# Wednesday January 29th

## Cauchy's Integral Formula

Theorem (Cauchy's Integral Formula)
: 	Let $f: \Omega \to \CC$ be holomorphic, so $f\in C^1(\bar \Omega)$.
	Then for any $z\in \Omega$,

	\begin{align*}
	f(z) = \frac{1}{2\pi i} \int_{\bd \Omega} \frac{f(\xi)}{\xi - z} ~d\xi
	.\end{align*}

	In general,
	\begin{align*}
	f^{(n)}(z) = \frac{n!}{2\pi i} \int_{\bd \Omega} \frac{f(\xi)}{\qty{\xi - z}^{n+1}} ~d\xi
	.\end{align*}

This implies that $f$ is analytic, i.e. 
$$
f(z) = \sum a_n (z-z_0)^n \quad\text{where}\quad a_n = \frac{f^{(n)}(z_0) }{n!}
.$$
Thus $f$ is holomorphic iff $f$ is analytic,

and

\begin{align*}
\int_{\bd \Omega} f = 0 \implies \int_{\bd \Omega_\gamma} \frac{f(\xi)}{\xi - z}~d\xi = 0
.\end{align*}

where $\Omega_r = \Omega\setminus D_r(z)$, and $\bd \Omega_r = \bd \Omega \union (-\bd D_r)$.

We can thus shrink integrals:

\begin{align*}
\int_{\bd \Omega} \frac{f(\xi)} {\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)} {\xi - z} ~d\xi
.\end{align*}

Proposition (Homotopy Invariance)
: 	Let $f\in C^1(\Omega)$ be holomorphic on $\Omega$.
	Let $\gamma_s(t)$ be a family of smooth curves in $\Omega$; then $\int_{\gamma_s} f$ is independent of $s$.

Proof
: 	Write $$\gamma_s(t) = \gamma(s, t): [a, b] \cross [0, 1] \to \Omega.$$
	
	We have $\gamma_s(0)  = \gamma_s(1)$ so $\dd{\gamma}{s}(s, 0) = \dd{\gamma}{s}(s, 1)$.
	Then 

	\begin{align*}
	\dd{\gamma}{s} 
	&= \int_0^1 \qty{ f'(r(s, t)) \dd{r}{s} \dd{r}{t} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial s \partial t} } ~dt \\
	&= \int_0^1 \qty{ f'(r(s, t)) \dd{r}{s} \dd{r}{t} + f(r(s, t)) \frac{\partial^2 \gamma}{\partial \mathbf{t} \partial \mathbf{s}} } ~dt \\
	&= \int_0^1 \dd{}{t} \qty{ f(\gamma(s, t))  \gamma_s } \\
	&= f(\gamma(s, 1))\gamma_s(s, 1) - f(\gamma(s, 0)) \gamma_s(s, 0) \\
	&= 0
	.\end{align*}

	where we can just take the paths $\gamma(s, t) = z_0 \in \Omega$ for all $s, t$.

Proposition (Pointwise Limit of Locally Uniform is Locally Uniform)
: 	Let $\Omega \subset \CC$ be open and $f_v: \Omega \to \CC$.
	Suppose that each $f_v$ is holomorphic, $f_v \to f$ pointwise, and *locally uniform*, i.e. $f_v \to f$ uniformly on every compact $K \subset \Omega$.
	Then $f$ is holomorphic in $\Omega$ and $f$ is locally uniform.

Proof 
: 	Given a compact set $K \subset \Omega$, pick an $O$ with smooth boundary such that $K \subset O \subset \bar O \subset \Omega$.
	We have

	\begin{align*}
	f_v(z) &= \frac{1}{2\pi i} \int_{\bd O} \frac{f_v(\xi)}{\xi - z}~d\xi \\
	f_v^{(n)}(z) &= \frac{n!}{2\pi i} \int_{\bd O} \frac{f_v(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\
	.\end{align*}

	Then on $\bd O$, we have uniform convergence

	\begin{align*}
	\frac{f_v(\xi)}{\qty{\xi - z}^{n+1}} \converges{u}\to
	\frac{f(\xi)}{\qty{\xi - z}^{n+1}}
	.\end{align*}

	By moving the limits inside, we obtain

	\begin{align*}
	f(z) &= \frac{1}{2\pi i} \int_{\bd O} \frac{f(\xi)}{\xi - z}~d\xi \\
	f^{(n)}(z) &= \frac{n!}{2\pi i} \int_{\bd O} \frac{f(\xi)}{\qty{ \xi - z}^{n+1} }~d\xi \\
	.\end{align*}

Theorem (Cauchy's Inequality)
: 	Given $z_0\in \Omega$, pick the largest disc $D_R(z_0) \subset \Omega$ and let $C_R = \bd D_R$.
	Using the integral formula, defining $\norm{f}_{C_R} = \max_{\abs{z-z_0} = R} \abs{f(z)}$

	\begin{align*}
	\abs{ f^{(n)}(z_0) } \leq \frac{n!}{2\pi} \int_0^{2\pi} \frac{\norm{f}_{C_R}}{R^{n+1}} R~d\theta = \frac{n! \norm{f}_{C_R}}{R^{n}}
	.\end{align*}

Corollary (Liouville's Theorem)
: 	If $f$ is entire and bounded, then $f$ is constant.

Proof
: 	For all $z_0 \in \CC$, there exists an $M$ such that $\abs{f(z)} \leq M$.
	Then $\abs{f'(z_0)} \leq \frac{M}{R}$ for any $R> 0$.
	Taking $R\to \infty$ yields $f'(z_0) = 0$, so $f$ is constant.

Corollary (Weak Fundamental Theorem of Algebra)
: 	Every non-constant polynomial $p(z) = a_0 + a_1z + \cdots a_n z^n$ has a root in $\CC$.

Remark: A general proof technique is when proving something for $f(z)$, consider $\frac{1}{f(z)}$ and $f(\frac 1 z)$.

Proof
: 	Suppose $p$ is nonconstant and does not have a root, $\frac 1 p$ is entire.
	Assume that $a_n \neq 0$, then

	\begin{align*}
	\frac{p(z)}{z^n} = a_n \qty{ \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}  } \definedas a_n + y
	\end{align*}

	We can note that $\lim_{z\to \infty} \frac{a_{n-k}}{z^k} \to 0$, so there exists an $R>0$ such that 

	\begin{align*}
	\abs{ \frac{p(z)}{z^n}  } \geq \frac 1 2 \abs{a_n} \quad \text{ for } \abs{z} > R \\
	\implies \abs{p(z)} \geq \frac 1 2 \abs{a_n} \abs{z}^n \geq \frac 1 2 \abs{a_n} R^n
	.\end{align*}

	Since $p(z)$ is continuous and has no root in the disc $\abs{z} \leq R$, $\abs{p(z)}$ is bounded from below in this disc.
	Since $p(z)$ is continuous on a compact set, it attains a minimum, and so $\abs{p(z)} \geq \min_{\abs z \leq R} \abs{p(z)} = c_2 \neq 0$.
	Then $\abs{p(z)} \geq A = \min(C_2, \frac 1 2 \abs{a_n}R^n)$, so $\frac{1}{p}$ is bounded.
	Then $f$ is constant, a contradiction.