1.7 Power Series · Preliminaries to Complex Analysis

[1.7.1]Definition(Complex Exponential and Trigonometric Functions)#

The complex exponential function is defined for all $z \in \bC$ by the power series:

e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}.

This series converges absolutely for every $z \in \bC$ , and converges uniformly in every closed diskin $\bC$ .

Using the complex exponential, the standard complex trigonometric functions are defined by:

\cos z = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = \frac{e^{iz} + e^{-iz}}{2},

\sin z = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} = \frac{e^{iz} - e^{-iz}}{2i}.

These definitions yield the Euler formulas and agree with their real counterparts for $z \in \bR$ .

[1.7.2]Theorem(Hadamard's Formula and Convergence)#

Given a power series $\sum_{n=0}^{\infty} a_n z^n$ , there exists a radius of convergence $R$ ( $0 \leq R \leq \infty$ ) such that:

If $|z| < R$ , the series converges absolutely.
If $|z| > R$ , the series diverges.

Under the conventions $1/0 = \infty$ and $1/\infty = 0$ , $R$ is determined by Hadamard's formula:

1/R = \limsup_{n \to \infty} |a_n|^{1/n}.

The open region $D_R(0) = \{z \in \bC : |z| < R\}$ is called the disk of convergence.

Proof.

Let $L = 1/R = \limsup_{n \to \infty} |a_n|^{1/n}$ .

If $|z| < R$ , then $L|z| < 1$ . Choose $\epsilon > 0$ sufficiently small such that $(L + \epsilon)|z| = r < 1$ . By the definition of the limit superior, we have $|a_n|^{1/n} \leq L + \epsilon$ for all sufficiently large $n$ . Thus:

|a_n||z|^n \leq [(L+\epsilon)|z|]^n = r^n.

Since $r < 1$ , absolute convergence follows by comparison with the convergent geometric series $\sum r^n$ .

If $|z| > R$ , then $L|z| > 1$ . A similar argument shows that $|a_n||z|^n > 1$ infinitely often, so the terms do not tend to $0$ , forcing divergence.

[1.7.3]Remark#

The behavior of a power series on its boundary $|z| = R$ is delicate; one can have either convergence or divergence depending on the specific coefficients (e.g., $\sum z^n$ , $\sum z^n/n^2$ , and $\sum z^n/n$ on the unit circle $\partial \mathbb{D}$ ).

[1.7.4]Theorem(Holomorphicity of Power Series)#

The power series $f(z) = \sum_{n=0}^{\infty} a_n z^n$ defines a holomorphic function in its disk of convergence. Furthermore, $f'(z)$ can be obtained by term-by-term differentiation:

f'(z) = \sum_{n=1}^{\infty} n a_n z^{n-1},

and this differentiated series has the same radius of convergence $R$ .

[1.7.5]Corollary(Infinite Complex Differentiability)#

A power series $f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$ is infinitely complex differentiable in its disk of convergence, and all higher-order derivatives are obtained by term-by-term differentiation.

[1.7.6]Definition(Analytic Functions)#

A complex-valued function $f$ defined on an open set $\Omega \subseteq \bC$ is said to be analytic (or to have a power series expansion) at $point$ $z_0 \in \Omega$ if there exists a power series $\sum_{n=0}^{\infty} a_n(z - z_0)^n$ with a positive radius of convergence such that:

f(z) = \sum_{n=0}^{\infty} a_n(z - z_0)^n \quad \text{for all } z \text{ in a neighborhood of } z_0.

If $f$ is analytic at every point in $\Omega$ , we say $f$ is analytic on $\Omega$ .

[1.7.7]Remark#

By Theorem 2.6, any analytic function on $\Omega$ is automatically holomorphic on $\Omega$ . A fundamental and deep result in complex analysis (the converse, which we prove in Chapter 2) is that every holomorphic function is analytic. Thus, we will use the terms holomorphic and analytic interchangeably.

1.7Power Series