Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$ Let $\lambda>1$ and show the equation $\lambda - z -e^{-z} = 0$ has exacly one solution in the half plane $\{z:Re(z)>0\}$. Show that this solution must be real. What happens to the solution as $\lambda \rightarrow 1 $?
 A: Apologies for my misunderstanding previously. we have that $|\lambda-z|=e^{-\Re{(z)}}<1$ holds, if there are any solutions at all. Then you can apply Rouche's Theorem to $f(z)=e^{-z}$ and $g(z)=\lambda-z$. Since $g$ has only one solution, $f+g$ does inside this curve. To prove it is real, you can simply apply the intermediate value theorem. It shouldn't be hard to make the function positive and then negative.
An alternative fun and interesting approach, if you already knew that the solution was real, is as follows:
Applying the intermediate value theorem tells you there must be at least one solution. To see uniqueness, any solution satisfies: $\lambda=z+e^{-z}$. Since $\lambda>1$, we can write $\lambda=\alpha+\beta$ where $\alpha\in\mathbb{N}$ and $0\leq\beta<1$. Partitioning $z=\gamma+\delta$ into an integer and fractional part as well, and setting $\alpha=\gamma$, tells us that any solution must satisfy $\beta=\delta+e^{-\alpha}e^{-\delta}$. But everything except for $\delta$ in this equation is known, so we have one equation with one unknown and so there are one, zero, or infinitely many solutions. Zero is impossible because the IVT gave us one, and infinite means that this equation is degenerate. By inspection it isn't, and so there is only one solution.
