Showing that for large $m$ and $n$ the equation $e^z=z+\lambda$ has exactly $m+n$ solutions in the horizontal strip $\{-2\pi*i m< Imz < 2i\pi n\}$ Could anyone give me a hint on how to approach this problem? I have little idea of how to go about it. 

For a fixed complex number $\lambda$, show that if $m$ and $n$ are large integers, then the equation $e^z = z + \lambda$ has exactly $m+n$ solutions in  the horizontal strip $\{-2i\pi m < Im z < 2i \pi n\}$.

Thanks!
 A: Let $f(z)=e^z-z-\lambda$ and $g(z)=e^z+1$. 
We know that $g(z)=0$ iff $z=(2k+1)\pi i$, i.e., $g$ does have exactly $n+m$ roots in the strip $-2\pi m<\Im z<2\pi n$. In fact, if $A,B$ are positve reals and  $K$ is the rectangle given by $-2\pi m<\Im z<2\pi n$ and $-A<\Re z<B$ then $g$ still has precisely $m+n$ roots in $K$ (and no root on $\partial K$).
We may try to use Rouché 's theorem to show that $f$ also has $n+m$ roots in $K$ under some mild restrictions on $n,m,A,B$.
To this end, we need to show $|f(z)-g(z)|<|f(z)|+|g(z)|$ for all $z\in\partial K$. 
As the triangle inequality already gives us $|f(z)-g(z)|\le |f(z)|+|g(z)|$, we need only show that it does not happen on $\partial K$ that $\frac{f(z)}{g(z)}\in(-\infty,0]$.
Consider first the horizontal boundary parts of $K$. There we have $g(z)=e^{\Re z}+1>0$ and $f(z)=e^{\Re z}-z-\lambda$. We need to exclude that $f(z)$ is non-positive real. For this it is sufficient to have $\Im z\ne\Im \lambda$, and this can easily be achieved for example by requiring
$$\tag1 n,m>\frac{|\lambda|}{2\pi}.$$
(This is very wasteful; in reality we need only exclude at most one value for $n$ or $m$, and that only if $\Im\lambda$ happens to be a multiple of $2\pi$).
Next consider the left vertical boundary part.
Here we have $|e^z|<1$ and hence $\Re g(z)>0$.
On the other hand $\Re f(z)>A-\Re\lambda-1$. Hence if we demand
$$\tag2A>\Re\lambda+1 $$
we find that $\frac{\Re f(z)}{\Re g(z)}>0$ and hence $\frac{f(z)}{g(z)}$ cannot be $\in(-\infty,0]$.
Finally consider the right vertical boundary part.
Here $|g(z)|\ge |e^z|-1 =e^B-1>B+\frac12B^2$ whereas $$\begin{align}|f(z)-g(z)|&=|z+\lambda+1|\\&\le |\Re z|+|\Im z|+|\lambda+1|\\&\le B+2\pi\max\{n,m\}+|\lambda+1|.\end{align}$$
Hence in order to have $|f(z)-g(z)|<|f(z)|+|g(z)|$ it is sufficient to have 
$$\tag3B>2\sqrt{2\pi\max\{n,m\}+|\lambda+1|} $$
We conclude: Given sufficiently large $n,m$ (i.e., obeying condition $(1)$), we have for all $A$ satisfying $(2)$ and all $B$ satisfying $(3)$ that $f$ has the same number of roots as $g$ (that is, $n+m$ roots) in the rectangle $K$ determined by these numbers. As this allows $A,B$ to become arbitrarily large for given $n,m$, we see that for $n,m$ satisfying $(1)$, the function $f$ also has exactly $n+m$ roots in the strip $-2\pi m<\Im z<2\pi n$.
