# Proving that $e^z = z+\lambda$ has exactly $m+n$ solutions $z$ such that $-2\pi m<\Im z<2\pi n$

I need to prove that for $\lambda\in \mathbb{C}$ and for $m,n\in\mathbb{Z}$
large enough, the equation:
$$e^z = z+\lambda$$
has exactly $m+n$ solutions $z$ such that $-2\pi m<\Im z<2\pi n$, where $\Im z$ denotes the imaginary part of $z$.

I thought about looking at the rectangle $\pm R + 2\pi in, \pm R – 2\pi im$ and try to use Rouche’s Theorem with:
$$f(z)=e^z-(z+\lambda)=P_{n+m}(z) + \sum_{k=m+n+1}^{\infty}\frac{z^k}{k!} – (z+\lambda)$$

where:
$$P_{n+m}(z) =\sum_{k=0}^{m+n}\frac{z^k}{k!}$$

is a polynomial of degree $m+n$, and all is left is to prove that:
$$|P_{n+m}(z)| > \left| \sum_{k=m+n+1}^{\infty}\frac{z^k}{k!} – (z+\lambda)\right|$$

in the boundary of the rectangle. But I don’t know how to do that (or even if this solution is the right way to go)

Thanks!