§2.3 Complex Power Series

2.3 Complex Power Series

Power series in real analysis can be generalized into complex series. Particularly, concepts such as uniform convergence are analogous to those in real analysis:

Definition 2.3.1 (Uniform Convergence).

For a set

𝑈 \subseteq ℂ

, a function sequence

{𝑓_{𝑛} (𝑧)}

uniformly converges to a function

𝑓 (𝑧)

𝑈

iff

\forall 𝜀 > 0

\exists 𝑁 \in ℕ

such that

\forall 𝑛 > 𝑁

\forall 𝑧 \in 𝑈

| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | < 𝜀

Remark.

The definition above is equivalent to the following definition.

For a set $𝑈 \subseteq ℂ$ , a function sequence ${𝑓_{𝑛} (𝑧)}$ uniformly converges to $𝑓 (𝑧)$ iff

lim_{𝑛 \to \infty} sup_{𝑧 \in 𝑈} | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | = 0 .

Informally, we will use the notation $𝑓_{𝑛} (𝑧) ⇉ 𝑓 (𝑧)$ to represent uniform convergence.

Theorem 2.3.2 (Cauchy Criterion).

For a set

𝑈 \subseteq ℂ

, a function sequence

{𝑓_{𝑛} (𝑧)}

uniformly converges to a function

𝑓 (𝑧)

iff

\forall 𝜀 > 0

\exists 𝑁 \in ℕ

such that

\forall 𝑛, 𝑚 > 𝑁

\forall 𝑧 \in 𝑈

| 𝑓_{𝑛} (𝑧) - 𝑓_{𝑚} (𝑧) | < 𝜀

Proof.

If $𝑓_{𝑛} (𝑧)$ uniformly converges to $𝑓 (𝑧)$ , then for $\forall 𝜀 > 0$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛, 𝑚 > 𝑁$ and $\forall 𝑧 \in 𝑈$ ,

| 𝑓_{𝑚} (𝑧) - 𝑓 (𝑧) | < \frac{𝜀}{2}, | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | < \frac{𝜀}{2} .

Then,

| 𝑓_{𝑚} (𝑧) - 𝑓_{𝑛} (𝑧) | \leq | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | + | 𝑓_{𝑚} (𝑧) - 𝑓 (𝑧) | < 𝜀 .

For the converse, refer to the analogous proof in Theorem 1.2.17 $Theorem 1.2.17 Theorem 1.2.17$ . ⁠ $□$

Function series are defined to be a sequence formed by the partial sums of function sequences. There are many ways to verify the uniform convergence of a function series. Perhaps the most widely known is the Weierstrass $𝑀$ -Test.

Theorem 2.3.3 (Weierstrass $𝑀$ -Test).

Let $𝑈 \subseteq ℂ$ be a region and ${𝑓_{𝑛}}$ be a function sequence on $𝑈$ .

If $\exists {𝑀_{𝑛}} \subset ℝ_{\geq 0}$ such that $\forall 𝑛 \in ℕ$ , $\forall 𝑧 \in 𝑈$ , $| 𝑓_{𝑛} (𝑧) | \leq 𝑀_{𝑛}$ and the series $\sum_{𝑛 = 1}^{\infty} 𝑀_{𝑛}$ converges, then the series $\sum_{𝑛 = 1}^{\infty} 𝑓_{𝑛} (𝑧)$ converges uniformly and absolutely on $𝑈$ .

Proof.

By the convergence of $\sum_{𝑛 = 1}^{\infty} 𝑀_{𝑛}$ , $\forall 𝜀 > 0$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑚 \geq 𝑛 > 𝑁$ ,

| 𝑀_{𝑚} + 𝑀_{𝑚 - 1} + \dots + 𝑀_{𝑛 + 1} | < 𝜀 .

Since $𝑀_{𝑗}$ bounds $𝑓_{𝑗} (𝑧)$ , it follows that

| 𝑓_{𝑚} (𝑧) + 𝑓_{𝑚 - 1} (𝑧) + \dots + 𝑓_{𝑛 + 1} (𝑧) | \leq | 𝑀_{𝑚} + 𝑀_{𝑚 - 1} + \dots + 𝑀_{𝑛 + 1} | < 𝜀,

and the result follows from Theorem 2.3.2 $Theorem 2.3.2 Theorem 2.3.2$ . ⁠ $□$

The concept of uniform convergence is generalized to improper integrals with parameters, and the same theorems from series have a corresponding counterpart.

In both complex and real analysis, the concept of power series, a unique type of function series, is of trivial importance. Similar to real power series, complex series have the form

\sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛},

where ${𝑎_{𝑛}}$ are constants.

Let $𝐷 (𝑎, 𝑟) = 𝐵^{1} (𝑎, 𝑟) = {𝑧 \in ℂ : | 𝑧 - 𝑎 | < 𝑟}$ denote the open disk centered at $𝑎$ with radius $𝑟$ . For simplicity, from now on we will have $𝔻$ denote the unit open disk, or $𝐷 (0, 1)$ . We will now observe the convergence of power series.

Theorem 2.3.4 (Abel’s Theorem).

For a power series $𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}$ , there exists a constant $𝑅 \in ℝ_{\geq 0} \cup {\infty}$ , known as the radius of convergence such that:

$𝑓$ absolutely converges on $𝐷 (0, 𝑅)$ , and $\forall 0 \leq 𝜌 < 𝑅$ , uniformly converges on $\overline{𝐷 (0, 𝜌)}$ .
$𝑓 (𝑧)$ diverges when $| 𝑧 | > 𝑅$ .
$𝑓$ is holomorphic over $𝐷 (0, 𝑅)$ and $𝑓^{'} (𝑧)$ can be obtained by termwise differentiation, or $𝑓^{'} (𝑧) = \sum_{𝑛 = 1}^{\infty} 𝑛 𝑎_{𝑛} 𝑧^{𝑛 - 1}$ , which also has a convergence radius of $𝑅$ .

The disk $| 𝑧 | < 𝑅$ is known as the disk of convergence, a direct generalization of the interval of convergence for real series. There are many ways to determine the radius of convergence:

Theorem 2.3.5 (Cauchy–Hadamard).

The radius of convergence of the power series in the form $\sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}$ can be determined by

𝑅 = \frac{1}{{lim sup}_{𝑛 \to \infty} \sqrt[𝑛]{| 𝑎_{𝑛} |}} . (2.3 . 1)

Of course, a convergence radius of $0$ implies that the series is divergent everywhere except for possibly at $0$ , and a convergence radius of $\infty$ means that the series absolutely converges everywhere.

Proof of Theorem 2.3.4 $Theorem 2.3.4 Theorem 2.3.4$ and Theorem 2.3.5 $Theorem 2.3.5 Theorem 2.3.5$ .

We will prove that the value in (2.3.1) $(2.3.1) (2.3.1)$ satisfies the criteria in Theorem 2.3.4 $Theorem 2.3.4 Theorem 2.3.4$ .

Assume $| 𝑧 | < 𝑅$ . Then $\forall 𝜌 \in (| 𝑧 |, 𝑅)$ , and consequently, $\frac{1}{𝜌} > \frac{1}{𝑅}$ . By Definition 1.2.18 $Definition 1.2.18 Definition 1.2.18$ and (2.3.1) $(2.3.1) (2.3.1)$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , $\sqrt[𝑛]{| 𝑎_{𝑛} |} < \frac{1}{𝜌}$ and $| 𝑎_{𝑛} | < \frac{1}{𝜌^{𝑛}}$ . It follows that $| 𝑎_{𝑛} 𝑧^{𝑛} | < \frac{{| 𝑧 |}^{𝑛}}{𝜌^{𝑛}} < 1$ for all $𝑛 > 𝑁$ . Then $\sum_{𝑛 = 0}^{\infty} | 𝑎_{𝑛} 𝑧^{𝑛} |$ converges.

Let $𝜌^{'} \in (𝜌, 𝑅)$ . Similarly, $\exists 𝑁^{'} \in ℕ$ such that $\forall 𝑛 > 𝑁^{'}$ , $\sqrt[𝑛]{| 𝑎_{𝑛} |} < \frac{1}{𝜌^{'}}$ , and $| 𝑎_{𝑛} | < \frac{1}{𝜌^{' 𝑛}}$ . Then $| 𝑎_{𝑛} 𝑧^{𝑛} | < | 𝑎_{𝑛} 𝜌^{𝑛} | < \frac{𝜌^{𝑛}}{𝜌^{' 𝑛}}$ . By the Weierstrass $𝑀$ -Test (Theorem 2.3.3 $Theorem 2.3.3 Theorem 2.3.3$ ), $\sum_{𝑛 = 0}^{\infty} | 𝑎_{𝑛} 𝑧^{𝑛} |$ is uniformly bounded for $𝑛 > 𝑁^{'}$ by the convergent series $\sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝜌^{𝑛}$ , and thus uniformly converges on $| 𝑧 | < 𝜌$ . This proves Part 1 $Part 1 Part 1$ .

Assume that $| 𝑧 | > 𝑅$ . For all $𝜌 \in (𝑅, | 𝑧 |)$ , $\frac{1}{𝜌} < \frac{1}{𝑅}$ . By Definition 1.2.18 $Definition 1.2.18 Definition 1.2.18$ , $\forall 𝑁 \in ℕ$ , $\exists 𝑛_{𝑁} > 𝑁$ such that $\sqrt[𝑛_{𝑁}]{| 𝑎_{𝑛_{𝑁}} |} > \frac{1}{𝜌}$ . It follows that $| 𝑎_{𝑛_{𝑁}} 𝑧^{𝑛_{𝑁}} | > \frac{| 𝑧^{𝑛_{𝑁}} |}{𝜌^{𝑛_{𝑁}}} > 1$ . Since $\forall 𝑁 \in ℕ$ ,

| \sum_{𝑘 = 0}^{𝑛_{𝑁}} 𝑎_{𝑘} 𝑧^{𝑘} - \sum_{𝑘 = 0}^{𝑛_{𝑁} - 1} 𝑎_{𝑘} 𝑧^{𝑘} | > 1,

by the Cauchy Criterion (Theorem 1.2.17 $Theorem 1.2.17 Theorem 1.2.17$ ), $\sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}$ is divergent. Thus, Part 2 $Part 2 Part 2$ is satisfied.

To prove Part 3 $Part 3 Part 3$ , first observe that $\sum_{𝑛 = 1}^{\infty} 𝑛 𝑎_{𝑛} 𝑧^{𝑛}$ and $\sum_{𝑛 = 1}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}$ have the same convergence radius since ${lim sup}_{𝑛 \to \infty} \sqrt[𝑛]{𝑛} = 1$ . For $𝑧 \in 𝐷 (0, 𝑅)$ , let $𝑓 (𝑧) = 𝑆_{𝑛} (𝑧) + 𝑅_{𝑛} (𝑧)$ , where

𝑆_{𝑛} (𝑧) = \sum_{𝑘 = 0}^{𝑛 - 1} 𝑎_{𝑘} 𝑧^{𝑘}, 𝑅_{𝑛} (𝑧) = \sum_{𝑘 = 𝑛}^{\infty} 𝑎_{𝑘} 𝑧^{𝑘} .

Let

𝑓_{1} (𝑧) = lim_{𝑛 \to \infty} 𝑆_{𝑛}^{'} (𝑧) = \sum_{𝑛 = 1}^{\infty} 𝑛 𝑎_{𝑛} 𝑧^{𝑛 - 1} .

Let $𝜌 < 𝑅$ be positive and $| 𝑧_{0} | < 𝜌$ . Then we aim to show that

lim_{𝑧 \to 𝑧_{0}} (\frac{𝑓 (𝑧) - 𝑓 (𝑧_{0})}{𝑧 - 𝑧_{0}} - 𝑓_{1} (𝑧)) = 0 .

By analyzing the difference,

\begin{aligned} \frac{𝑓 (𝑧) - 𝑓 (𝑧_{0})}{𝑧 - 𝑧_{0}} - 𝑓_{1} (𝑧) & = [\frac{𝑆_{𝑛} (𝑧) - 𝑆_{𝑛} (𝑧_{0})}{𝑧 - 𝑧_{0}} - 𝑆_{𝑛}^{'} (𝑧)] \\ \begin{matrix}  \end{matrix} + 𝑆_{𝑛}^{'} (𝑧) - 𝑓_{1} (𝑧) + \frac{𝑅_{𝑛} (𝑧) - 𝑅_{𝑛} (𝑧_{0})}{𝑧 - 𝑧_{0}} . (2.3 . 2) \end{aligned}

Since $𝑆_{𝑛}^{'} (𝑧) \to 𝑓_{1} (𝑧)$ as $𝑛 \to \infty$ , it follows that $\forall 𝜀 > 0$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , $| 𝑆_{𝑛}^{'} (𝑧) - 𝑓_{1} (𝑧) | < \frac{𝜀}{3}$ . Since

\frac{𝑅_{𝑛} (𝑧) - 𝑅_{𝑛} (𝑧_{0})}{𝑧 - 𝑧_{0}} = \sum_{𝑘 = 𝑛}^{\infty} 𝑎_{𝑘} (𝑧^{𝑘 - 1} + 𝑧^{𝑘 - 2} 𝑧_{0} + \dots + 𝑧_{0}^{𝑘 - 1})

for $𝑧 \neq 𝑧_{0}$ , with $| 𝑧 | < 𝜌 < 𝑅$ ,

| \sum_{𝑘 = 𝑛}^{\infty} 𝑎_{𝑘} (𝑧^{𝑘 - 1} + \dots + 𝑧_{0}^{𝑘 - 1}) | \leq \sum_{𝑘 = 𝑛}^{\infty} | 𝑎_{𝑘} | (| 𝑧^{𝑘 - 1} | + \dots + | 𝑧_{0}^{𝑘 - 1} |) < \sum_{𝑘 = 𝑛}^{\infty} | 𝑎_{𝑘} | 𝑘 𝜌^{𝑘 - 1} .

Since $\sum_{𝑘 = 1}^{\infty} 𝑘 | 𝑎_{𝑘} | 𝜌^{𝑘 - 1}$ is absolutely convergent, $\sum_{𝑘 = 𝑛}^{\infty} | 𝑎_{𝑘} | 𝑘 𝜌^{𝑘 - 1}$ is the remainder term of a convergent series. Then, $\exists 𝑁^{'} \in ℕ$ such that $\forall 𝑛 > 𝑁^{'}$ , $\sum_{𝑘 = 𝑛}^{\infty} | 𝑎_{𝑘} | 𝑘 𝜌^{𝑘 - 1} < \frac{𝜀}{3}$ .

Finally, for a fixed $𝑛 > max (𝑁, 𝑁^{'})$ , $\exists 𝛿 > 0$ such that $\forall 𝑧 \in 𝐷 (𝑧_{0}, 𝛿) ∖ {𝑧_{0}}$ ,

| \frac{𝑆_{𝑛} (𝑧) - 𝑆_{𝑛} (𝑧_{0})}{𝑧 - 𝑧_{0}} - 𝑆_{𝑛}^{'} (𝑧) | < \frac{𝜀}{3} .

From (2.3.2) $(2.3.2) (2.3.2)$ , we get

| \frac{𝑓 (𝑧) - 𝑓 (𝑧_{0})}{𝑧 - 𝑧_{0}} - 𝑓_{1} (𝑧) | < 𝜀,

confirming Part 3 $Part 3 Part 3$ . ⁠ $□$

Obviously, a substitution of $𝑧 = 𝜁 - 𝑎$ where $𝑎 \in ℂ$ translates the disk of convergence to $𝐷 (𝑎, 𝑅)$ . The subsequent results on uniform convergence hold for complex functions:

Theorem 2.3.6 (Uniform Limit).

Let

{𝑓_{𝑛} (𝑧)}_{𝑛}

be a sequence of continuous functions on

𝑈 \subseteq ℂ

and uniformly convergent to

𝑓 (𝑧)

. Then

𝑓 (𝑧)

is continuous on

𝑈

Proof.

Fix $𝜀 > 0$ and $𝑧_{0} \in 𝑈$ . By uniform convergence, $\exists 𝑛 \in ℕ$ such that $\forall 𝑧 \in 𝑈$ , $| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | < \frac{𝜀}{3}$ and $| 𝑓_{𝑛} (𝑧_{0}) - 𝑓 (𝑧_{0}) | < \frac{𝜀}{3}$ . By continuity, $\exists 𝛿 > 0$ such that $\forall 𝑧 \in 𝐷 (𝑧_{0}, 𝛿) \cap 𝑈 \subseteq 𝑈$ , $| 𝑓_{𝑛} (𝑧) - 𝑓_{𝑛} (𝑧_{0}) | < \frac{𝜀}{3}$ .

By the triangle inequality,

| 𝑓 (𝑧) - 𝑓 (𝑧_{0}) | \leq | 𝑓 (𝑧) - 𝑓_{𝑛} (𝑧) | + | 𝑓_{𝑛} (𝑧) - 𝑓_{𝑛} (𝑧_{0}) | + | 𝑓_{𝑛} (𝑧_{0}) - 𝑓 (𝑧_{0}) | < 𝜀 .

for all $𝑧 \in 𝐷 (𝑧_{0}, 𝛿) \cap 𝑈$ . Then the continuity of $𝑓$ is satisfied. ⁠ $□$

Lastly, the sufficient criteria to pass a limit through an integral:

Theorem 2.3.7.

Let $𝛾$ be a rectifiable curve on which the function sequence ${𝑓_{𝑛}}_{𝑛 \in ℕ}$ is continuous. If ${𝑓_{𝑛} (𝑧)}$ uniformly converges to $𝑓$ , then

lim_{𝑛 \to \infty} \int_{𝛾} 𝑓_{𝑛} (𝑧) d 𝑧 = \int_{𝛾} 𝑓 (𝑧) d 𝑧 .

Proof.

Since ${𝑓_{𝑛} (𝑧)}$ uniformly converges to $𝑓 (𝑧)$ on $𝛾$ , $\forall 𝜀 > 0$ , there exists $𝑁 \in ℕ$ such that for all $𝑛 > 𝑁$ ,

| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | < \frac{𝜀}{length (𝛾)}, \forall 𝑧 \in 𝛾 .

Since each $𝑓_{𝑛}$ is continuous and $𝛾$ is rectifiable, each integral $\int_{𝛾} 𝑓_{𝑛} (𝑧) d 𝑧$ is convergent and well-defined.

Then $\forall 𝑛 > 𝑁$ ,

\begin{aligned} | \int_{𝛾} 𝑓_{𝑛} (𝑧) d 𝑧 - \int_{𝛾} 𝑓 (𝑧) d 𝑧 | & = | \int_{𝛾} [𝑓_{𝑛} (𝑧) - 𝑓 (𝑧)] d 𝑧 | \\ \leq \int_{𝛾} | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | | d 𝑧 | \\ < \int_{𝛾} \frac{𝜀}{length (𝛾)} | d 𝑧 | = 𝜀 . \end{aligned}

Therefore,

lim_{𝑛 \to \infty} \int_{𝛾} 𝑓_{𝑛} (𝑧) d 𝑧 = \int_{𝛾} 𝑓 (𝑧) d 𝑧 . □

Remark.

For a uniformly convergent series $\sum_{𝑛 = 1}^{\infty} 𝑓_{𝑛} (𝑧)$ , the commutation between the limit and the integral becomes a summation-integral switch:

\sum_{𝑛 = 1}^{\infty} \int_{𝛾} 𝑓_{𝑛} (𝑧) d 𝑧 = \int_{𝛾} \sum_{𝑛 = 1}^{\infty} 𝑓_{𝑛} (𝑧) d 𝑧 .