Let \(f(z)=\sum _{n=0}^{\infty }a_nz^n\). Assume that \(a_0=0\) and \(a_1=1\). Prove that \(f\) is one to one on the unit disk \(U=\{z\in \mathbb C:|z|<1\}\) if \(\sum _{\ell =2}^{\infty }|a_\ell |\le 1\). (Hint: Show that for every \(z_0\in U\) the function \(z\to f(z)-f(z_0)=z-z_0+\sum _{n=2}^{\infty }a_n(z^n-z_0^n)\) has a simple zero in \(U\).)
Question
Math 6210 • Real Analysis
Fall 1994 • Question 6