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Proof of function zeroes.

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data
    a in Reals, a>1.

    f(z) = a + z - exp(z)

    a) Show that f has exactly one zero in the left half-plane {z in C : Re(z) < 0}

    b) Show that this zero is on the real line

    3. The attempt at a solution

    Well, I havnt had much progress on the problem as of yet. I'm trying to use Rouches theorem letting:

    h(z) = -exp(z)
    g(z) = z + a

    and then showing that restricting z to the left half-plane
    |h(z)|<|g(z)| implying that z+a has as many zeroes as f(z) and therefore showing that f(z) has one zerio in the left half plane (since z+a has one since a>1).


    I'm unsure how to get the inequality to show |h(z)|<|g(z)| any insight into this would be appreciated.
  2. jcsd
  3. May 7, 2007 #2
    (I cant edit my main post?!)

    For part b) I did it the same way I wanted to approach part A) and found that the zero must reside in the real interval [-a+1,0).

    For part A) I have the following inequality which troubles me.
    z= x+iy


    My problem here is how can I bound 1 by a function (z+a) that equates to 0 at -a. It makes me think I need to find another way to approach the problem that I'm not seeing.

    EDIT: Nvm just re-read Rouche's theorem. I was using it wrong.
    Last edited: May 7, 2007
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