# Friday January 17th

## Antiholomorphic Derivative

> Reference:
> See Lang's Complex Analysis, there are plenty of solution manuals.
> Note: look for 13 statements equivalent to holomorphic: Springer GTM Lipman.

Let $f; \Omega \to \CC$ be a complex-valued function.
Recall that $f$ is *complex differentiable* iff the usual ratio/limit exists.
Note that $h = x+iy$ and $h\to 0 \iff x,y\to 0$.

We can write $$f'(z) = \dd{f}{x} = \frac 1 i \dd{f}{y}.$$
This follows from Cauchy-Riemann since $u_x = v_y$ and $u_y = -v_x$.


We want to define $\del, \bar \del$ operators.
We have the identities

\begin{align*}
x = \frac{z + \bar z}{z} \quad y = \frac{z - \bar z}{iz}
.\end{align*}

We can then write

\begin{align*}
dz &= dx + idy \\
d\bar z &= dx - i dy
.\end{align*}


We define the dual operators by $\inner{\dd{}{z}}{dz} = 1$ and similarly $\inner{ \dd{}{\bar z}  }{d\bar z} = 1$.

By the chain rule, we can write

\begin{align*}
f_z &= \dd{f}{x} \dd{x}{z} +  \dd{f}{y} \dd{y}{z} \\
&= \frac 1 2 \dd{f}{x} + \dd{f}{y} \frac{1}{2i} \\
&= \frac 1 2 \qty{\dd{}{x} + i \dd{}{y} }f 
,\end{align*}

and similarly 
\begin{align*}
f_{\bar z} 
&=  \dd{f}{x} \dd{x}{\bar z}  + \dd{f}{y}  \dd{z}{\bar z} \\
&= \frac 1 2 \qty{ \dd{}{x} - \frac{1}{2i} \dd{}{y}  }f
.\end{align*}

We thus find $\del_x = \del_z + \del_{\bar z}$ and $\del_y = i\qty{ \del_z - \del_{\bar z}  }$,
so define

\begin{align*}
\del f &\definedas \dd{f}{z} dz \\
\bar{\del} f &\definedas \dd{f}{\bar z} d\bar z \\
\implies df &= \dd{f}{z} dz + \dd{f}{\bar z} d\bar z
.\end{align*}

Definition (Holomorphic and Antiholomorphic Derivatives)
: 
	\begin{align*}
	\bd f &= \frac 1 2 \qty{\dd{}{x} + i \dd{}{y} }f\\
	\bar \bd f &= \qty{ \dd{}{x} - \frac{1}{2i} \dd{}{y}  }f
	.\end{align*}


Proposition (Holomorphic Functions have vanishing antiholomorphic derivatives)
: 	$f$ is holomorphic iff $\bar \del f = 0$.

This means that $f$ depends on $z$ alone and not $\bar z$.

Proof
: $\bar \del f = 0$ iff $\frac 1 2 (f_x + if_y) = 0$, so $(u_x  - v_y) + i (v_x + u_y) = 0$.

Application to PDEs:
we can write $$u_{xx} = v_{xy} \quad u_{yy} = v_{yx}$$ and so $$u_{xx} + u_{yy} = 0 = v_{xx} + v_{yy}.$$

Thus $\Delta f = 0$, sp $f$ satisfies Laplace's equation and is said to be *harmonic*.

Corollary (Holomorphic Functions Have Harmonic Components)
: 	If $f$ is analytic, then $u, v$ are both harmonic functions.

Theorem (Chain Rule)
: 	Let $w = f(z)$ and $g(w) = g(f(z))$.
	Then

	\begin{align*}
	h_z &= g_w f_z + g_{\bar w} \bar f_z \\
	h_{\bar z} &= g_w f_{\bar z} + g_{\bar w} \bar f_{\bar z}
	.\end{align*}

	If $f, g$ are holomorphic, $f_{\bar z} = g_{\bar w} = 0$, so $h_{\bar z} = 0$ and $h$ is holomorphic and $$h_z = g_w f_z.$$

Example
: 	Given a power series $f=\sum a_n (z- z_0)^n$.
	Then

	1. There exists a radius of convergence $R$ such that $f$ converges precisely on $D_R(z_0)$.
	2. $f$ is continuous on $D_R(z_0)^\circ$.
	3. By the root test, $R = (\limsup \abs{a_n}^{1/n})\inv = \liminf \abs{a_n/a_{n+1}} = (\limsup \abs{a_{k+1}/a_k})\inv$.

Recall the **ratio test**: 
$$\sum \abs{a_k} <\infty \iff \limsup \abs{a_{k+1} / a_k} < 1$$

Theorem (Holomorphic series can be differentiated term-by-term)
: 	If $f(z) = \sum_{n=0} a_n z^n$ is holomorphic on $\abs{z} < R$ for $R> 0$ then $$f'(z) = \sum_{n=1} a_n n z^{n-1}.$$

Proof
: 	Given $\abs z < R$, fix $r>0$ such that $\abs {z} < r < R$.
	Suppose that $\abs{w-z} < r - \abs{z}$, so $\abs w < r$.

	![Image](figures/2020-01-17-14:14.png)\

	We want to show 
	\begin{align*}
	\abs{S} = \abs{\frac{f(w) - f(z)}{w - z} - \sum_{n=1} a_n n z^{n-1}} \to 0 \quad \text{as } w\to z
	.\end{align*}

	Idea: write everything in terms of power series.
	Use the fact that $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots)$, and so $\abs{(w^k-z^k)/(w-z)} \leq k r^{k-1}$.

	\begin{align*}
	S 
	&= \sum_{n=1} a_n \qty{ \frac{w^n - z^n}{w-z} - n z^{n-1}  } \\
	&= \sum a_n \qty{ w^{n-1} + w^{n-2}z + \cdots + z^{n-1} + nz^{n-1} } \\
	&=  \sum a_n \qty{ (w^{n-1} - z^{n-1}) + (w^{n-2} - z^{n-2})z + \cdots + (w-z) z^{n-2} }
	&= \sum a_n (w-z) \qty{ \cdots + z^{n-2} }\\
	&\leq \sum_{n=2} \abs{a_n} \frac 1 2 n(n-1) r^{n-2} \abs{z-w}
	.\end{align*}

Exercise
: Show $\lim_n n^{\frac 1 n} =  1$.
	
	Also tricky: show $\lim \sin(n)$ doesn't exist, and $\sin(n)$ is dense in $[-1, 1]$.
	
	Proof
	: Consider $\limsup \abs{a_n n}^{\frac 1 n}$.

Note that an analytic function is holomorphic in its domain of convergence, so analytic implies holomorphic.
The converse requires Cauchy's integral formula.

Next time: trying to prove holomorphic functions are analytic.