Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering if I could get some help with the following problem. I almost solved it, but I don't see how to justify using Rouché's theorem. The problem is this: Let $f(z) = z+a - e^z$ where $a$ is a real number satisfying $1 < a < \infty$. Show that $f(z)$ has exactly one zero in the half-plane $Re(z) < 0$ and moreover, show that this zero lies on the real axis.

As I said, I almost solved it, and here is my "solution". The boundary of the half-plane $Re(z) < 0$ is the line $Re(z) = 0$. On this boundary we have $|z+a| \geq a > 1$ while $|e^z| = 1$ so that by Rouché's theorem $z+a$ and $f(z)$ have the same number of solutions in $Re z < 0$. But $z+a$ clearly has only one zero in this half-plane, so it follows that so does $f(z)$. For the second part, consider the restriction of $f(z)$ to the real axis, i.e. to $(-\infty, 0)$. It's easy to see that as $x \rightarrow -\infty$, $f(x) \rightarrow -\infty$ and as $x \rightarrow 0^-$, $f(x) \rightarrow a-1 > 0$. Hence the intermediate value theorem tells us that there is a zero of $f(z)$ on the real axis $(-\infty, 0)$. and since the first part of the problem shows that $f$ has only one zero in the given half-plane, it follows that the only zero of $f$ in the given half-plane lies on the real axis.

My question here is this: why can we use Rouché's theorem here? Rouché's theorem (at least the version of it that I know) is only applicable to open bounded sets whose boundary is a simple closed contour. In this problem the half-plane $Re(z) < 0$ is certainly not bounded. So why can we apply Rouché's theorem as I did above? Or perhaps my solution is incorrect? Please let me know. I appreciate your help.

share|improve this question
3  
Your demonstration that there is a zero on the negative real line is valid (continuity of restrictions of holomorphic functions to $\mathbb{R}$ can be taken for granted, I suppose, allowing IMT to kick in), but your invocation of Rouche's is not because the open left half-plane is not a bounded. It might be possible to use Rouche's on some kind of specified bounded shape (e.g. hemisphere or rectangle) in the left-half plane with size as a parameter, and note that it works for arbitrary (sufficiently) large such domains and subsequently extends to the entire unbounded half-plane. –  anon Aug 6 '11 at 2:10
    
@anon It seems like that works. –  Dylan Moreland Aug 6 '11 at 4:01
    
@anon: Dear anon, thanks a lot for your comments. I was able to fix this argument following your suggestion. If you post your comment as an answer, I will accept it. Once again, thank you. I appreciate your help. –  algebra_fan Aug 6 '11 at 8:04

1 Answer 1

up vote 2 down vote accepted

[Comment reposted as answer.]

Your demonstration that there is a zero on the negative real line is valid (continuity of restrictions of holomorphic functions to $\mathbb{R}$ can be taken for granted, I suppose, allowing IMT to kick in), but your invocation of Rouche's is not because the open left half-plane is not a bounded. It might be possible to use Rouche's on some kind of specified bounded shape (e.g. hemisphere or rectangle) in the left-half plane with size as a parameter, and note that it works for arbitrary (sufficiently) large such domains and subsequently extends to the entire unbounded half-plane

share|improve this answer
2  
You can apply Rouché to the ball $B_1(-a)$ and $g(z) = z + a$ to see that the zero is located in there. To show that there is no zero outside that ball in the left half plane can be achieved with a rather straightforward estimate. –  t.b. Aug 6 '11 at 8:12
    
@Theo Buehler: Theo, thanks a lot for your comment. I really appreciate your help! –  algebra_fan Aug 6 '11 at 21:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.