Taylor series of 1/(z²+4) about z = i

Taylor's series of f(z) about z = a (f analytic in |z − a| < R): f(z) = Σ aₙ(z − a)ⁿ, with aₙ = f⁽ⁿ⁾(a)/n!.

Here f(z) = 1/(z² + 4) has poles at z = ±2i. The nearest one to z = i is 2i, at distance 1. So the series is valid for |z − i| < 1.

Partial fractions

1/(z² + 4) = (1/4i) [1/(z − 2i) − 1/(z + 2i)]

Let t = z − i, so z = i + t.

• z − 2i = t − i = −i(1 + it), so 1/(z − 2i) = i · 1/(1 + it) = i Σ (−it)ⁿ = i Σ (−i)ⁿ tⁿ (valid for |t| < 1)
• z + 2i = t + 3i = 3i(1 − it/3), so 1/(z + 2i) = (−i/3) · 1/(1 − it/3) = (−i/3) Σ (i/3)ⁿ tⁿ (valid for |t| < 3)

Combine

f(z) = (1/4i) [ i Σ (−i)ⁿ tⁿ + (i/3) Σ (i/3)ⁿ tⁿ ]
     = (1/4) Σ (−i)ⁿ tⁿ + (1/12) Σ (i/3)ⁿ tⁿ

Answer:

f(z) = Σ_{n=0}^{∞} (1/4) [ (−i)ⁿ + iⁿ/3ⁿ⁺¹ ] (z − i)ⁿ, |z − i| < 1.

First few terms

a₀ = (1/4)(1 + 1/3)        = 1/3
a₁ = (1/4)(−i + i/9)       = −2i/9
a₂ = (1/4)(−1 − 1/27)      = −7/27
a₃ = (1/4)(i − i/81)       = 20i/81

f(z) = 1/3 − (2i/9)(z − i) − (7/27)(z − i)² + (20i/81)(z − i)³ + …

Check: f(i) = 1/(−1 + 4) = 1/3, and f′(i) = −2i/(3)² = −2i/9, which match a₀ and a₁.