Argument principle for non-compact domains

Question

Let $\Omega\subseteq\mathbb{C}$ be open, bounded and simply connected, with the boundary $\partial\Omega$ consisting of a single smooth curve. Let $z_0\in\Omega$, and let $f$ be a holomorphic function defined on a neighborhood of $\Omega$ which has a zero of multiplicity $1$ at $z_0$, and no other zeroes. The argument principle then tells us that $$\Delta_{\mathrm{arg}}:=\int_{\partial\Omega}\frac{f'(z)}{f(z)}\,dz=2\pi i.$$ What about the case where $\Omega$ is unbounded, but still open and simply connected, with $\partial\Omega$ still consisting of a single smooth curve? Can we say anything about $\Delta_{\mathrm{arg}}$ in this case?

Answer 1

For simplicity, we look at some examples where the smooth curve in cause is a vertical line, so the domain is a half-plane.

First you need some conditions on $f$ to have $\int_{\Re z=c}f'/fdz$ convergent even if only in the principal sense - for example if $f(z)=ze^z, \frac{f'}{f}=1+\frac{1}{z}$ and on any vertical line avoiding zero the integral doesn't exist

If now $f(z)=z$ it is easy to see that the integral of $\frac{f'}{f}=\frac{1}{z}$ on any vertical line avoiding zero is $\pm \pi i$ (depending if we are to the right or the left of zero while the integral is taken as the principal value so the limit of $\int_{-T}^{T}$) so half the bounded case value which basically expresses that at zero we have a $2\pi i$ jump or if you want that in one of the domains bounded by it there is no zero, in one there is a zero and we cannot distinguish between them by boundness as in the usual Jordan case.

Similar considerations apply in general and basically the only general thing we can say is that if $f'/f$ is nice and we can integrate it on vertical lines, we change the value of the integral by $\pm 2\pi i$ every time we pass through a simple zero/pole of $f$ but the overall result depends on the shape of the function in the plane (or at the boundary if extendable only on some domain containing some vertical lines)