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Thread: complex number, horizontal strip

  1. #1
    Junior Member
    Mar 2009

    complex number, horizontal strip

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

    The hint says to consider $\displaystyle f(z)=e^z -z-\lambda$ and sketch the image of the boundary of $\displaystyle \{ -R< x<R, -2 \pi m < y < 2 \pi n \}$. Keep in mind $\displaystyle |\lambda | << m, n << R$.

    I do not see how to prove this now. I need some help on this one. Thank you.
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  2. #2
    Super Member
    Aug 2008
    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 $\displaystyle f(z)=e^z-z-\lambda$ in terms of the Lambert-W function:

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

    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:

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

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