# 3423/3424

Revision as of 13:26, 3 June 2012; view current revision

3423/3424: Complex Analysis I/II

Lecturer: Dr. Dmitri Zaitsev

## Contents

### The course

The course focuses around several main theorems, both proof (given below) and applications. There is on average four assignments between both semesters. The exam is worth a 100% of the course.

## Outline of Riemann Mapping Theorem

Just learn the bold points for a minimal outline.

• Reduce to case $\Omega$ is bounded, using fact $\Omega$ is not all of $\mathbb{C}$

As $\Omega \neq \mathbb{C}$ $\exists b \in \mathbb{C}, b \not \in \Omega$, so $g(z) = z-b \neq 0$ on $\Omega$, so there is a square root $h(z)^2 = g(z)$, injective as $g$. We can show there is $z_0$ such that $B_\delta(z_0) \subset h(\Omega)$ but $B_\delta(-z_0) \not \subset h(\Omega)$, let $H(z) = 1/(h(z) +z_0)$, then $H(\Omega)$ is bounded (by $1/\delta$).

• Create a family $\mathcal{F}$ of injective maps taking $a\in \Omega$ to 0

Consider $\Omega$ bounded, let $\mathcal{F} = \lbrace f: \Omega \rightarrow \Delta | f(a) = 0, f$ injective $\rbrace$. Let $A = \sup_{f \in \mathcal{F}} |f^\prime(a)|$, there exists a sequence $f_n$ such that $|f_n^\prime(a)| \rightarrow A$, obviously $|f_n|\leq 1$ so $f_n$ uniformly bounded, so...

• Apply Montel's theorem to find a compactly convergent subsequence in $\mathcal{F}$ converging to $f$ with the maximum possible $|f^\prime(a)|$

... Montel's theorem implies there is a compactly convergent subsequence $f_{n_k}$ tending to $f$ such that $|f^\prime(a)| = A$.

• Show $f$ is injective (using Rouche's theorem)

Assume $f$ not injective, let $z_1 \neq z_2$ but $f(z_1) = f(z_2) = w$, let $h(z) = f(z) - w$, $h_{n_k}(z) = f_{n_k}(z) - f_{n_k}(z_2)$. Look at a disc $\Delta_\varepsilon(z_1)$ not containing $z_2$, suppose $h$ is non-zero on boundary of disc and note can choose $k$ so that $|h_k(z) - h(z)| < \delta = \min_{\Delta_\varepsilon(z_1)} |h|$ (by uniform convergence of $h_k$ to $h$), Rouche then implies $h_k(z)=0$ in disc, injectivity of $f_k$ then says $z_2$ in disc, contradiction.

• Show $f$ is surjective, by supposing not and then constructing a function $F \in\mathcal{F}$ satisfying $|F^\prime(a)| > |f^\prime(a)|$

If no surjective, can find $b \in \Delta$ not in image of $f$. Consider $g(z) = \varphi_b \circ f$, non-zero on $\Omega$ so has a square root $h$, let $h(a) = c$, consider $F(z) = \varphi_c \circ h$, sends $a$ to zero and is injective, an explicit calculation shows $|F^\prime(a)|>|f^\prime(a)|$, a contradiction.