$\lambda-z-e^{-z}=0$ tiene una solución en el medio plano derecho

Question

Deje$\lambda > 1$, desea mostrar que la ecuación$$\lambda-z-e^{-z}=0$$ 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 -xe ^ {- x}$ has a zero in $ (0, \ lambda) $ para concluir que la solución es real. es esto aceptable? Gracias por tu ayuda.

Answer 1

Una pista.

Si$\operatorname{Re} z > 0$ y$\lambda - z - e^{-z} = 0$, entonces

$$ | \ lambda - z | = e ^ {- \ operatorname {Re} z} <1. $$

En otras palabras, si la ecuación tiene alguna solución en el semiplano derecho, entonces se encuentran en el disco abierto$|z-\lambda|<1$.