Taylor series of (2z³+1)/(z²+z) about z = 1

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.

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

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 ✓.