# Proof of function zeroes.

1. May 7, 2007

### moo5003

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).

Problem:

I'm unsure how to get the inequality to show |h(z)|<|g(z)| any insight into this would be appreciated.

2. May 7, 2007

### moo5003

(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

abs[-exp(z)]=exp(x)=exp(-abs(x))<1

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