Let \(\Delta=\{z\in\mathbb C:|z|<1\}\).
State Rouche’s Theorem.
Let \(\varphi : \Omega \to \mathbb{C}\) be holomorphic on an open convex set \(\Omega\). Show that for \(z, w \in \Omega\)
\[|\varphi(z) - \varphi(w)| \leq \max_{\xi \in L} |\varphi'(\xi)| \cdot |z - w|,\]
where \(L\) is the straight line segment from \(z\) to \(w\).
- Use the above to prove the following: suppose
\[f(z) = z + \sum_{n=2}^\infty a_n z^n\]
where
\[\sum_{n=2}^\infty n|a_n| \leq 1.\]
Show that \(f(z)\) is a 1-1 holomorphic function on \(\Delta\).
Solution
1. Part (a): Rouché’s theorem.
Let \(C\) be a positively oriented, piecewise smooth simple closed contour and suppose \(F,G\) are holomorphic on an open set containing \(C\) 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.
Indeed, \(F\) has no zero on \(C\), and for \(0\le t\le1\) the functions \(F+tG\) also have no zero on \(C\), since \(|tG|<|F|\). Therefore the integer \[ \frac1{2\pi i}\int_C\frac{F'(z)+tG'(z)}{F(z)+tG(z)}\,dz \] is constant in \(t\) by continuity and the argument principle [@SS03]. Its values at \(t=0\) and \(t=1\) are the zero counts of \(F\) and \(F+G\).
2. Part (b): integrate \(\varphi'\) along the line segment.
Because \(\Omega\) is convex, the segment \[ L=\{w+t(z-w):0\le t\le1\} \] lies in \(\Omega\). The fundamental theorem of calculus applied to \(t\mapsto\varphi(w+t(z-w))\) gives \[ \varphi(z)-\varphi(w) =(z-w)\int_0^1\varphi'(w+t(z-w))\,dt. \] Hence \[ |\varphi(z)-\varphi(w)| \le |z-w|\int_0^1|\varphi'(w+t(z-w))|\,dt \le |z-w|\max_{\xi\in L}|\varphi'(\xi)|. \] The maximum exists because \(L\) is compact and \(\varphi'\) is continuous.
3. Part (c): the nonlinear tail is a strict contraction on every segment in \(\Delta\).
Write \[ h(z)=f(z)-z=\sum_{n=2}^\infty a_nz^n. \] On every closed disk of radius \(r<1\), the coefficient hypothesis gives convergent majorants for both series: \[ \sum_{n=2}^\infty |a_n|r^n\leq\sum_{n=2}^\infty n|a_n|<\infty, \qquad \sum_{n=2}^\infty n|a_n|r^{n-1}\le\sum_{n=2}^\infty n|a_n|<\infty. \] Thus the series defines a holomorphic function on \(\Delta\) and can be differentiated term by term [@SS03], giving \[ h'(z)=\sum_{n=2}^\infty n a_nz^{n-1}. \] Take distinct \(z,w\in\Delta\) and let \(L\) be their segment. Compactness of \(L\) and openness of the disk give \[ r:=\max_{\xi\in L}|\xi|<1. \] For every \(\xi\in L\), since \(n-1\geq1\), \[ |h'(\xi)| \le\sum_{n=2}^\infty n|a_n|r^{n-1} \leq r\sum_{n=2}^\infty n|a_n| \leq r<1. \] Applying part (b) to \(h\) on the convex disk gives \[ |h(z)-h(w)|\leq r|z-w|<|z-w|. \] Therefore \[ |f(z)-f(w)| =|(z-w)+(h(z)-h(w))| \ge |z-w|-|h(z)-h(w)|>0. \] Hence \(f(z)\ne f(w)\) whenever \(z\ne w\), so \(f\) is one-to-one on \(\Delta\).