Let \(c \in \mathbb{C}\) with \(|c| < \frac{1}{3}\). Show that on the open half-plane \[ \Omega = \{z \in \mathbb{C} \mid \operatorname{Re}(z) < 1\}, \] the function \(f(z) = c e^z\) has exactly one fixed point.
Solution
Goal: Prove that \(f(z) = c e^z\) has a unique fixed point in \(\Omega = \{\operatorname{Re} z < 1\}\) by applying Rouché’s Theorem on bounded truncated domains \(\Omega_R\).
1. Reformulation as a root-finding problem:
1.1. A point \(z \in \Omega\) is a fixed point of \(f\) if and only if \(f(z) = z\), which is equivalent to \(g(z) = 0\), where\[g(z) = c e^z - z.\]
1.2. Decompose \(g(z) = F(z) + G(z)\), where \(F(z) = -z\) and \(G(z) = c e^z\).
2. Bounded domain \(\Omega_R\) and its boundary:
2.1. For \(R > 1\), define the bounded domain\[\Omega_R = \{z \in \mathbb{C} \mid |z| < R, \, \operatorname{Re}(z) < 1\}.\]
2.2. The boundary \(\partial \Omega_R\) consists of two parts:- The vertical line segment \(L_R = \{1 + i t \mid -\sqrt{R^2 - 1} \le t \le \sqrt{R^2 - 1}\}\).- The circular arc \(C_R = \{R e^{i\theta} \mid \cos\theta \le 1/R\}\).
3. Rouché estimate on \(\partial \Omega_R\):
3.1. Estimate on the vertical segment \(L_R\):- For \(z = 1 + i t \in L_R\), \(\operatorname{Re}(z) = 1\).- \(|G(z)| = |c e^{1 + i t}| = |c| e^1 < \frac{1}{3} \cdot e = \frac{e}{3}\).- Since \(e < 3\), \(\frac{e}{3} < 1\).- \(|F(z)| = |z| = |1 + i t| = \sqrt{1 + t^2} \ge 1\).- Thus \(|G(z)| < \frac{e}{3} < 1 \le |F(z)|\) on \(L_R\).
3.2. Estimate on the circular arc \(C_R\):- For \(z \in C_R\), \(|z| = R\) and \(\operatorname{Re}(z) \le 1\).- \(|G(z)| = |c e^z| = |c| e^{\operatorname{Re}(z)} \le |c| e^1 < \frac{e}{3} < 1\).- \(|F(z)| = |z| = R > 1\).- Thus \(|G(z)| < \frac{e}{3} < 1 < R = |F(z)|\) on \(C_R\).
3.3. Therefore, \(|G(z)| < |F(z)|\) holds strictly at every point \(z \in \partial \Omega_R\).
4. Counting zeros:
4.1. Both \(F(z) = -z\) and \(G(z) = c e^z\) are holomorphic on and inside the bounded domain \(\Omega_R\).
4.2. By Rouché’s Theorem, \(g(z) = F(z) + G(z) = c e^z - z\) and \(F(z) = -z\) have the same number of zeros inside \(\Omega_R\), counted with multiplicity.
4.3. \(F(z) = -z\) has exactly one simple zero inside \(\Omega_R\) (at \(z = 0\), since \(|0| = 0 < R\) and \(\operatorname{Re}(0) = 0 < 1\)).
4.4. Thus \(g(z)\) has exactly one zero inside \(\Omega_R\).
5. Extension to the full half-plane \(\Omega\):
5.1. If \(z_0 \in \Omega\) is any zero of \(g(z) = c e^z - z\), then \(|z_0| = |c e^{z_0}| = |c| e^{\operatorname{Re}(z_0)} < \frac{e}{3} < 1\).
5.2. Thus any zero \(z_0 \in \Omega\) must satisfy \(|z_0| < 1\), and therefore \(z_0 \in \Omega_R\) for every \(R > 1\).
5.3. Since \(g(z)\) has exactly one zero in \(\Omega_R\) for any \(R > 1\), that single zero is the unique zero of \(g(z)\) in all of \(\Omega = \{\operatorname{Re}(z) < 1\}\).
5.4. Hence \(f(z) = c e^z\) has exactly one fixed point in \(\Omega\).
6. Conclusion:
$f(z) = c e^z$ has a unique fixed point in $\{\operatorname{Re}(z) < 1\}$.