## Analytic function

### Set

 context $\mathcal O\subset \mathbb C$ definiendum $f\in \mathrm{it}$ inclusion $f:\mathcal O\to\mathbb C$ for all $c$ … series in $\mathbb C$
todo, roughly

$\exists c.\ \forall z.\ f(z)=\sum_{n=-\infty}^\infty c_n\,z^n$

#### Discussion

Picture a continuous function $f:\mathbb R^2\to\mathbb R$ as a surface given by $f(x,y)$ and imagine drawing a circle of radius $1$ around the point at the origin and is parametrized by $\langle \cos\theta,\sin\theta,h\rangle$ where $\theta\in[0,2\pi)$ and $h\in[0,f(\cos\theta,\sin\theta))$. See the picture below. It looks like a cylinder cut of at height $f(\cos\theta,\sin\theta)$. Let's call it a “fence”. What is the surface of that fence? Clearly, it's given by the integral $\int_0^{2\pi}\mathrm{d}\theta$ of $f(\cos\theta,\sin\theta)$. And so the average height (if we count negative height as negative contributions) of the fence is

$\frac{1}{2\pi}\int_0^{2\pi}f(\cos\theta,\sin\theta)\,\mathrm{d}\theta$

For example, a parabola $x^2+y^2$ has average height of $1$ and a tilted plane like $7x+3y$ always has average height of $0$.

As a remark, we can trivially extend the definition to compute the fence height of a fence with radius $R$ at a point $p=\langle p_x,p_y\rangle$ by shifting an scaling the integral

$\frac{1}{2\pi R}\int_0^{2\pi}f(R\cos\theta+p_x,R\sin\theta+p_y)\,\mathrm{d}\theta$

A complex function $f(z)$ consists of two real functions, so it's fence height is just given by the sum of the fence heigh of $\mathrm{Re}\,f(z)$ and $i\,\mathrm{Im}\,f(z)$. Let's consider the function $f(z):=z=r\,\cos\theta+i\,r\,\sin\theta$. We have

$z^n=r^n\,\mathrm{e}^{i\,n\,\theta}=r^n\,\cos(n\,\theta)+i\,r^n\,\sin(n\,\theta)$

Both real and imaginary part oscillate along $\theta$. So from the plots below alone it is obvious that the average fence height for $n\neq 0$ must be zero

$n\neq 0\implies\frac{1}{2\pi}\int_0^{2\pi}z^n\mathrm{d}\theta=0$

and if $n=1$, then it’s clearly $1$.

Note that we can compute the real and the imaginary fence height at once. The circle is parametrized by $z:=\mathrm{e}^{i\,\theta}$ and so we can express the infinitesimal length in $\mathbb R^2$ as $\mathrm{d}\theta=\frac{1}{i}\frac{\mathrm{d}z}{z}$. The factor $\frac{1}{z}$ corrects the orientation of $\mathrm{d}z$, it cancels out the complex mixing of components introduced by walking along the complex plane. The fence height of $z\cdot f(z)$ is called the residue and equals $\frac{1}{2\pi\,i}\oint_\gamma f(z)\ \mathrm d z$. In this language, the picture saying that only the fence height of the constant function $z^0$ isn't zero is the message that $\frac{1}{z}$ is the special function with non-vanishing line integral.

Now consider an analytic function, i.e. a function which can be written by a countable series with coefficients $c_n\equiv a_n+i\,b_n$

$f(z)=\sum_{n=-\infty}^\infty c_nz^n=\sum_{n=-\infty}^\infty \left(a_n+i\,b_n\right)r^n\,\mathrm{e}^{i\,n\,\theta}$

Explicitly separating real and imaginary parts, this reads

$\mathrm{Re}\,f(z)=\sum_{n=-\infty}^\infty \left(a_n\cos(n\,\theta)+b_n\sin(n\,\theta)\right)r^n$

$\mathrm{Im}\,f(z)=\sum_{n=-\infty}^\infty \left(b_n\cos(n\,\theta)+a_n\sin(n\,\theta)\right)r^n$

Now it’s clear that analyticity is a strong restriction: They are just a two real functions where, for every $n\neq 0$, their two coefficients oscillate along $\theta$, and even more, they even all have fence height zero. This implies that $f$'s coefficient $c_0$ is already the fence height of all of $f$. In fact, this implies that $f(z)=\sum_{n=-\infty}^\infty c_nz^n$ itself is determined by the fence heights of of the functions $z^k\,f(z)$! So for analytic functions we have $c_n = \frac{1}{2\pi\, i} \oint_\gamma \frac{f(z)}{z^n}\, \frac{\mathrm dz}{z}$, or by shifting the fences to $p$ we get

##### Cauchy's integral formula

$\frac{1}{n!}f^{(n)}(p) = \frac{1}{2\pi\, i} \oint_\gamma \frac{f(z)}{(z-p)^{n+1}}\, \mathrm dz$

Roughly, the Laplace transform uses this for a re-encoding of a functions $f:\mathbb R^+\to\mathbb R$ with Taylor expansion $f(t)=\sum_{n=0}^\infty a_n t^n$, namely by mapping $t^n$ to $s^{-n}\cdot \frac{1}{s}$.