Let $\lambda > 1$ , want to show that the equation $$\lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane $\{z:Re(z)>0\}$. Moreover, the solution must be real.
I tried to use Rouche's theorem on $g(z)=\lambda – z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $\gamma$ such that this will work.
for the second part I used the IVT to show that $\lambda -x-e^{-x}$ has a zero in $(0,\lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.
Best Answer
A hint.
If $\operatorname{Re} z > 0$ and $\lambda - z - e^{-z} = 0$ then
$$ |\lambda - z| = e^{-\operatorname{Re} z} < 1. $$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-\lambda|<1$.