Use Rouché’s theorem and the Schwarz lemma to prove the following: if \(f:\DD\to \bar{ \DD}\) is holomorphic with \(f(0) = 0\), then there are exactly \(m\) solutions (counted with multiplicity) to \(f(z) = (2z)^m\) in the disc \(\abs{z} < 1/2\).
Solution
First note that the image of \(f\) is in fact \(\DD\) rather than \(\bar{\DD}\), using the open mapping theorem and that the domain \(\DD\) is open. So Schwarz applies to \(f\). Write \(g(z) \da f(z) - (2z)^m\), the claim is that \(g\) has \(m\) zeros. Toward applying Rouché, identify
- The big part of \(g\): \(M(z) \da -(2z)^m\)
- The small part of \(g\): \(m(z) \da g(z) - M(z) = f(z)\).
Now \[ \abs{m(z)} \da \abs{f(z)} \leq \abs{z} = {1\over 2} < \abs{m(z)} = \abs{2z}^m = 1 \qquad \text{on } \abs{z} = {1\over 2} ,\] so Rouché applies: \(\size Z_m = \size Z_M\) on \(\abs{z} < {1\over 2}\), where \(m(z) = f(z)\) and \(M(z) = -(2z)^m\) which has exactly \(m\) zeros.
Note that this works with \(g(z) \da (cz)^m\) and \(R = {1\over c}\).