What is holomorphic function?

Example 16.3

Polynomials in $z$, rational functions off their poles, and — by Year 2’s term-by-term differentiation theorem for power series, whose proof works identically over $\C$ — every sum of a power series $\sum a_n(z - a)^n$ inside its disc of convergence: holomorphic, with derivative $\sum na_n(z - a)^{n-1}$ (same radius). In particular $\exp z = \sum z^n/n!$ is entire (holomorphic on $\C$) with $\exp' = \exp$. On the other hand $z \mapsto \bar z$, $\abs z$, $\operatorname{Re}z$ are nowhere holomorphic (Cauchy–Riemann fails everywhere): holomorphy is orientation-and-angle-preserving rigidity, not smoothness.

Example 16.11 (Singularities dictate radii)

Why does the innocent real function $\frac1{1 + x^2}$ have a Taylor series at $x = 3$ converging only for $\abs{x - 3} < \sqrt{10}$, when nothing goes wrong on the real line? Because the theorem above makes the radius of convergence at $a$ equal to the distance from $a$ to the nearest point where holomorphy fails. Here $f(z) = \frac1{1 + z^2}$ is holomorphic exactly on $\C\setminus\{\pm\iu\}$, so the expansion at $a = 3$ converges on the largest disc avoiding $\pm\iu$, of radius $\abs{3 - \iu} = \sqrt{10}$ — and cannot converge on a larger one, since the sum would extend $f$ holomorphically to a neighborhood of $\pm\iu$, where $\abs f \to \infty$. The real theory sees the mysterious radius $\sqrt{10}$; the complex plane sees two poles. This is the practical rule: to find a radius of convergence, locate the singularities — e.g. the Taylor series of $\tan$ at $0$ has radius $\frac\pi2$ (nearest zeros of $\cos$), and the Bernoulli generating function $\frac z{\eu^z - 1}$ (Problem 16.1, Part VI) has radius $2\pi$ (nearest nonzero zeros of $\eu^z - 1$: $\pm2\iu\pi$).