For \(k=1,2,\cdots, n\), suppose \(\abs{a_k} < 1\) and \[ f(z) \da \qty{z - a_1 \over 1 - \bar a_1 z} \qty{z-a_2 \over 1 - \bar a_2 z} \cdots \qty{z - a_n \over 1 - \bar a_n z} .\] Show that \(f(z) = b\) has \(n\) solutions in \(\abs{z} < 1\).
Solution
Note that \(f\) is holomorphic on \(\DD\) and \(S^1\), since the poles are at \(1/\bar{a_k}\) and if \(\abs{a_l} < 1\) then \(\abs{\bar{a_k}} > 1\). Fix \(b\), then define \(g_w(z) \da f(z) - w\) and form the solution counting function \[ F(w) \da {1\over 2\pi i}\oint_{S^1} \logd g_w(z) \dz = {1\over 2\pi i} \oint_{S^1} {f'(z) \over f(z)-w}\dz .\] Start by computing \(F(0)\). \[ F(0) &= {1\over 2\pi i }\oint_{S^1} \logd \prod_{1\leq k\leq n} \psi_{a_k}(z) \dz \\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} \logd \psi_{a_k}(z) \dz \\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} \qty{1-\abs{a_k}^2 \over (1-\bar{a_k} z)^2} \inverseof{\qty{z-a_k \over 1-\bar{a_k} z}} \dz \\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} {1-\abs{a_k}^2 \over (z-a_k)( 1-\bar{a_k}z) } \dz \\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} \oint_{S^1} {1-\abs{a_k}^2 \over (z-a_k)( 1-\bar{a_k}z) } \dz \\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} 2\pi i \\ &= n ,\] where we’ve used that the integrand has a simple pole at \(a_k\) since \(1/\bar{a_k}\in \DD^c\). So the equation \(f(z) = 0\) has \(n\) solutions. Now use that \(F\) is a continuous function of \(w\) on \(\DD\) and integer valued, thus constant. So \(F(w) = n\) for any \(w\), meaning \(f(z) = w\) has \(n\) solutions in \(\DD\) for every \(w\).
Alternative: \(F\) continuously depends on the \(a_k\), so send them all to zero to get \(f(z) = z^n\), which has exactly \(n\) zeros at the origin.