Taylor series of 1/(1-z) about z = 3i

Taylor series about z = a: f(z) = Σ aₙ (z − a)ⁿ with aₙ = f⁽ⁿ⁾(a)/n!.

f(z) = 1/(1 − z) has its only singularity at z = 1. The distance from the centre 3i to 1 is |1 − 3i| = √10, so the series is valid for |z − 3i| < √10.

Expansion

Let t = z − 3i, so z = 3i + t and

1 − z = 1 − 3i − t = (1 − 3i)[1 − t/(1 − 3i)].

f(z) = 1/(1 − 3i) · 1/[1 − t/(1 − 3i)]
     = 1/(1 − 3i) · Σ_{n=0}^{∞} [t/(1 − 3i)]ⁿ        (|t| < |1 − 3i| = √10)
     = Σ_{n=0}^{∞} (z − 3i)ⁿ / (1 − 3i)ⁿ⁺¹

Answer: 1/(1 − z) = Σ_{n=0}^{∞} (z − 3i)ⁿ/(1 − 3i)ⁿ⁺¹, |z − 3i| < √10.

Since 1/(1 − 3i) = (1 + 3i)/10, the first terms are

f(z) = (1 + 3i)/10 + [(1 + 3i)/10]² (z − 3i) + [(1 + 3i)/10]³ (z − 3i)² + …

(Check: f(3i) = 1/(1 − 3i) = (1 + 3i)/10 ✓.)