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

[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.