- Find all entire functions f such that \(| f ( z ) | = 1\) whenever \(| z | = 1\) . Give explicit formulas for the functions and give a proof for your answer. (An entire function is a holomorphic function on C.)
Solution
The functions are exactly \[ \boxed{f(z)=\alpha z^n,\qquad |\alpha|=1,\quad n=0,1,2,\ldots.} \]
1. Removing all zeros in the unit disk leaves a constant-modulus quotient.
The boundary condition implies that \(f\) is not identically zero and has no zero on the unit circle. It has finitely many zeros in the closed unit disk: otherwise compactness would give an accumulation point in \(\mathbb C\), contradicting the identity theorem [@SS03]. List its interior zeros as \(a_1,\ldots,a_n\), repeated according to multiplicity. The list may be empty. Define \[ P(z)=\prod_{j=1}^n(z-a_j),\qquad Q(z)=\prod_{j=1}^n(1-\overline{a_j}z), \] with empty products equal to one. Since \(|a_j|<1\), the polynomial \(Q\) has no zero in the closed unit disk. The quotient \[ H(z)=\frac{f(z)Q(z)}{P(z)} \] extends holomorphically across every \(a_j\), because \(P\) has exactly the same multiplicity there as \(f\). The extended values are nonzero. Thus \(H\) is holomorphic and nonzero on a neighborhood of the closed unit disk.
For \(|z|=1\), direct expansion gives \(|z-a_j|=|1-\overline{a_j}z|\). Hence \(|P(z)|=|Q(z)|\) there, and \(|H(z)|=1\). Apply the maximum modulus principle to both \(H\) and \(1/H\): throughout the disk, \(|H|\leq1\) and \(|1/H|\leq1\), so \(|H|=1\) [@SS03]. The same principle makes \(H\) a constant \(\alpha\) with \(|\alpha|=1\). Consequently \(fQ=\alpha P\) on the disk, and then on all of \(\mathbb C\) by the identity theorem.
2. The entire-plane hypothesis forces every \(a_j\) to be zero.
If some \(a_j\ne0\), the point \(b=1/\overline{a_j}\) is a zero of \(Q\) with \(|b|>1\). But every zero of \(P\) has modulus less than one, so \(P(b)\ne0\). Evaluating the entire identity \(fQ=\alpha P\) at \(b\) would give \(0=\alpha P(b)\ne0\), a contradiction.
Thus all \(a_j\) are zero. It follows that \(P(z)=z^n\) and \(Q(z)=1\), so \(f(z)=\alpha z^n\) everywhere. Conversely every such monomial is entire and has modulus one on the unit circle. This proves both necessity and sufficiency, including the constant case \(n=0\).