Prove that if \(f:U\to V\) is holomorphic and injective then \(f'(z)\neq 0\) on \(U\).
Solution
Suppose, toward a contradiction, that \(f'(z_0)=0\) at some \(z_0\in U\). Since an injective holomorphic function is nonconstant, there is a least \(m\ge2\) such that \(f^{(m)}(z_0)\ne0\). Thus \[ f(z)-f(z_0)=(z-z_0)^m g(z) \] for a holomorphic function \(g\) with \(g(z_0)\ne0\).
Choose \(r>0\) so small that the closed disk \(\overline{D(z_0,r)}\) lies in \(U\), \(g\) has no zero there, and \(z_0\) is the only zero of \(f'\) in that disk. The latter is possible because \(f'\) is holomorphic and not identically zero.
Set \[ M=\min_{|z-z_0|=r}|f(z)-f(z_0)|>0. \] For any \(w\) with \(0<|w-f(z_0)|<M\), Rouché’s theorem on \(|z-z_0|=r\) shows that \[ f(z)-w \quad\text{and}\quad f(z)-f(z_0) \] have the same number of zeros in \(D(z_0,r)\), counted with multiplicity. The second function has exactly \(m\) zeros there, counted with multiplicity, so \(f(z)=w\) has exactly \(m\) solutions counted with multiplicity.
None of those solutions is \(z_0\), because \(w\ne f(z_0)\). By the choice of \(r\), every other point in the disk has \(f'(z)\ne0\), so every solution of \(f(z)=w\) is simple. Hence there are exactly \(m\ge2\) distinct solutions in \(D(z_0,r)\), contradicting injectivity of \(f\).
Therefore \(f'(z)\ne0\) for every \(z\in U\).