Problem 2. Fix a real number \(\alpha > 1\) . Show that the equation \(z - \alpha = e ^ { - z }\) has precisely one solution in the half plane \(\mathrm { R e } ( z ) > 0\) and that this solution must be real.
Solution
1. Every right-half-plane solution lies in the disk \(|z-\alpha|<1\).
If \(z\) satisfies \[ z-\alpha=e^{-z} \] and \(\operatorname{Re}z>0\), then \[ |z-\alpha|=|e^{-z}|=e^{-\operatorname{Re}z}<1. \] Thus every required solution lies in \[ B=\{z:|z-\alpha|<1\}. \] Conversely, because \(\alpha>1\), every point of the closed disk \(|z-\alpha|\le1\) has \[ \operatorname{Re}z\ge\alpha-1>0. \] Hence every zero found in \(B\) is automatically in the required half-plane.
2. Rouché’s theorem gives exactly one solution in \(B\).
Let \[ F(z)=z-\alpha-e^{-z}. \] On the boundary \(|z-\alpha|=1\), one has \(\operatorname{Re}z\ge\alpha-1\), and therefore \[ |e^{-z}|=e^{-\operatorname{Re}z} \le e^{-(\alpha-1)}<1=|z-\alpha|. \] Thus Rouché’s theorem shows that \(F\) and \(z-\alpha\) have the same number of zeros in \(B\), counted with multiplicity. The latter has exactly one simple zero. Therefore \(F\) has exactly one zero in \(B\), counted with multiplicity. Together with step 1, this is precisely one solution in the right half-plane.
3. The unique solution is real.
2. Uniqueness therefore gives\[ z_0=\overline{z_0}, \]which means \(z_0\) is real.