Show that the equation $\lambda - z - e^{-z} = 0$ has exactly one solution in the half plane. 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.
 A: [For completeness sake]
Since $\lambda > 1$, note that $ \gamma \doteq \partial B(\lambda,1) $ is in the right semi-plane. 
Consider the functions $f(z) = e^z + z - \lambda $ and $g(z) = \lambda - z$, as above.
Notice that at $\{\gamma\}$ we have:
$$
|f(z) + g(z)| = |e^z| < 1 = |g(z)| < |f(z)| + |g(z)|
$$
then by Rouché, since $f$ and $g$ don't have any poles, $f$ must have exactly one zero  $\beta \in B(\lambda,1)$, just like $g$.
On the other hand, let $\alpha $ in the right semi-plane be a root of $f$, 
$$
|\lambda - \alpha| =  
|e^\alpha| < 1 
\implies 
\alpha \in B(\lambda,1) 
\implies 
\alpha = \beta 
$$
then $\alpha$ is unique.  
A: 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.$\square$
