- How many solutions does the equation
\[ z + e ^ { - z } = 2 + i \]
have in the half-plane Re \(z > 0 ?\) Prove that your answer is correct.
Solution
There is exactly \(\boxed{1}\) solution in the right half-plane, and it is a simple zero of \(z+e^{-z}-(2+i)\).
1. Every right-half-plane solution lies in \(B=\{z:|z-(2+i)|<1\}\).
If \(z\) satisfies the equation and \(\operatorname{Re}z>0\), then \[ |z-(2+i)|=|e^{-z}|=e^{-\operatorname{Re}z}<1. \] Thus \(z\in B\). Conversely every point of \(B\) has real part greater than one, so all zeros in \(B\) lie in the required half-plane. It remains to count the zeros there.
2. There is precisely one zero in \(B\), counted with multiplicity.
For \(|z-(2+i)|=1\) one has \(\operatorname{Re}z\geq1\). Consequently \[ |e^{-z}|=e^{-\operatorname{Re}z}\leq e^{-1}<1=|z-(2+i)|. \] Both \(z-(2+i)\) and \(e^{-z}\) are entire, so Rouché’s theorem on this circle gives the same number of zeros inside for \(z+e^{-z}-(2+i)\) as for \(z-(2+i)\) [@SS03]. The latter has exactly one simple zero. Thus the former has one zero counted with multiplicity, which means that there is one distinct zero and its multiplicity is one. By step 1 no right-half-plane solution lies outside \(B\). This gives the asserted complete count.