# Friday February 14th

## Defining Residues

> Interesting open problems: dynamical systems on $\CC^2$.

If $f$ is holomorphic in $\Omega$ with $f(z_0) = 0$ then there exists a disc on which $f(z) = \sum a_n (z-z_0)^n$ where $a_0 = f(z_0) = 0$.
There is then a minimal $k$ such that $f(z) = (z-z_0)^k g(z)$ where $g(z_0)\neq 0$; this $k$ is the *order* of the zero $a_0$.

Recall the definition of a *pole*:
A function defined in a deleted neighborhood of $z_0$ has a *pole* at $z_0$ iff $F = \frac 1 f$ with $F(z_0) \definedas 0$ is holomorphic in a full neighborhood of $z_0$.

Theorem (Extraction of Holomorphic Part)
: 	If $f$ has a pole at $z_0$, then there exists a holomorphic function $h$ and a unique $k$ such that $f(z) = (z-z_0)^{-k} h(z)$.

Proof
: 	Write $$\frac 1 f = (z-z_0)^k g(z)$$ with $g(z_0) \neq 0$.
	Then there is an $r$ such that $\abs{g(z)} \geq \frac 1 2 \abs{g(z_0)}$ in a disc about $z_0$.
	Then $$f(z) = \frac{1}{\qty{z-z_0} g(z)} \definedas (z-z_0)^{-k} h(z)$$ where $h = 1/g$.

	We can then write $$f(z) = \qty{ \sum_{i=0}^{k-1} b_k (z-z_0)^{-k} } + b_k + \sum_{i=1}^\infty b_{k+i} (z-z_0)^i$$ for some fixed $k$, where $\sum b_i (z-z_0)^i$ is the power series expansion of $h$.
	Write this as $P(z) + G(z)$ where $G(z) = \sum_{i=0}^\infty b_{i+k}(z-z_0)^i$.
	Denote $P$ *the principal part* of $f$ at the pole $z=z_0$.

	Note that $$\int_{D_r(z_0)} f = \int_{D_r(z_0)} P(z) = 2\pi i ~a_{-1}.$$

Definition (Residue)
: 	The coefficient $a_{-1}$ is referred to as the *residue* of $f$ at $z=z_0$.


## Residues

Note that $$\int \frac{1}{\qty{z-z_0}^k} = \begin{cases} 2\pi i &k = 1 \\ 0 & \text{else}\end{cases}.$$

Residues can be computed using the following formula:
\begin{equation}
a_{-1} = \frac 1 {2\pi i} \int_{D_r(z_0)} f
.\end{equation}

Theorem (Residue Formula)
:
	\begin{align*}
	\Res_{z=z_0} f = \lim_{z\to z_0} \frac{1}{\qty{k-1}!} \qty{\dd{}{z}}^{k-1} (z-z_0)^k f(z)
	.\end{align*}

Proof
: 	Expand in power series, direct check.

A useful special case: if $z_0$ is a pole of order 1, then

\begin{align*}
\Res_{z=z_0} f = \lim_{z\to z_0} (z-z_0) f(z)
.\end{align*}

A useful formula:

\begin{align*}
\frac 1 {2\pi i} \int_{\Gamma(z_0)} f = \Res_{z=z_0} f
.\end{align*}

Theorem (Integral Residue Theorem)
: 	Suppose that $f$ is holomorphic in an open set containing a toy contour $\gamma$ and its interior except for finitely many poles $\theset{z_i}$.
	Then

	\begin{align*}
	\frac 1 {2\pi i} \int_\gamma f = \sum \Res_{z=z_i} f(z)
	.\end{align*}

Proof
: 	Omitted to cover some material needed for homework.


Note that if $f$ has a pole of order $k$, we can expand it in *Laurent series* as

\begin{align*}
\sum_{n=-k}^1 a_n (z - z_0)^n + \sum_{n=0}^\infty a_n(z-z_0)^k
.\end{align*}

How to determine the radius of convergence\index{Radius of convergence} of a Laurent series:

\begin{align*}
\sum_{-\infty}^\infty a_n z_n
&= \sum_{n\in \NN} c_n z^n + \sum_{n\in \NN} d_n z^{-n} 
.\end{align*}

Applying the root test,

\begin{align*}
\limsup_n \abs{c_n (z-a)}^{1/n} &< 1 \\
\iff \limsup_n \abs{c_n}^{1/n} \abs{z-z_0}^n &< 1 \\
\iff \abs{z-a} &\leq \frac{1}{\limsup_n \abs{c_n}^{1/n}} \definedas \rho_1
.\end{align*}


Similarly, we need 

\begin{align*}
\rho_2 \definedas \limsup_n \abs{d_n}^{1/n} < \abs {z-a}
.\end{align*}

If $\rho_1> \rho_2$, this will converge on an annulus.