1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook