Suppose \(f:\DD\to\DD\) is analytic, has a single zero of order \(k\) at \(z=0\), and satisfies \(\lim_{\abs z \to 1} \abs{f(z)} = 1\). Give with proof a formula for \(f(z)\).
Solution
Since \(f\) has a zero of order \(k\) at \(0\), the quotient \[ g(z)={f(z)\over z^k} \] extends holomorphically across \(0\). Because \(0\) is the only zero of \(f\), this extension is nowhere zero on \(\DD\).
Moreover, \[ |g(z)|={|f(z)|\over |z|^k}\longrightarrow1 \qquad (|z|\to1). \] We claim \(|g|\le1\) on \(\DD\). Fix \(z_0\in\DD\) and \(\varepsilon>0\). By the boundary limit there is \(r_0<1\) such that \[ |g(z)|<1+\varepsilon \qquad(r_0<|z|<1). \] Choose \(r\) with \(\max\{r_0,|z_0|\}<r<1\). The maximum modulus principle on \(|z|\le r\) gives \[ |g(z_0)|\le \max_{|z|=r}|g(z)|<1+\varepsilon. \] Since \(\varepsilon\) is arbitrary, \(|g(z_0)|\le1\).
Apply the same argument to the holomorphic function \(1/g\). Since \[ |1/g(z)|\longrightarrow1 \qquad(|z|\to1), \] we get \(|1/g|\le1\), hence \(|g|\ge1\). Therefore \(|g|\equiv1\) on \(\DD\).
By the open mapping theorem, a holomorphic function with image contained in \(S^1\) is constant. Thus \[ g(z)=\lambda, \qquad |\lambda|=1, \] and consequently \[ f(z)=\lambda z^k. \]