Let \(f\) be analytic in a domain \(D\) and fix \(z_0 \in D\) with \(w_0 \definedas f(z_0)\). Suppose \(z_0\) is a zero of \(f(z) - w_0\) with finite multiplicity \(m\). Show that there exists \(\delta >0\) and \(\eps > 0\) such that for each \(w\) such that \(0 < \abs{w-w_0} < \eps\), the equation \(f(z) - w = 0\) has exactly \(m\) distinct solutions inside the disc \(\abs{z-z_0} < \delta\).
Solution
Write \[ f(z)-w_0=(z-z_0)^m g(z), \qquad g(z_0)\ne0. \] Then \[ f'(z)=(z-z_0)^{m-1}h(z) \] for a holomorphic function \(h\) with \(h(z_0)=m g(z_0)\ne0\). Choose \(\delta>0\) so small that the closed disk \(\overline{D(z_0,\delta)}\) lies in \(D\), \(z_0\) is the only zero of \(f-w_0\) there, and \(h\) has no zero there. Thus \(z_0\) is the only critical point of \(f\) in this closed disk.
Set \[ M=\min_{|z-z_0|=\delta}|f(z)-w_0|>0 \] and choose \(0<\varepsilon<M\). If \(0<|w-w_0|<\varepsilon\), then on the boundary circle \[ |(w-w_0)|<|f(z)-w_0|. \] Rouché’s theorem therefore implies that \[ f(z)-w=(f(z)-w_0)-(w-w_0) \] has exactly \(m\) zeros in \(D(z_0,\delta)\), counted with multiplicity.
None of these zeros is \(z_0\), because \(w\ne w_0\). At every zero \(z\) in the disk we also have \(f'(z)\ne0\), since \(z_0\) is the only critical point there. Hence all \(m\) zeros are simple. Therefore they are \(m\) distinct solutions.