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Taylor series expansion

Short answer

Let w=z-i. 1/z = 1/(i+w) = -i/(1-iw) = -i Σ_{n=0}^∞ (iw)^n = -i Σ_{n=0}^∞ i^n (z-i)^n, converging for |z-i|<1.

Expand the function f(z) = 1/z about z = i in Taylor's series.

Let w = z − i, so z = i + w. The only singularity of 1/z is z = 0, at distance 1 from i. So the series converges for |z − i| < 1.

1/z = 1/(i + w) = (1/i) · 1/(1 + w/i)
    = −i · 1/(1 − i w)          [since 1/i = −i and w/i = −iw]
    = −i Σ_{n=0}^∞ (i w)ⁿ        for |w| < 1

Answer:

f(z) = −i Σ_{n=0}^∞ iⁿ (z − i)ⁿ = −i + (z − i) + i (z − i)² − (z − i)³ − ..., for |z − i| < 1.

Check: f'(i) = −1/i² = 1, which matches the coefficient of (z − i).

✓ Answer checked by Claude Opus 5.5 (high effort)

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