Problem 1786 (difficulty: 7/10)

Assume that \(\displaystyle f\in O(|z|<1)\) has image \(\displaystyle \re z>0\), and \(\displaystyle f(0)=1\). Show that

\(\displaystyle \frac{1-|z|}{1+|z|}\le |f(z)|\le \frac{1+|z|}{1-|z|}.\)

Solution:

The linear fraction \(\displaystyle g(w)=\frac{1-w}{1+w}\) maps the right half plane to the unit disk such that \(\displaystyle g(1)=0\).

By applying Schwarz's lemma to the function \(\displaystyle w\mapsto g(f(w))\), we have \(\displaystyle |g(f(z))|\le |z|\).

The function \(\displaystyle g^{-1}(w)=\frac{1-w}{1+w}\) maps the disk \(\displaystyle |w|\le|z|\) to the closed disk with diameter \(\displaystyle \left(\frac{1-|z|}{1+|z|},\frac{1+|z|}{1-|z|}\right)\), so \(\displaystyle f(z)=g^{-1}(g(f(z))\) lies in that disk.

Obviously, the disk lies in \(\displaystyle \frac{1-|z|}{1+|z|}\le|w|\le \frac{1+|z|}{1-|z|}\).

Give me another random problem!