# Friday February 28th

## Rouche, Open Mapping, and Maximum Modulus

Theorem (Rouche's Theorem)
:   Suppose $f, g$ are holomorphic in $D_r(z_0)$ and $\abs{f(z)} > \abs{g(z)}$ on $\bd D_r(z_0)$.
    Then $f$ and $f+g$ have the same number of zeros in $D_r(z_0)$.

Proof
: Let $f_t(z) = f(z) + tg(z)$ and use the argument principle.


Theorem (Open Mapping Theorem (Stein 4.4))
: If $f$ is holomorphic and nonconstant then $f$ is an open map.

Proof
: Let $w_0 \in \im(f)$ and say $f(z_0) = w_0$.
  We want to show that all $w$ near $w_0$ are also in $\im(f)$.
  Define $g(z) = f(z) - w = f(z) - w_0 + w_0 - w \definedas F(z) + G(z)$.

  Now choose $\delta>0$ such that $D_\delta(z_0) \subset \Omega$ and $f(z) \neq w_0$ on $\bd D_\delta(z_0)$.
  We then select $\delta$ such that $\abs{f(z) - w_0} \geq \eps > 0$ on $\bd D_\delta(z_0)$.
  We have $\abs{F(z)} = \abs{f(z) - w_0} \geq \eps$.
  
  Now choose $w$ such that $\abs{G(z)} = \abs{w-w_0} < \eps$, nothing that $G(z)$ is a constant function (?).
  Then $\abs{F(z)} \geq \eps > \abs{w-w_0} = \abs{G(z)}$.
  So apply Rouche's theorem and conclude that there exists $z\in D_\delta(z_0)$ such that $f(z) = w$.

> Qual alert, some questions related to the Open Mapping Theorem.

Theorem (Stein 4.5: Maximum Modulus)
: If $f$ is holomorphic and nonconstant on $\Omega$, then $\abs{f}$ can not attain a maximum in $\Omega$.

Proof
:   Suppose toward a contradiction that $\abs{f}$ attains a maximum in $\Omega$, say at $z_0$. 
    Since $f$ is holomorphic, it is an open mapping, and therefore if $D_\delta(z_0) \subset \Omega$ then $f(D_\delta(z_0))$ contains a disc.
    Thus there exists a $z\in D_\delta(z_0)$ such that $\abs{f(z)} > \abs{f(z_0)}$.
    But this contradicts maximality of $f$ at $z_0$.
  
Corollary
:   If $\abs{f(z)} = 0$ on $\bd U$ and is nonconstant, then $f$ has a zero in $U$.

Proof
:   Let $c = \abs{f(z)}$ for $z\in \bd U$.
    Suppose that $f(z)$ has no zeros in $U$.
    Then $g(z) = \frac{1}{f(z)}$ is continuous and holomorphic in $U$.
    Then for all $z_0 \in U$, $\abs{g(z)} = \frac{1}{\abs{f(z)}} = \frac{1}{\abs{f(z_0)}} > \frac 1 c$, since $c = \abs{f(z)}$ for $z\in \bd U$ implies $\abs{f(z_0)} < C$.
    But this contradicts the maximum principle.

> Proof technique: use the fact that the reciprocal is holomorphic.
> Note that this is stronger than $f$ just being smaller in the interior, the modulus actually takes on the smallest value.

## The Complex Logarithm

For $x>0$, we define $\log(x) = \int_1^x \frac 1 t ~dt$, which is the inverse of $e^x$.
For $z\neq 0$, we'd like to define $\log(z) = \log(re^{i\theta}) = \log(r) + i \theta$, but the argument $\theta$ is not uniquely defined.

Theorem (Existence of Logarithm)
:   Suppose $\Omega$ is simply connected with $1\in \Omega$ and $0\not\in\Omega$.
    Thin in $\Omega$, there is a branch of the logarithm $F(z) = \log_\Omega(z)$ such that

    1. $F(z)$ is holomorphic on $\Omega$.
    2. $e^{F(z)}  = z$ for all $z\in \Omega$
    3. $F(r) = \log(r)$ for all $r>0\in \RR$ near 1.

Proof
:   **Part 1**:

    We define $F(z)$ as a primitive of the function 

    \begin{align*}
    F(z) = \int_\gamma \frac 1 w ~dw
    .\end{align*}

    where $\gamma$ is any curve in $\Omega$ connecting $1$ and $z$.
    We have 

    \begin{align*}
    \frac{dF}{dz} = \frac 1 z = \lim_{h\to 0} \frac{F(z+h) - F(z)}{h}
    .\end{align*}

    Noting that $F(z+h) - F(z) = \int_\eta \frac 1 w ~dw$, we can parameterize $\eta$ as $w = (1-s) z + s(z+h) = z + sh$.
    Then

    \begin{align*}
    \int\eta \frac{1} {w} ~dw &= \int_0^1 \frac{1}{z+sh} ~hds \\
    \implies \frac{1} {h} \int_0^1 \frac {1} {w} ~dw 
    &= \int_0^1 \frac{1}{z + sh} ~ds \\
    &= \int_0^1 \qty{\frac {1} {z} + \frac{1}{z + sh} - \frac {1} {z} }~ds \\
    &= \frac {1} {z} - \frac {h} {z} \int_0^1 \frac{d}{z+sh} ~ds \\
    &\converges{h\to 0}\to \frac {1} {z}
    .\end{align*}

    **Part 2**:

    Note that $\qty{ z e^{F(z)} }' = e^{F(z)}  +ze^{-F(z)} \qty{- \frac 1 z } = 0$.

    Part 3:

    To do.
  
Next time: once we have the log we can say more about the argument principle.