Koebe Theorem Note 3

We have seen from the post that about the Cartan Theorem, with the univalent funtion’s Laurent expansion as:

$f(z)=z+a_2 z^2+\cdots$,

where $z\in \Omega$, which is the unit disk.

We call the family of univalent functions with above expansion is $\mathcal{S}$ family.

First, we prove that if the normalized function:

$f(z)=z+a_2 z^2+\cdots \in \mathcal{S}$, then

$|a_2|\le 2$.

Proof:

Consider $h(z)=\displaystyle \frac{f(z)^2}{z^2}$, then

$h(z)=1+a_2z^2+a_3 z^4+\cdots+a_n z^{2n-2}+\cdots$

is nonzero in $\Omega$ and is an even function.

Thus we can choose one branch of $\sqrt{h(z)}$ such that $\sqrt{h(0)}=1$. Also we have

$\phi=\sqrt{h(z)}$ is an even function, consider $f_1(z)=z\phi(z)$, then $f_1$ is odd and univalent in $\Omega$.

Consider the expansion of $h(z)$, we have

$\displaystyle f_1(z)=z+\frac{a_2}{2}z^2+\cdots$, thus $f_1\in \mathcal{S}$.

Assume that $\displaystyle g_0(z)=[f_1(\frac{1}{z})]^{-1}$, then

$g_0(z)=z-\displaystyle \frac{a_2}{2}z^{-1}+\cdots$, where $z\in \Omega$.

Because $g_0(z)$ is univalent, then we have:

$g_0$ maps $\partial B(0,r)$ into a close Jordan curve in complex plane. Consider the area then:

$|a_2|\le 2$.

$\Box$

Let’s prove the Koebe $\frac{1}{4}$ Theorem:

Consider

$K(z)=\displaystyle\frac{z}{(1-z)^2}=z+2z^2+\cdots+nz^n+\cdots$

which is univalent in $\Omega$, which maps $\Omega\rightarrow \mathbb{C}-(-\infty,-\frac{1}{4}]$.

Then $K(z)$ satisfies the theorem.

Now we start with our function $f(z)$, which is univalent in unit disk $\Omega$, and $f(0)=0, f'(0)=1$.

then think about $F(z)=\displaystyle\frac{cf(z)}{c-f(z)}$, where $c\,\,\overline{\in} f(\Omega)$. Then we have that $F(z)\in \mathcal{S}$, which means $\displaystyle|a_2+\frac{1}{c}|\le 2$.

And use the clue that $|a_2|\le 2$. We have $\displaystyle |c|\ge \frac{1}{4}$.

$\Box$