Show that the equation $\lambda - z - e^{-z} = 0$ has exactly one solution in the half plane.

Question

I know that this question has been asked before, but I'm not satisfied with the answers provided. I will post here my complete solution to this question, but I'm not very sure about my (a) and (c) parts.

Let $\lambda > 1$ and show that (a) the equation $\lambda - z - e^{-z} = 0$ has exactly one solution in the half plane $\{z: \Re(z) > 0\}$. (b) Show that this solution must be real. (c) What happens to the solution as $\lambda \to 1$?

Let f(z) = $\lambda - z - e^{-z}$

(a) The equation $f(z) = 0$ has exactly one solution in the half plane $\{z: \Re(z) > 0\}$.

Assume that $\exists z_0 \in G = \{z:\Re(z) > 0 \}$ such that $f(z_0) = 0$, we will show that this is the only solution in $G$.

$$f(z_0) = 0 \Rightarrow \lambda - z_0 = e^{-z_0} \Rightarrow |z_0 - \lambda| = e^{-\Re(z_0)}$$

But $Re(z_0) > 0$, thus

$$|z_0 - \lambda| < 1$$

Consider $h(z) = -e^{-z}, g(z) = \lambda - z$,

$$|h(z) + g(z)| = | \lambda - z - e^{-z} | \leq |e^{-z}| + |\lambda - z| = |h(z)| + |g(z)|$$

Hence, by Rouche's Theorem

$$Z_h + P_h = Z_g + P_g$$

But $P_h = P_g = 0$. Since $g$ has only one zero in $|z - \lambda| < 1$, $h + g = f(z)$ has only one zero also. $\square$

(b) Show that the solution is real

Consider the real function

$$f(x) = \lambda - x - e^{-x}$$

Then $f(0) = \lambda - 1 > 0$ and $f(\lambda) = -e^{-\lambda} < 0$, thus by the Intermediate value theorem,

$$\exists z_0 \in (0,\lambda), \mbox{ such that } f(z_0) = 0.$$

But this solution is unique by part (a), hence $f$ has a unique real solution in $G$.$\square$

(c) What happen to the solution $\lambda \to 1$?

$$ \begin{align*} \lim_{\lambda \to 1}f(z_0) & = \lim_{\lambda \to 1}0 \\ 1 - z_0 - e^{-z_0} & = 0 \\ z_0 + e^{-z_0} = 1 \end{align*} $$

Hence, $z_0 \to 0$ as $\lambda \to 1$. $\square$

My problem is that I can not see how we apply here the rouche theorem for make the claim about the roots of $f$ when we only consider $h$ and $g$ and not $f$. That point is not very clear to me.

Answer 1

Part (b) of this problem was not correct. I figured out how to solve my own problem by doing the following:

Consider $g(z) = -\lambda + z$,

$$ |f(z) + g(z)| = | \lambda - z - e^{-z} + -\lambda + z | = |-e^{-z}| $$

Then

$$ |f(z)| + |g(z)| = |\lambda - z - e^{-z}| + |-e^{-z}| $$

Thus,

$$ |f(z) + g(z)| = |-e^{-z}| < |f(z)| + |g(z)| = |f(z)| + |e^{-z}| $$

Hence, by Rouche's Theorem

$$ Z_f + P_f = Z_g + P_g $$

But $P_f = P_g = 0$. Since $g$ has only one zero in $|z - \lambda| < 1$, then $f(z)$ has only one zero also.$\square$