Complex Power Series: A Comprehensive Guide

If you find yourself in a hole, stop digging, Will Rogers

Introduction

Definition. Complex sequence A sequence of complex numbers is a function $a: \mathbb{N} \to \mathbb{C}$. We usually denote it by $(a_n)_{n \in \mathbb{N}}$ or simply $(a_n)$, where $a_n := a(n)$. The value $a_1$ is called the first term of the sequence, $a_2$ the second term, and in general $a_n$ the n-th term of the sequence.

Definition. Convergent complex sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is said to converge to a complex number $L \in \mathbb{C}$ if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n \geq N$ one has $|a_n - L| < \varepsilon$. In this case we write $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ as $n \to \infty$, and L is called the limit of the sequence $(a_n)_{n \in \mathbb{N}}$.

Definition. Cauchy sequence. A complex sequence $(a_n)_{n \in \mathbb{N}}$ is called a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N \in \mathbb{N}$ such that for all $n, m \geq N$ one has $|a_n - a_m| < \varepsilon$.

Definition. Series and partial sums.Let $(a_n)_{n \in \mathbb{N}}$ be a complex sequence. For each n $\in \mathbb{N}$, the finite sum $s_n := a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ is called the n-th partial sum of the (infinite) series $\sum_{k=1}^\infin a_k$ which we also denote simply by $\sum a_n$ when the index is clear from the context.

Definition. Convergent series. The series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $s \in \mathbb{C}$ if the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ defined by $s_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k$ converges to s, that is, $\lim_{n \to \infty} s_n = s$. In this case we write $s := \sum_{n=1}^\infin a_n$. If the sequence $(s_n)_{n \in \mathbb{N}}$ does not converge, we say that the series $\sum_{n=1}^{\infty} a_n$ diverges (or does not converge).

Complex Power Series: A Comprehensive guide

Definition. A complex power series centered at 0 in the variable z is a series of the form $a_0 + a_1z + a_2z^2 + \cdots = \sum_{n=0}^\infty a_n z^n$ with coefficients $a_i \in \mathbb{C}$

Definition. A complex power series centered at a complex number $a \in \mathbb{C} $ is an infinite series of the form: $\sum_{n=0}^\infty a_n (z - a)^n,$ where each $a_n \in \mathbb{C}$ is a coefficient, z is a complex variable, and $(z - a)^n$ is the nth power about the center.

Example. The geometric series $\sum_{n=0}^\infty z^n = 1 + z + z^2 + \cdots $ is a power series centered at $z_0 = 0$ with coefficients $a_n = 1 $ for all n.

$(1 - z)(1 + z + z^2 + \cdots + z^n) = 1 - z^{n + 1} \leadsto 1 + z + z^2 + \cdots + z^n = \frac{1 - z^{n + 1}}{1 - z}, \forall z \ne 1$.

If ∣z∣ < 1, then $\lim_{n \to \infin} z^{n + 1} = 0$. Therefore, the geometric series $\sum_{n=0}^\infty z^n$ converges to $\frac{1}{1-z}$ inside the unit disk.

On the unit circle, ∣z∣ = 1, there are two options:

1. z = 1, the series becomes $1 + 1 + \cdots + 1$ which clearly diverges to ∞.
2. z = $e^{i\theta} \text{ where } \theta \ne 0$ (any other point on the unit circle), $|z^{n + 1}| = |e^{i(n+1)\theta}| = 1$ for all n and the terms do not shrink (approach zero, i.e., the series diverges because the terms don’t satisfy the necessary condition for convergence: $\lim_{n \to \infin} a_n = 0$), they keep rotating around the unit circle. Therefore, the series diverges for |z| = 1.

Outside the unit disk, ∣z∣ > 1, $|z^{n + 1}| = |z|^{n+1} \to \infin$. The terms “blow up”, so the series diverges. The unit circle is the event horizon for $\sum_{n=0}^\infty z^n$, everything inside shrink (gets sucked into 0) and the series converges, everything on or outside stays large (or blow up) and the series diverges.

Theorem. Given a power series $\sum_{n=0}^\infty a_n z^n$, there exists a unique value R, $0 \le R \le \infin$ (called the radius of convergence) such that:

1. For any z with |z| < R (inside the circle), the series $\sum_{n=0}^\infty a_n z^n$ converges absolutely (this is a “green light” zone).
2. For any z with |z| > R, the series diverges (this is a “red light” zone).

   On the Circle (|z| = R), this theorem gives no information. This is the yellow light zone —the series could converge or diverge.

Proof.

Part 1: Proving Absolute Convergence for |z| < R.

Let R be the supremum (the least upper bound) of all non-negative numbers $r \ge 0$ such that the sequence of terms $|a_n|r^n$ is bounded (this means that if we are inside $R$, the terms are manageable; if we are outside, they are obviously not), S := { $r\ge 0 : \sup_{n\ge 0} |a_n|r^n < \infty$}.

Pick any complex number z such that |z| < R (inside the circle). Since |z| is less than the supremum R, we can always find another number, $r_1 \ge 0$, that is “sacrificed” to be between them. Let’s choose an $r_1$ such that $|z| \lt r_1 \le R$, so the sequence of terms $|a_n|r_1^n$ is bounded.

Since the sequence is bounded, there must be a constant M such that for all n: $|a_n| r_1^n \le M$.

Now we look at the absolute value of the terms of our original series, $|a_n z^n|, |a_n z^n| = |a_n| |z|^n = (|a_n| r_1^n) \cdot \left( \frac{|z|^n}{r_1^n} \right) \le M \cdot \left( \frac{|z|}{r_1} \right)^n$

Let’s define $\rho = \frac{|z|}{r_1}$. $|z| < r_1 \leadsto + \rho < 1$. This makes the series $\sum_{n=0}^\infty M \rho^n$ a convergent geometric series (a constant $M$ times a series with a ratio less than 1). By the Comparison Test, since every term $|a_n z^n|$ is less than or equal to the corresponding term of a convergent series, our series $\sum |a_n z^n|$ must also converge. This proves that the original series converges absolutely.

Part 2: Proving Divergence for |z| > R.

We aim to show that if z is outside the circle, |z| > R, the series $\sum a_n z^n$ diverges.

Since |z| is greater than the supremum R, |z| cannot be in our set (the set of radii that keep the terms bounded, $|z|\notin S$). This means the sequence $|a_n| |z|^n$ is unbounded, hence cannot tend to 0. A series whose terms do not tend to 0 diverges.

The terms of our series are $a_n z^n$ and their magnitude is $|a_n z^n| = |a_n| |z|^n$, and we know this sequence of magnitudes is unbounded (it cannot possibly converge to 0). If the terms of a series do not tend to zero, the series must diverge.

Theorem. For a complex power series $\sum a_n (z - z_0)^n$, the set of points $z \in \mathbb{C}$ where the series converges forms a disk in the complex plane. The disk is defined by: (1) Interior: All points z satisfying $|z - z_0| < R$; (2) Boundary: Points z with $|z - z_0| = R$; (3) Exterior: Points z with $|z - z_0| > R$ where $R \geq 0$ is the radius of convergence.

For a power series $\sum a_n (z - z_0)^n$, there are two common ways to determine the radius of convergence R:

1. The Cauchy-Hadamard Formula (Root Test), $R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}$ with the conventions: If limsup = 0, then R = ∞ (the series converges everywhere). If limsup = ∞, then R=0 (converges only at z = z₀)
2. Ratio Test $R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|.$

The Cauchy-Hadamard Formula (Root Test)

Proof:

(For simplicity, we’ll analyze the proof for $\sum |a_nz^n|$ centered at $z_0 = 0$).

Convergence when ∣z∣ < R = 1/L where $L = \limsup_{n \to \infty} |a_n|^{1/n}$. Choose a number q such that $L < q < \frac{1}{|z|}$ (possible since ∣z∣ < 1/L). By the definition of limsup, eventually (for all large n): $|a_n|^{1/n} \le q \leadsto |a_n| \le q^n \leadsto |a_nz^n| \le (q|z|)^n$

Since q∣z∣<1 by construction, $\sum(q ∣z∣)^n$ is a convergent geometric series. By the Comparison Test, $\sum |a_nz^n|$ converges, then $\sum |a_nz^n|$ converges absolutely.

Divergence when ∣z∣ > 1/L. Choose a number p such that $\frac{1}{|z|} \lt p \lt L$ (possible since ∣z∣ > 1/L). By the definition of limsup, infinitely many n satisfy: $|a_n|^{1/n} \ge p \leadsto |a_n| \ge p^n \leadsto$

$|a_nz^n| \ge (p|z|)^n \gt[\text{We have chosen p > 1/|z|, which means the value p|z| is greater than 1. This means that for infinitely many $n$, the terms of our series satisfy:}] 1$. The terms do not approach zero (they are greater than 1 infinitely often), $|a_nz^n| ↛ 0$, which direly implies $a_nz^n ↛ 0$, hence the series $\sum a_nz^n$ diverges.

$|a_nz^n| ↛ 0$, which direly implies $a_nz^n ↛ 0$. That’s because convergence to $0 in \mathbb{C}$ requires both the real and imaginary parts to vanish, which is equivalent to the modulus tending to 0.

For a general center $z_0$, replace $z^n$ by $(z-z_0)^n$; the same proof goes through with $|z-z_0|$ in place of |z|.

• Examples

1. Geometric Series $ \sum_{n=0}^\infty z^n $: $ a_n = 1$, so using Ratio Test $R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{1}{1} \right| = 1$. It converges for |z| < 1, diverges for |z| > 1. At |z| = 1, the series diverges (If |z| = 1, the magnitude of the terms is $|z^n| = |z|^n = 1^n = 1$ and the terms are not approaching zero).
2. Exponential Function $e^z$: Power series: $ \sum_{n=0}^\infty \frac{z^n}{n!}$. Ratio Test, $\left| \frac{a_n}{a_{n+1}} \right| = \left| \frac{1/n!}{1/(n+1)!} \right| = \frac{(n+1)!}{n!} = \frac{(n+1) \cdot n!}{n!} = n+1$. Radius $R = \lim_{n \to \infty} (n+1) = \infty $ (converges for all $ z \in \mathbb{C}$).
3. Logarithm Series $\log(1 + z)$: Power series: $\sum_{n=1}^\infty (-1)^{n+1} \frac{z^n}{n}$. Root Test, $|a_n|^{1/n} = \left| \frac{1}{n} \right|^{1/n} = \frac{1}{n^{1/n}} = \frac{1}{\sqrt[n]{n}}$. Take the limit (L): $L = \limsup_{n \to \infty} \frac{1}{\sqrt[n]{n}} = \frac{1}{1} = 1$ (Recall $\lim_{n \to \infty} \sqrt[n]{n} = 1$). Radius R = 1/L = 1/1 = 1. Converges for |z| < 1.

Boundary Behavior (when $|z|=1$)

Case: z = 1. The series becomes $\sum_{n=1}^\infty (-1)^{n+1} \frac{(1)^n}{n} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$. This is the alternating harmonic series ($1 - \frac{1}{2} + \frac{1}{3} - \dots$). By the Alternating Series Test, this series converges (it converges conditionally to $\ln 2$).

Alternating Series Test (Leibniz Criterion). Consider a series of the form $\sum_{n=1}\infty (-1){n+1} b_n \text{ or } \sum_{n=1}^\infty (-1)^n b_n$, where $b_n \ge 0$. The series converges if: (1) Monotonic decrease: $b_{n+1} \le b_n$ for all sufficiently large n (e.g., $b_{n+1} = \frac{1}{n+1} \le \frac{1}{n}$); (2) $\lim_{n\to\infty} b_n = 0$ (e.g., $\lim_{n\to\infty} \frac{1}{n} = 0$).

Case: z = -1. The series becomes $\sum_{n=1}^\infty (-1)^{n+1} \frac{(-1)^n}{n} = \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n}$. Since 2n+1 is always odd, $(-1)^{2n+1} = -1 \leadsto \sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n} = \sum_{n=1}^\infty \frac{-1}{n} = -1 \cdot \sum_{n=1}^\infty \frac{1}{n}$. This is the negative of the harmonic series, which famously diverges.

So, this series converges for |z| < 1 and at z = 1, but diverges at z = -1 (and all other points on the circle |z| = 1).