For \(k=1,2,\cdots, n\), suppose \(\abs{a_k} < 1\), and suppose \(|b|<1\). Let \[ f(z) \definedas \qty{z - a_1 \over 1 - \bar a_1 z} \qty{z-a_2 \over 1 - \bar a_2 z} \cdots \qty{z - a_n \over 1 - \bar a_n z} .\] Show that \(f(z) = b\) has \(n\) solutions in \(\abs{z} < 1\).
Solution
For \(|z|=1\) and each \(k\), \[ |z-a_k|=|1-\bar a_k z|, \] so every Blaschke factor has modulus \(1\) on the unit circle. Hence \[ |f(z)|=1>|b| \qquad(|z|=1). \] The poles of \(f\) are at \(1/\bar a_k\), all outside the closed unit disk, so \(f\) is holomorphic on a neighborhood of \(\overline\DD\).
Apply Rouché’s theorem on \(|z|=1\) to \(f\) and \(f-b\). They have the same number of zeros in \(\DD\), counted with multiplicity. The zeros of \(f\) in \(\DD\) are exactly \(a_1,\ldots,a_n\), with repetitions counted according to multiplicity, so there are \(n\) of them. Therefore \(f-b\) also has exactly \(n\) zeros in \(\DD\), counted with multiplicity. Equivalently, the equation \[ f(z)=b \] has \(n\) solutions in the unit disk, counted with multiplicity.