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

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