Let \(p(z)\) be a nonconstant polynomial of even degree with all roots inside the disc \(|z|<R\). Show that there is a holomorphic function \(f(z)\) on the annulus \(|z|>R\) with \(f(z)^2=p(z)\).
Solution
Write \[ p(z)=a z^{2m}\prod_{j=1}^N\left(1-\frac{\alpha_j}{z}\right)^{\nu_j}, \] where \(|\alpha_j|<R\) and \(\sum_j\nu_j=2m\). On \(|z|>R\), every factor \(1-\alpha_j/z\) lies in the disk centered at \(1\) of radius \(<1\), hence admits a holomorphic logarithm. Choose a square root \(\sqrt a\) of the leading coefficient and define \[ f(z)=\sqrt a\,z^m \exp\!\left( \frac12\sum_{j=1}^N\nu_j \Log\left(1-\frac{\alpha_j}{z}\right) \right). \] This is holomorphic for \(|z|>R\), and squaring gives exactly \(p(z)\).
Equivalently, because the degree is even, the winding number of \(p\) around \(0\) along a large circle is even, so the nonvanishing holomorphic function \(p\) on the annulus admits a holomorphic square root.