State Rouche’s theorem.
Let \(f\) be analytic in a neighborhood of \(0\), and satisfying \(f'(0) \neq 0\). Use Rouche’s theorem to show that there exists a neighborhood \(U\) of \(0\) such that \(f\) is a bijection in \(U\).
Solution
Rouché’s theorem. Let \(C\) be a positively oriented simple closed contour, and suppose \(f\) and \(g\) are holomorphic on a neighborhood of \(C\) and its interior. If \[ |g(z)|<|f(z)|\qquad(z\in C), \] then \(f\) and \(f+g\) have the same number of zeros inside \(C\), counted with multiplicity.
For the local injectivity statement, replace \(f\) by \(f-f(0)\), so that \(f(0)=0\) and still \(f'(0)\ne0\). Then \[ f(z)=az+z\varepsilon(z), \qquad a=f'(0)\ne0, \qquad \varepsilon(z)\to0. \] Choose \(r>0\) so small that \(f\) is holomorphic on \(\overline{B(0,r)}\) and \[ |\varepsilon(z)|<\frac{|a|}{2} \qquad (|z|\le r). \] Then on \(|z|=r\), \[ |f(z)-az|=|z\varepsilon(z)|<|az|. \] Rouché’s theorem shows that \(f\) and \(az\) have the same number of zeros in \(B(0,r)\), namely one, counted with multiplicity. Thus \(0\) is the unique zero of \(f\) in this disk.
To prove injectivity, shrink \(r\) if necessary so that \[ |f'(z)-a|<\frac{|a|}{2} \qquad (|z|<r). \] For distinct \(z,w\in B(0,r)\), the line segment joining them lies in the disk, and \[ f(z)-f(w)=\int_0^1 f'(w+t(z-w))(z-w)\,dt. \] Hence \[ \left|\frac{f(z)-f(w)}{z-w}-a\right| \le \frac{|a|}{2}, \] so \((f(z)-f(w))/(z-w)\ne0\) and therefore \(f(z)\ne f(w)\). Thus \(f\) is injective on \(U=B(0,r)\). Since a nonconstant holomorphic map is open, it is a bijection from \(U\) onto the open set \(f(U)\).