Show that if \(f\) is holomorphic on \(\Omega\) and injective, then \(f'(z)\) is nonvanishing on \(\Omega\).
Solution
By contradiction: after translating, suppose \(0\in\Omega\), \(f(0) = 0\) and \(f'(0)=0\). An injective \(f\) is nonconstant, so \(0\) is a zero of \(f\) of some finite order \(m\), and \(m\geq2\) since \(f(0)=f'(0)=0\). Write \(f(z)=z^mg(z)\) with \(g\) holomorphic near \(0\) and \(g(0)\neq0\). On a small disk \(D\) about \(0\), \(g\) is nonvanishing and has a holomorphic \(m\)-th root \(h\), so \(f=\varphi^m\) with \(\varphi(z)\da zh(z)\). Since \(\varphi'(0)=h(0)\neq0\), \(\varphi\) maps a neighborhood \(U\subseteq D\) of \(0\) biholomorphically onto a disk \(\abs w<\delta\). For \(0<\eps<\delta\) and \(\zeta=e^{2\pi i/m}\neq1\), the distinct points \(w_1=\eps\) and \(w_2=\zeta\eps\) satisfy \(w_1^m=w_2^m\), so their preimages under \(\varphi\) are distinct points of \(U\) with the same image under \(f\), contradicting injectivity.