3.1 Power Series

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One useful way of defining functions of a complex variable is via power series. The theory of power series over complex numbers is similar to the theory of power series over the real numbers. In this section we will carefully go over this theory again because power series are of central importance in complex analysis.

Definition (Power Series)

A power series is an expression of the form $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n=a_0+a_1(z-b)+a_2(z-b{)}^2+a_3(z-b{)}^3+\cdots$$ where $b$ is a fixed complex number and $a_1,a_2,a_3,\dots$ is a fixed sequence of complex numbers.

Note that the index of summation starts at zero! This is to allow for the inclusion of a constant term $a_0$.

Given any power series we can consider the set $$\{z\in \mathbb{C}:\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n\text{ converges}\}$$ of complex numbers $z$ where the series $$\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ converges and the function $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ is defined.

Example (The Geometric Power Series)

The geometric series $$\sum _{n=0}^{\mathrm{\infty }}{z}^n=1+z+z^2+z^3+\cdots$$ is a power series, with $b=0$ and $a_n=1$ for all $n$. The algebraic identity $$1+z+z^2+\cdots +z^N=\frac{1-z^{N+1}}{1-z}$$ holds for all $z\ne 1$ and all $N\in \mathbb{N}$. As we saw in the previous section, it implies that the geometric series converges when $|z|<1$ and diverges for $|z|>1$. Thus the set $$\{z\in \mathbb{C}:\sum _{n=0}^{\mathrm{\infty }}{z}^n\text{ converges}\}$$ contains the open ball $\mathsf{B}(0,1)=\{z\in \mathbb{C}:|z|<1\}$ and is contained in the closed ball $\{z\in \mathbb{C}:|z|\le 1\}$.

The main abstract result on convergence of power series is that the geometric series is typical: for every power series there is $R\in [0,\mathrm{\infty })\cup \{\mathrm{\infty }\}$ such that the power series converges whenever $|z-b|<R$ and diverges whenever $|z-b|>R$. Formally, we have the following theorem.

Theorem

Given a power series $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ put $$R=sup\{r>0:\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n\text{ converges for some }|z-b|=r\}$$ which is $\mathrm{\infty }$ when the set is not bounded.

1. The series converges absolutely for all $z\in \mathbb{C}$ with $|z-b|<R$.
2. The series diverges for all $z\in \mathbb{C}$ with $|z-b|>R$.

Proof:

(We give the proof in the case $b=0$.) Let $z\in \mathbb{C}$ be such that $|z|<R$. Choose $w\in \mathbb{C}$ such that $|z|<|w|\le R$ and such that $${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n{w}^n}$$ converges. It follows that $|a_n{w}^n|\to 0$ as $n\to \mathrm{\infty }$. Thus $|a_n{w}^n|$ is a bounded sequence; that is, there exists $K>0$ such that $|a_n{w}^n|<K$ for all $n$. Let $q=|z|/|w|$. As $|z|<|w|$, we have that $q<1$. Now $$|a_n{z}^n|=|a_n{w}^n|{\left|\frac{z}{w}\right|}^n<K{q}^n.$$ for all $n\in \mathbb{N}$. Hence $${\displaystyle \sum _{n=0}^{\mathrm{\infty }}|a_n{z}^n|}$$ converges by comparison with the geometric series. Since absolute convergence implies convergence, we are done.

It follows immediately from the definition of $R$ that the series diverges whenever $|z|>R$. Indeed, if $|z|>R$ and the series converges at $z$ then $R$ was not the supremum in the first place. $◻$

Definition (Radius of Convergence)

We call $R$ the radius of convergence of the power series.

The theorem tells us that the set of complex numbers $z$ where a given power series converges cannot be too complicated. It always contains the open ball $\{z\in \mathbb{C}:|z-z_0|<R\}$ and is always contained in the closed ball $\{z\in \mathbb{C}:|z-z_0|\le R\}$. The theorem does not tell us about convergence on the circle $\{z\in \mathbb{C}:|z-b|=R\}$.

For specific examples, the ratio and root tests can be used to determine radii of convergence.

Example

Determine the radius of convergence of the power series $${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{n}}$$ using the ratio test.

Solution:

Fix $z\in \mathbb{C}$ non-zero. From $$\frac{{\displaystyle \frac{z^{n+1}}{n+1}}}{{\displaystyle \frac{z^n}{n}}}=z\frac{n}{n+1}$$ we get $$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{{\displaystyle \frac{z^{n+1}}{n+1}}}{{\displaystyle \frac{z^n}{n}}}\right|=|z|\underset{n\to \mathrm{\infty }}{lim}\frac{n}{n+1}=|z|$$ implying that the series converges absolutely when $|c|<1$ and diverges when $|c|>1$. The radius of convergence of the power series must then be 1.