# Friday January 31st

## Fundamental Theorem of Algebra

Recall that if $f$ is holomorphic, we have Cauchy's integral formula.

Corollary (Weak Fundamental Theorem of Algebra)
: 	If $P(z)$ is a polynomial in $\CC$ then $P$ has a root in $\CC$.

Proof
: 	See previous notes.

Corollary (Fundamental Theorem of Algebra)
: 	Every polynomial of degree $n$ has precisely $n$ roots in $\CC$.

Proof
: 	By induction on the degree of $P$.
	From the first corollary, $P$ has a root $w_1$, so write $z = z-w_1 + w_1$.
	Then

	\begin{align*}
	p(z) 
	&= p(z - w_1 + w_1) \\
	&= \sum_k^n a_k(z -w_1 + w_1)^k \\
	&= \sum_k^n a_k \sum_{j}^k {k\choose j} w_1{k-j} (z-w_1)^j \\
	&= \sum_k^n \sum_j^k a_k {k\choose j} w_1^{k-j} (z-w_1)^j \\
	&= \sum_j^n \qty{\sum_{k\geq j} a_k {k\choose j}}(z-w_1)^j \\
	&= b_0 + b_1(z-w_1) + \cdots + b_n(z-w_1)^n 
	.\end{align*}

	Since $P(w_1) = 0$, we must have $b_0 = 0$, and thus this equals

	\begin{align*}
	b_1(z-w_1) + \cdots + b_n(z-w_1)^n 
	&= (z-w_1) \qty{ b_1 + \cdots + b_n (z-w_1)^{n-1}  } \\
	&\definedas  (z-w_1) \phi(z)
	,\end{align*}

	where $\phi(z)$ is degree $n-1$, which has $n-1$ roots by induction.

Definition (Characterizations of Limit Points)
: 	For a sequence $\theset{z_n}$, TFAE

	1. $z$ is a limit point.
	2. There exists a subsequence $\theset{z_{n_k}}$ converging to $z$.
	3. For every $\eps > 0$, there are infinitely many $z_i$ in $D_\eps(z)$. 

Theorem (Only the zero function vanishes on a sequence in a domain (Stein 4.8))
: 	Suppose $f$ is holomorphic on a bounded connected region $\Omega$ and $f$ vanishes on a sequence of distinct points with a limit point in $\Omega$. 
		Then $f$ is identically zero.

Proof
: 	WLOG by restricting to a subsequence, suppose that $\theset{w_k} \in \Omega$ with $f(w_i) = 0$ for all $i$ and $z_0$ is a limit point of $\theset{w_i}$.
	Let $U = \theset{ z\in \Omega \suchthat f(z) = 0 }$.
	Then

	1. $U$ is nonempty since $f(w_k) = f(z_0) = 0$.
	2. Since holomorphic functions are continuous, if $w_k \to z$ then $z\in U$, so $U$ is closed. 
	3. (To prove) $U$ is open.

	Since $U$ is closed and open, $U = \Omega$.

	We will first show that $f(z) \equiv 0$ in a disk containing $z_0$.
	Choose a disc $D$ containing $z_0$ and contained in $\Omega$.
	Since $f$ is holomorphic on $D$, we can write $$f(z) = \sum a_nn (z-z_0)^n.$$
	Since $f(z_0) = 0$, we have $a_0 = 0$.

	Suppose $f\not\equiv 0$.
	Then there exists a smallest $n\in \ZZ^+$ such that $a_n \neq 0$, so $f(z) = a_n(z-z_0)^n + \cdots$.
	Since $a_n \neq 0$, we can factor this as $a_n(z-z_0)^n \qty{ 1 + g(z-z_0)  }$ where $$g(z-z_0) = \sum_{k=n+1}^\infty \frac{a_k}{a_n} (z-z_0)^{k-n}.$$
	Note that $g$ is holomorphic, and $g(z_0 - z_0) = 0$.

	Choose some $w_k$ such that $f(w_k) = 0$ and $\abs{g(w_k - z_0)} \leq \frac 1 2$ by continuity of $g$.
	Then $$\abs{1 + g(w_k - z_0)} > 1 - \frac 1 2 = \frac 1 2.$$
	
	So $$\abs{f(w_k)} = \abs{ a_n(w_k - z_0)^n \qty{1 + g(w_k - z_0)  }  } > \abs{a_n} \abs{w_k - z_0}^n \frac 1 2 > 0,$$ a contradiction.
	So $U$ is open, closed, and nonempty, so $U = \Omega$.

Corollary
: 	Suppose $f, g$ are holomorphic in a region $\Omega$ with $f(z_k) = g(z_k)$ where $\theset{z_k}$ has a limit point.
	Then $f(z) \equiv g(z)$.

Theorem (Mean Value)
: 	Let $z_0$ be a point in $\Omega$ and $C_\gamma$  the boundary of $D_r(z_0)$. 
	Then

	\begin{align*}
	f(z_0)
	&= \frac{1}{2\pi i} \int_{C_\gamma} f(z)/(z-z_0) dz \\
	&= \frac {1}{2\pi i} \int_0^{2\pi} f(z_0 + re^{i\theta})/re^{i\theta} r i e^{i\theta} ~d\theta \quad\text{by } z = z_0 + re^{i\theta} \\
	&= \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})~d\theta \\
	&= \frac{1}{2\pi r} \int_0^{2\pi} f(z_0 + re^{i\theta})~rd\theta \\
	&= \frac{1}{\abs{C_\gamma}} \int_0^{2\pi} f(z) ~ds
	,\end{align*}

	which is the average value of $f$ on the circle.

Note that there is another formula that averages over the disc (see book for derivation?)

\begin{align*}
f(z_0) &= \frac{1}{D_s(z_0)}  \int_{P_s} \int_{D_s} f(z) ~dA
.\end{align*}

These imply the maximum modulus principle, since the average can not be the max or min unless $f$ is constant.
Note that $\abs{f(z)}$ is continuous!

Next time: maximum modulus principle.