# Justification for applying Rouché's theorem for an unbounded domain

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.

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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

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
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