Solution of $e^{-z}+z=\lambda$

Question

Let $\lambda\in\mathbb{C}$ with $\Re(\lambda)>1$ be given. I want to show, that $e^{-z}+z=\lambda$ has exactly one solution in $U=\{z\in\mathbb{C}:\Re(z)>0\}$.

I think that exercise can be solved by using Rouché's theorem (symmetric version). Can you tell me how to apply that theorem please?

Thank you!

Answer 1

Let $R>|\lambda|+1$. Take for $\gamma$ the segment of the $y-$axis from $-iR$ to $iR$ and the half circle of radius $R$, centered at the origin, that is in $U$. Put $f(z)=\exp(-z)+z-\lambda$, $g(z)=z-\lambda$. We have $|f(z)-g(z)|=|\exp(-z)|\leq 1$ on $\gamma$.

On the segment, we get $|g(z)|=|z-\lambda|\geq |{\rm Re}(z-\lambda)|={\rm Re}(\lambda)>1$. On the half-circle, $|g(z)|\geq |z|-|\lambda|>1$. So we can apply Rouché, and it is easy to finish.

For the second question, note that the conjugate of $\exp(-z)+z-\lambda$ is $\exp(-\overline{z})+\overline{z}-\overline{\lambda}$.

Answer 2

The number of solutions of $e^{-z}+z=\lambda$ in the right half-plane is given by:

$$ \lim_{R\to +\infty}\frac{1}{2\pi i}\oint_{C_R} \frac{-e^{-z}+1}{e^{-z}+z-\lambda}\,dz =\lim_{R\to +\infty}\frac{1}{2\pi i}\oint_{C_R}\frac{e^z-1}{1+(z-\lambda)e^z}\,dz$$ where $C_R$ is the semicircular contour that joins $Ri$ to $-Ri$ with a line segment and $-Ri$ to $Ri$ with a semicircle in the right half-plane. The integrand function can be written as: $$ \frac{1}{z-\lambda}\left(\frac{(z-\lambda)e^{z}-(z-\lambda)}{1+(z-\lambda)e^{z}}\right)=\frac{1}{z-\lambda}\color{blue}{-\frac{2}{(z-\lambda)+(z-\lambda)^2 e^z}}$$ and the contribute given by the blue term on the semicircle clearly vanishes as $R\to +\infty$.
So it is enough to prove that $$ \lim_{R\to+\infty}\int_{-iR}^{+iR}\frac{dz}{(z-\lambda)+(z-\lambda)^2 e^z}=0$$ to have that the number of solutions of $e^{-z}+z=\lambda$ in the right half plane is the same as the number of solutions of $z=\lambda$, i.e. $1$.

It is also interesting to point out that by Picard's Great Theorem $e^{-z}+z=\lambda$ has an infinite number of solutions in the complex plane, since $e^{-z}+z$ is a non-constant entire function with an essential singularity at infinity, hence it attains every complex value (with at most one exception) infinitely often. However, if $\lambda$ is exceptional, $\lambda+2\pi i$ is exceptional too, so there are no exceptional points in this case.