Personal notes on power series in complex analysis.

Notes taken while taking a course [1].

## Power series

A *power series* centered at \(z_0 \in \Complex\) is a series of the form

\[ \sum_{k=0}^\infty a_k (z-z_0)^k \]

Power series are also known as Taylor series.

## Convergence

### Absolute convergence

A series \(\sum_{k=0}^\infty\) *converges absolutely*
if the series \(\sum_{k=0}^\infty |a_k|\) converges.

If \(\sum_{k=0}^\infty a_k\) converges absolutely, then it also converges, and \(\left|\sum_{k=0}^\infty a_k\right| \leq \sum_{k=0}^\infty |a_k|\).

### Radius of convergence

For a power series \(\sum_{k=0}^\infty a_k (z-z_0)^k\), there exists a number \(0 \leq R \leq \infty\) such that the series converges absolutely in \(\{|z-z_0| < R\}\) and diverges in \(\{|z-z_0| > R\}\). The convergence is uniform in \(|z-z_0| \leq r\) for any \(0 \leq r < R\).

#### Ratio test

If \(\left\{\left|\frac{a_k}{a_{k+1}}\right|\right\}\) has a limit as \(k \rightarrow \infty\), then the radius of convergence \(R\) for \(\sum_{k=0}^\infty a_k (z-z_0)^k\) is

\[ R = \lim_{k \rightarrow \infty} \left|\frac{a_k}{a_{k+1}}\right| \]

\(\infty\) is considered a valid limit value in this context.

#### Root test

If \(\left\{\sqrt[k]{|a_k|}\right\}\) has a limit as \(k \rightarrow \infty\), then the radius of convergence \(R\) for \(\sum_{k=0}^\infty a_k (z-z_0)^k\) is

\[ R = \frac{1}{\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|}} \]

If \(\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|} = 0\), then \(R=\infty\), and if \(\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|} = \infty\), then \(R=0\).

#### Cauchy-Hadamard criterion

The radius of convergence \(R\) for \(\sum_{k=0}^\infty a_k(z-z_0)^k\) is

\[ R = \frac{1}{\limsup_{k \rightarrow \infty} \sqrt[k]{|a_k|}} \]

## Analyticity of power series

If \(f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k\) has radius of convergence \(R>0\), \(f(z)\) is analytic in \(\{|z-z_0|<R\}\).

The series can be differentiated term by term.

\[\begin{align*} \frac{df}{dz}(z) & = \sum_{k=1}^\infty a_k k(z-z_0)^{k-1} \\ \frac{d^2 f}{dz^2}(z) & = \sum_{k=2}^\infty a_k k(k-1)(z-z_0)^{k-2} \\ & \vdots \end{align*}\]

The coefficient \(a_k\) can be derived using the \(k\)-order derivative.

\[ a_k = \frac{1}{k!} \frac{d^k f}{dz^k}(z_0) \]

### Power series representation

If \(f : U \rightarrow \Complex\) is analytic and \(r\) is such that \(\{|z-z_0| < r\} \subset U\), then \(f\) has a power series representation in this disk.

\[ f(z) = \sum_{k=0}^\infty a_k (z-z_0)^k \]

for \(|z-z_0|<r\) and where

\[ a_k = \frac{1}{k!} \frac{d^k f}{dz^k}(z_0) \]

The radius of convergence is at least \(r\).

### Determination by derivatives

If \(f\) and \(g\) are analytic in \(D=\{|z-z_0|<r\}\) and \(\frac{d^k f}{dz^k}(z_0) = \frac{d^k g}{dz^k}(z_0)\) for all \(k \geq 0\), then \(f(z)=g(z)\) for all \(z \in D\).

## Integration of power series

If \(\sum_{k=0}^\infty a_k(z-z_0)^k\) has radius of convergence \(R\), then for any \(w\) such that \(|w-z_0| < R\),

\[\begin{align*} \int_{z_0}^w \sum_{k=0}^\infty a_k(z-z_0)^k \, dz &= \sum_{k=0}^\infty a_k \int_{z_0}^w (z-z_0)^k \, dz \\ &= \sum_{k=0}^\infty \frac{a_k}{k+1}(w-z_0)^{k+1} \end{align*}\]