# Monday February 17th

> See Hans Lewy 1957 Annals, Folland and Stein 1973. Does a linear system of PDEs with analytic functions have an analytic solution? What about just $C^\infty$?

## Getting a Holomorphic Function from a Laurent Series

We can write a formal series 
\begin{align*}
f(z) &= \sum_{n\in \ZZ}a_n (z-a)^n \\
&= \sum_{n\geq 0} a_n (z-z_0)^n + \sum{n\leq -1} a_n (z-z_0)^n \\
&\definedas A(z) + B(z)
.\end{align*}

Part $A$ converges for $$\abs{z-a} < R_1 = \qty{\limsup \abs{x_n}^{1/n}} \inv.$$
Part $B$ converges for $$\abs{z-a} > R_2 = \limsup \abs{c_{-n}}^{1/n}.$$

If $R_1 < R_2$, this does not converge.
Note that if $R_1 > R_2$, then $f$ converges and defines a holomorphic function on the annulus $R_2 < \abs{z-a} < R_1$.
Moreover, $f$ converges uniformly on any compact subset of this annulus, so it can be differentiated term-by-term, and the derivative has the same region of convergence.

Note that if $f$ equals its Laurent expansion, then $$c_n = \frac{f^{(n)}(a)}{n!} = \frac 1 {2\pi i} \int_\gamma \frac{f(\xi)}{\qty{\xi - a}^{n+1}} ~dz$$ where $\gamma$ is contained in the annulus of convergence, and $c_{n\leq -1}  0$.

We also have

\begin{align*}
\frac 1 {2\pi i} \int_\gamma \frac{f(z)}{(z-a)^m} ~dz 
&= \sum_{n\in \ZZ} \frac{c_n}{2\pi i} \int_\gamma \frac{1}{(z-a)^{m-n}}~dz \\
&=c_{m-1}
,\end{align*}

since
$$\int_\gamma \frac{1}{\qty{z-a}^k}~dz = \begin{cases} 2\pi i &k = 1 \\ 0 & \text{else}\end{cases},$$
we have the following formula for the coefficients:


\begin{align}
c_m &= \frac 1 {2\pi i} \int_\gamma \frac{f(z)}{\qty{z-a}^{m+1}} ~dz
.\end{align}

So we can start with a series and get a holomorphic function on some region.

## Obtaining a Laurent Series from a Holomorphic Function

We can also start with a holomorphic function and get a Laurent series.
Suppose $f$ is holomorphic on an annulus $R_2 < \abs {z} < R_1$.
We can then write

\begin{align*}
f(z) = \frac 1 {2\pi i} \int_{\abs{w-a} = R_1} \frac{f(w)}{w-z} ~dw - \int_{\abs{w-z} = R_2} \frac{f(w)}{w-z} ~dw
\end{align*}

![Image](figures/2020-02-17-14:07.png)\

Since $\abs{z-a}/\abs{w-a} < 1$, we have

\begin{align*}
\frac {1}{2\pi i} \int_{\abs{w-a} = R_1} \frac{f(w)}{w-z} ~dz 
&= \frac {1}{2\pi i} \int_{\abs{w-a} = R_1} \frac{f(w)}{(w-a) - (z-a)} ~dz \\
&= \frac {1}{2\pi i} \int_{\abs{w-a} = R_1} \frac{f(w)}{(w-a)} \sum_{n\in \NN} \frac{(z-a)^n}{(w-a)^n} ~dz \\
&= \sum_{n\in \NN} (z-a)^n \frac 1 {2\pi i} \int_{\abs{w-a} = R_1} \frac{f(w)}{(w-a)^{n+1}} ~dw \\
&= \sum_{n\in \NN} c_n (z-a)^n
.\end{align*}

Similarly,

\begin{align*}
-\frac 1 {2\pi i} \int_{\abs{w-a} = R_2}  \frac{f(w)}{w-z} ~dw  
&= -\frac 1 {2\pi i} \int_{\abs{w-a} = R_2} \frac{f(w)}{(w-a) - (z-a)} ~dw  \\ 
&= -\frac 1 {2\pi i} \frac {1}{z-a} \int_{\abs{w-a} = R_2} \frac {f(w)} {\frac{w-a}{z-a} - 1} ~dw  \\
&= \frac 1 {2\pi i} \frac {1}{z-a} \int_{\abs{w-a} = R_2}  \frac{f(w)} {1 - \frac{w-a}{z-a} } ~dw  \\
&= \frac 1 {2\pi i} \frac {1}{z-a} \int_{\abs{w-a} = R_2} f(w) \sum_{n\in \NN} \frac{(w-a)^n}{(z-a)^n} ~dw \\
&= \sum_{n\in\NN} \frac 1 {2\pi i} \frac {1}{(z-a)^{n+1}} \int_{\abs{w-a} = R_2} f(w) (w-a)^n ~dw  \\ 
&= \sum_{n=-\infty}^{-1} c_n (z-a)^n
.\end{align*}

This yields a formula

\begin{equation}
c_m = \frac {1}{2\pi i} \int_\gamma \frac{f(z)}{(z-a)^{m+1}} ~dz
.\end{equation}

In practice, we don't use this formula for extracting coefficients.

Example
: 	Let $f(z)  = \frac{1}{z(z-1)}$.
	This has four Laurent series.

	Let $C(a, R_1, R_2)$ be the annulus centered at $a$.
	Then at $C(0, 0, 1) = \DD\setminus \theset{0}$, we have 

	\begin{align*}
	f(z) = \frac 1 z \frac 1 {1-z} = -\frac 1 z \sum_{k\in \NN} z^k
	.\end{align*}

	In $C(1, 1, 0) = \DD(1, 1) \setminus\theset{1}$, we have

	\begin{align*}
	f(z) 
	&= \frac 1 {z-1} \frac 1 z \\
	&= \frac 1 {z-1} \frac 1 {1 + (z-1)} \\
	&= \frac 1 {z-1} \sum_{k\in \NN} (-1)^k (z-1)^k
	.\end{align*}

	In $C(0, 1, \infty)$, we can write

	\begin{align*}
	f(z) 
	&= \frac 1 {z^2} \frac 1 {1 - \frac 1 z} \\
	&= \frac 1 {z^2} \sum_{k\in \NN} \frac 1 {z^k}
	.\end{align*}

	And in $C(1,1,\infty)$ we have

	\begin{align*}
	f(z) &= \frac 1 {z-1} \frac{1}{z-1+1}
	.\end{align*}