expansion of f(z) = (2z³ + 1)/(z² + z) about the point z = 1. ## Short answer After partial fractions and expansion around z=1, f(z) = 3/2 + (3/4)(z−1) + (9/8)(z−1)² − (17/16)(z−1)³ + …, converging for |z−1| < 1. ## Answer **Taylor series** about z = 1: f(z) = Σ aₙ (z − 1)ⁿ. f(z) = (2z³ + 1)/(z² + z) = (2z³ + 1)/(z(z + 1)). Poles at z = 0 and z = −1; the nearest to z = 1 is z = 0, at distance 1. So the series converges for |z − 1| < 1. #### Step 1: divide (improper fraction) 2z³ + 1 = (z² + z)(2z − 2) + (2z + 1) so f(z) = 2z − 2 + (2z + 1)/(z(z + 1)). #### Step 2: partial fractions (2z + 1)/(z(z + 1)) = A/z + B/(z + 1): A = 1/1 = 1, B = (−2 + 1)/(−1) = 1. f(z) = 2z − 2 + 1/z + 1/(z + 1). #### Step 3: expand about z = 1 Let t = z − 1. Then 2z − 2 = 2t, z = 1 + t, z + 1 = 2 + t. ```text 1/z = 1/(1 + t) = Σ (−1)ⁿ tⁿ (|t| < 1) 1/(z + 1) = 1/(2 + t) = (1/2) · 1/(1 + t/2) = Σ (−1)ⁿ tⁿ / 2ⁿ⁺¹ (|t| < 2) ``` f(z) = 2t + Σ_{n=0}^{∞} (−1)ⁿ [ 1 + 1/2ⁿ⁺¹ ] tⁿ #### Result ```text a₀ = 1 + 1/2 = 3/2 a₁ = −(1 + 1/4) + 2 = 3/4 a₂ = 1 + 1/8 = 9/8 a₃ = −(1 + 1/16) = −17/16 ``` **f(z) = 3/2 + (3/4)(z − 1) + (9/8)(z − 1)² − (17/16)(z − 1)³ + …, |z − 1| < 1** General term: aₙ = (−1)ⁿ(1 + 2^{−(n+1)}) for n ≠ 1, and a₁ = 3/4. Check: f(1) = 3/2 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must we divide before using partial fractions on (2z³+1)/(z²+z)? A. The denominator has a repeated root B. The denominator is zero at the centre z=1 C. It is an improper fraction (numerator degree is higher than denominator) D. The numerator vanishes at z=0 Answer: C. Degree 3 over degree 2, so division gives 2z-2 plus a proper fraction. (Key point: First step) 2. What is the radius of convergence about z=1? A. 2, because the nearest pole is z = -1 B. 1, because the nearest pole is z = -1 C. Infinite, since the function is a rational function D. 1, because the nearest pole is z = 0 Answer: D. Poles are 0 and -1; z=0 is at distance 1 from z=1. (Key point: Radius) 3. What is f(z) after division and partial fractions? A. 2z - 2 + 1/z + 1/(z+1) B. 2z - 2 + 1/z - 1/(z+1) C. 2z + 2 + 1/z + 1/(z+1) D. 2z - 2 + 2/z + 1/(z+1) Answer: A. The remainder (2z+1)/(z(z+1)) has A = 1 and B = 1. (Key point: Decomposition) 4. Why is a_1 = 3/4 and not just the general formula value? A. Because 1/z has no t^1 term B. The polynomial part 2z-2 = 2t adds 2 to the t^1 coefficient: -(1+1/4) + 2 C. Because 1/(z+1) converges only for |t|<1 D. Because the series has a sign error at n=1 Answer: B. 2z-2 = 2t contributes an extra 2 at n=1, giving -5/4 + 2 = 3/4. 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