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Taylor series expansion

Short answer

Let w=z-4i. 1/(z-3)=1/(w+(4i-3)) = -∑_{n=0}^∞ (z-4i)^n/(3-4i)^(n+1), valid for |z-4i|<5 (distance to the pole at z=3).

Find the Taylor series expansion of f(z) = 1/(z − 3) centered at the point z = 4i.

Let w = z − 4i. Then z − 3 = w + (4i − 3). Put a = 4i − 3, so |a| = 5.

f(z) = 1/(z − 3) = 1/(a + w) = (1/a) · 1/(1 + w/a)
     = (1/a) Σ_{n=0}^∞ (−w/a)ⁿ         for |w| < |a| = 5
     = Σ (−1)ⁿ (z − 4i)ⁿ / (4i − 3)^(n+1)

Since (4i − 3)^(n+1) = (−1)^(n+1) (3 − 4i)^(n+1), this simplifies to

f(z) = − Σ_{n=0}^∞ (z − 4i)ⁿ / (3 − 4i)^(n+1)
     = −(3 + 4i)/25 − (3 + 4i)²/625 · (z − 4i) − ...

Answer: f(z) = − Σ (z − 4i)ⁿ/(3 − 4i)^(n+1), valid for |z − 4i| < 5 (the distance from 4i to the singularity z = 3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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