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Homework Help: Argument principle for a rectangle

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data

    . I want to prove that there is one solution for e^z-z in every shifted copy of the fundamental strip by applying the argument principle to the boundary of a rectangle −M≤Rez≤M , 2kπi≤Imz≤2(k+1)πi for large M and integer k . I need help in using the argument principle here.

    2. Relevant equations

    Argument principle.

    3. The attempt at a solution
    I tried to compute the integral of the logarithmic derivative along the boundary of the square,but i'm stuck with it.
  2. jcsd
  3. Jun 11, 2013 #2


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    Gold Member

    Is some of the statement of the problem missing here?
  4. Jun 11, 2013 #3


    Staff: Mentor

    What is the "argument principle?"
  5. Jun 12, 2013 #4
    See attachment for the argument principle
    Last edited: Jun 12, 2013
  6. Jun 12, 2013 #5


    Staff: Mentor

    There is no attachment.
  7. Jun 12, 2013 #6
    attached here

    Attached Files:

    • 001.jpg
      File size:
      35.6 KB
  8. Jun 12, 2013 #7


    Staff: Mentor

    It would be nice if you expended the effort to attach an image that one wouldn't need to rotate to be able to read.
  9. Jun 12, 2013 #8
    I tried to attach from wikipedia but it is an invalid file.Could you look at wikipedia?
  10. Jun 12, 2013 #9


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    What do you mean by a "solution for e^z-z"? Was this supposed to be an equation? [itex]e^z- z= 0[/itex] or [itex]e^x= z[/itex]?
  11. Jun 12, 2013 #10


    Staff: Mentor

  12. Jun 12, 2013 #11

    The Argument Principle right? That's what Riemann used to analyze the zeros to the zeta function: Go around a closed contour and accumulate the change in argument of the function. The number of zeros and poles inside the contour are then a function of the change in argument. Look that relation up. Simple.

    So how about for now just let k=1 and let M go to infinity. Let's start on the lower horizontal contour, [itex]z=x+2\pi i[/itex] then [itex]f(x+2\pi i)=e^x-x-2\pi i[/itex]. As you vary x from -infinity to infinity, [itex]f(x+2\pi i)[/itex] travels along a horizontal line at [itex]z=p+2\pi i[/itex] as p goes from -infinity to infinity. Ain't the change in argument then of f(z) along that contour just -pi? What about on the top line? Haven't looked at the two vertical contours. Might not be so easy. You try though. Ain't that what Hall said?
  13. Jun 13, 2013 #12
    I will look at it,thank you
  14. Jun 13, 2013 #13
    come on, come on, what ya' waitin' for? Top one is the same except the argument change is the negative of the bottom so we got zero accumulated argument change over top and bottom or is that -pi thing up there not too easy to see? Look at those later. Anyway, gotta' find [itex]\Delta \arg=2\pi[/itex] somewhere else (cus' there's a zero in there, check the theorem). How about the left contour? Got [itex]z=-m+iy[/itex] so we end up with:

    [tex]f(-m+iy)=\left(e^{-m} \cos(y)+m\right)+i\left(e^{-m}\sin(y)-y\right)[/tex]

    When m is very large, say in the limit as m goes to infinity, the real part tends to a very large number (m), and the imaginary component goes from 0 to [itex]-2\pi i[/itex]. But in the limit as m goes to infinity, that change in argument is an infinistessimal so the change in argument there is zero.

    You know the zeta function is really hard to do this to. Anyway, must be the last contour. Can we find, in the limit as m goes to infinity, an accumulated change in argument of 2pi since what, we got one zero in there?
    What if we analyze:

    [tex]f(m+iy)=\left(e^{m} \cos(y)-m\right)+i\left(e^{m}\sin(y)-y\right)[/tex]

    for very large m and y as it goes from 0 to 2pi (k=0)? How could you prove the change in argument of [itex]f(m+iy)[/itex] over that contour as m goes to infinity tends to [itex]2\pi[/itex]?

    Edit: When I say the top, bottom, left and right contour I mean the box contour with top, bottom, left, and right legs with left bottom corner at (-m,0) and top right corner at (m,2pi).
    Last edited: Jun 13, 2013
  15. Jun 14, 2013 #14
    if we consider ,instead ,the rectangle −M≤Rez≤M ,- 2mπi≤Imz≤2nπi for large M and integers m and n,then the image would be a new rectangle around the origin,transversed possibly several times.Is it true that it is transversed m+n times?
    Last edited: Jun 14, 2013
  16. Jun 14, 2013 #15
    Afraid that's not exactly correct: The image of the lower and upper legs of the contour remains in the fourth quadrant. The plot below is the image of [itex]f(z)=e^z-z[/itex] for m=4 and k=4. Red and green contours are the images of the upper and lower legs of the rectangle. m has to be large enough for the image to encircle the origin. If m=2 and k=4, the image remains in the lower half-plane. So I assume that is why the problem stipulates for large (enough) m.

    Sorry I posted before adequately studying the problem.

    Attached Files:

  17. Jun 14, 2013 #16
    The image is as shown in the attached,so the change in argument is zero,am i right?

    Attached Files:

    • 001.jpg
      File size:
      10 KB
  18. Jun 14, 2013 #17
    Know what, looks like it's not happening for me neither and that's why I originally posted; it wasn't happening with the others I felt. Here's in my opinion, the best demonstration of the Argument Principle as it relates to the argument change of a function across a circular contour:


    Now, being good programmers like you and me, we simply download that code, insert [itex]f(z)=e^{z}-z[/itex], change what it already complicated code to include an arbitrary square contour, and we'd get a pretty good idea what the image of this function is doing across the contours you specified above, and it's argument change across that contour. We can do that huh? Not sure I want to work on it though, but just to say, it's another front.
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