Complex Power Series#

Definition: Complex Power Series

A complex power series is an expression of the form

\[ \sum_{n\in \mathcal{D}} a_n (z-c)^n, \]

where \((a_n)_{n \in \mathcal{D}}\) is a complex sequence and \(c \in \mathbb{C}\).

Intuition

By plugging in a concrete value \(z_0 \in \mathbb{C}\) for \(z\), one obtains the complex series

\[ \sum_{n \in \mathcal{D}} a_n (z_0 - c)^n \]

Convergence#

Definition: Convergence and Divergence of Power Series

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z-c)^n\) be a complex power series and let \(z_0 \in \mathbb{C}\).

We say that \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z-c)^n\)

is convergent or converges for \(z_0\) if the resultant complex series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is convergent;
is absolutely convergent or converges absolutely for \(z_0\) if the resultant complex series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is absolutely convergent;
is divergent or diverges for \(z_0\) if the resultant complex series \(\sum_{n \in \mathcal{D}} a_n (z_0 - c)^n\) is divergent;

Theorem: Convergence Disks

If the complex power series \(\sum_{n \in \mathcal{D}} a_n (z-c)^n\) converges for \(w \in \mathbb{C}\), then it also converges for all \(w' \in \mathbb{C}\) with \(|w' - c| \lt |w - c|\).

Proof

Since the series \(\sum_{n \in \mathcal{D}} a_n (w-c)^n\) converges, we know that the limit of \(a_n (w - c)^n\) must be \(0\):

\[ \lim_{n \to \infty} a_n (w-c)^n = 0 \]

Since \(a_n (w - c)^n\) is convergent, it must be bounded, i.e. there exists \(B \in \mathbb{R}_{\gt 0}\) such that

\[ |a_n (w-c)^n| \le B \]

for all \(n \in \mathcal{D}\).

For \(w'\) we have

\[ |a_n (w' - c)^n| = \left|a_n(w' - c)^n \frac{(w-c)^n}{(w-c)^n}\right| = \left|a_n(w - c)^n\right|\left| \frac{w'-c}{w-c}\right|^n \]

Since \(|a_n (w-c)^n| \le B\) for all \(n\in\mathcal{D}\), we know that

\[ |a_n (w' - c)^n| \le B \left| \frac{w'-c}{w-c}\right|^n \]

for all \(n \in \mathcal{D}\). Let \(q = \left| \frac{w'-c}{w-c}\right|\), i.e.

\[ |a_n (w' - c)^n| \le B q^n \]

Since \(|w'-c| \lt |{w-c}|\), we know that \(q \lt 1\). Therefore, \(\sum_{n \in \mathcal{D}} B q^n\) is convergent (\(\sum_{n\in\mathcal{D}}q^n\) is a geometric series). The convergence of \(\sum_{n \in \mathcal{D}} a_n(w'-c)^n\) is thus guaranteed by the majorant convergence criterion.

Theorem: Radius of Convergence

For each complex power series \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z - c)^n\), there exists some \(r \in \mathbb{R}_{\ge 0} \cup \{\infty\}\) such that:

If \(r = 0\), then the power series converges only for \(x = c\).
If \(r \in (0; \infty)\), then the power series converges absolutely for all \(z \in \mathbb{C}\) with \(|z - c| \lt r\) and diverges for all \(z \in \mathbb{C}\) with \(|z - c| \gt r\). The behavior for \(z \in \mathbb{C}\) with \(|z| = r\) must be examined separately.
If \(r = \infty\), then the power series converges absolutely for all \(z \in \mathbb{C}\).

Definition: Radius of Convergence

We call \(r\) the radius of convergence.

Proof

TODO

Theorem: Radius of Convergence via Quotient Limit

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z - c)^n\) be a complex power series.

If the limit of \(\left|\frac{a_n}{a_{n+1}}\right|\) exists or is \(\infty\), then it is equal to the radius of convergence \(r\):

\[ r = \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| \in \mathbb{R}_{\ge 0} \cup \{\infty\} \]

Proof

TODO

Example: \(\sum_{n = 0}^{\infty} z^n\)

We examine the complex power series

\[ \sum_{n = 0}^{\infty} z^n \]

To see that it is indeed a complex power series, we rewrite it a bit:

\[ \sum_{n = 0}^{\infty} a_n(z - 0)^n \qquad a_n = 1 \]

We thus have the following:

\[ r = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1 \]

Therefore, \(\sum_{n = 0}^{\infty} z^n\) is absolutely convergent when \(|z| \lt 1\) and divergent when \(|z| \gt 1\). Moreover, we have noticed that \(\sum_{n = 0}^{\infty} z^n\) is a geometric series for each \(z \in \mathbb{C}\). Therefore, we know that it is also divergent when \(|z| = 1\).

Cauchy-Hadamard Theorem

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z - c)^n\) be a complex power series.

The limit superior of the sequence \(\sqrt[n]{|a_n|}\) can be used to find the radius of convergence:

If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} = 0\), then \(R = \infty\).
If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} \in (0; \infty)\), then \(R = \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}}\).
If \(\limsup_{n\to \infty} \sqrt[n]{|a_n|} = \infty\), then \(R = 0\).

Proof

TODO

Example: \(\sum_{k = 0}^{\infty} z^{k^2}\)

We examine the following complex power series:

\[ \sum_{k = 0}^{\infty} z^{k^2} = 1 + z + z^4 + z^9 + z^{16} + \cdots \]

To see that this is indeed a complex power series, we need to rewrite it a bit:

\[ \sum_{n = 0}^{\infty} a_n(z-0)^{n} \qquad a_n = \begin{cases}1 & \text{if } n = k^2, k \in \mathbb{N}_0 \\ 0 & \text{otherwise}\end{cases} \]

Therefore, we have:

\[ \sqrt[n]{|a_n|} = \begin{cases}1 & \text{if } n = k^2, k \in \mathbb{N}_0 \\ 0 & \text{otherwise}\end{cases} \]

The sequence \(\sqrt[n]{|a_n|}\) is divergent, but its limit superior is \(1\):

\[ \limsup_{n\to \infty} \sqrt[n]{|a_n|} = 1 \]

Therefore, the radius of convergence is \(1\):

\[ r = \frac{1}{\limsup_{n\to \infty} \sqrt[n]{|a_n|}} = \frac{1}{1} = 1 \]

Theorem: Radius of Convergence via \(n\)-th Root Limit

Let \(\displaystyle \sum_{n \in \mathcal{D}} a_n (z - c)^n\) be a complex power series.

If the limit \(\lim_{n \to \infty} \sqrt[n]{|a_n|} = L\) exists or is \(\infty\), then the radius of convergence \(r\) is given by:

\[ r = \begin{cases} 0 & \text{if } L = \infty \\ \infty & \text{if } L = 0 \\ \frac{1}{L} & \text{otherwise} \end{cases} \]

Proof

TODO