# Math Help - complex number, horizontal strip

1. ## complex number, horizontal strip

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 $\{ -2\pi i m < \text{Im}(z)<2 \pi i n \}$.

The hint says to consider $f(z)=e^z -z-\lambda$ and sketch the image of the boundary of $\{ -R< x. Keep in mind $|\lambda | << m, n << R$.

I do not see how to prove this now. I need some help on this one. Thank you.

2. I'd like to try and tackle this if I may although not via Rouche': First, it's easy to solve for the roots of $f(z)=e^z-z-\lambda$ in terms of the Lambert-W function:

$z_n=-\lambda-\textbf{W}\left(-e^{-\lambda}\right)$

and keep in mind the W-function is infinitely valued and looks a bit like the log multi-sheet so we should actually write it as:

$z_n=-\lambda-\textbf{W}\left(n,-e^{-\lambda}\right)$

where $n$ is the sheet number. The top left plot is the contour for $n=2, m=1, R=4, \lambda=1+i$ and the zeros $z_n$ as the black dots. As you can see, the contour is enclosing three zeros. The lower-left is the image of $f(z)$ over that contour. On the brown contour we have: $e^{R+it}-(R+it)-(1+i)$ and if $R$ is large, the expression is dominated by $e^{R+it}$ giving rise to the circular contours around the origin but note the remaining contours complete the circuit around the origin giving rise to a winding number of $n+m=3$. This is primarily due I think to the blue contour over which we have $e^{-R+iy}-(-R+iy)-(a+ib)$ and since the $e^{-R}$ term is small, the real part is dominated by $R-a$ and if this is positive, then the circular contour complete the circuit around the origin $n+m$ times as shown by the lower left plot. However, in the case of $R=4$ and $\lambda=7+i$ as the second set shows, then $R-a=-3$ which causes a detour around the origin and gives rise to a winding number of $n+m-1=2$. This is reflected in the top right plot showing only two zeros enclosed by the contour . . . I'm thinkin' B at best.