Lesson 6 · deriving (not asserting) that the radius of convergence of f(z) = 1/(z − 3) at z0 = 0 equals the distance to the nearest singularity
The z-plane f(z) = 1/(z−3), z₀ = 0
r′ = — · z = — · |z| = —
—
Half 1 — R ≥ 3Neither cn = f(n)(0)/n! nor the Cauchy integral (1/2πi)∮γr′f(w)/wn+1dw mentions r′ on the right — it's one fixed series no matter which admissible circle certified it. Any z with |z| < 3 gets certified by some r′ strictly between |z| and 3, so the series converges on all of |z| < 3.cn and the Cauchy integral never mention r′ — one fixed series, any admissible circle. Every |z| < 3 has one, so the series converges there.
Blow-up as z → 3 real axis
3
2
—= |f(z)| = 1/(3−z)
z = — · distance to 3 = —
Half 2 — R ≤ 3If R were > 3, the series (= f wherever it converges) would have to stay bounded on a disk reaching past z = 3. It doesn't: |f(z)| has no finite ceiling as z → 3. So R can't exceed 3. General relation for real z < 3: |f(z)| = 1/|z−3| = 1/(3−z).If R > 3, f would stay bounded past z = 3. It doesn't — |f(z)| → ∞. So R ≤ 3.