Analytic Functions — Mathematics Knowledge System

Overview

A complex function f is analytic (equivalently, holomorphic) at a point z0 if it is complex-differentiable in some open neighborhood of z0, not merely at z0 itself; f is analytic on an open set if it is analytic at every point of that set. This single requirement is dramatically stronger than real differentiability: an analytic function is automatically infinitely differentiable and equals its own Taylor series in a disk around each point of analyticity. Analyticity is equivalent to satisfying the Cauchy–Riemann equations with continuous partials throughout a neighborhood, and it is the central hypothesis underlying essentially every deep theorem of complex analysis (Cauchy's theorem, the identity theorem, Liouville's theorem).

Intuition

Requiring a derivative to exist at just one complex point is nearly meaningless — the real and imaginary parts can still be badly behaved nearby. But requiring differentiability throughout a whole neighborhood is a much stronger, almost rigid demand: it forces the function to look like a rotation-and-scaling map at every nearby point simultaneously, and this local consistency propagates outward, forcing f to be smooth to every order and to be perfectly captured by its own power series. That is why 'differentiable at a point' in real calculus and 'analytic at a point' in complex analysis are such different beasts — the complex version bakes in an entire neighborhood's worth of structure.

Formal Definition

Definition

A function f: U → ℂ on an open set U ⊆ ℂ is analytic (holomorphic) on U if the complex derivative exists at every point of U:

f'(z_0) = \lim_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h} \text{ exists for every } z_0 \in U

Complex differentiability on U

f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!}

Local power series (Taylor) expansion

f \text{ analytic} \iff u,v \text{ satisfy Cauchy--Riemann with continuous partials on } U

Equivalent CR characterization

Worked Examples

f(z)=z² is a polynomial, so its complex derivative f'(z)=2z exists for every z in ℂ, and this holds on an open neighborhood of every point.
$f'(z) = 2z \text{ exists for all } z \in \mathbb{C}$
Since f is differentiable on all of ℂ (an open set), f is analytic everywhere — it is entire.
$f \text{ is entire}$
Its Taylor series about 0 is simply itself, since it is already a polynomial in z.
$f(z) = 0 + 0\cdot z + 1\cdot z^2 + 0 + \cdots = z^2$

Answer: f(z) = z² is entire, and its Taylor series about z0 = 0 is exactly z² (all higher coefficients vanish).

Practice Problems

Difficulty 4/10

Is f(z) = 1/z analytic on U = ℂ \ {0}?

Difficulty 5/10

If f is analytic on a domain and f'(z) = 0 for every z in that domain, what can you conclude about f?

Difficulty 6/10

Explain why an analytic function's real part u(x,y) automatically satisfies Laplace's equation ∇²u = 0.

Quiz

Summary

f is analytic (holomorphic) at z0 if it is complex-differentiable throughout an open neighborhood of z0, not just at the point itself.
Analyticity is equivalent to the Cauchy-Riemann equations holding with continuous partials, and it forces f to be infinitely differentiable and locally equal to its Taylor series.
Polynomials and e^z are entire (analytic on all of ℂ); z̄ and |z|² are analytic nowhere or only at isolated points.

References

WebsiteWikipedia — Holomorphic function