# Asymptotic behavior of complex function

So I have a function $f(x)$ which I know behaves like $\alpha \sqrt{x} + \frac{\beta}{\sqrt{x}}$ for large $x$. I want to extend $f(x)$ to $f(x + i y) = f(z)$, and I was hoping that the asymptotic behavior would just be $\alpha \sqrt{z} + \frac{\beta}{\sqrt{z}}$, but it isn't. I don't really understand why that is, because the way I found the asymptotic behavior of $f(x)$ was to just consider large $x$ and sort of chip away terms until those two were left. I could write the complex parameter as $z=r e^{i \theta}$, so I would have thought that considering large $x$ was the same as considering large $r$. The only other way I can think of doing this would be to explicitly write out the real and imaginary parts of $f(z)$ and then find the asymptotic behavior of each of them. I sort of started doing that, but it's turning into a nightmare so I was wondering if there was an easier way to do this, or if it actually should just be $\alpha \sqrt{z} + \frac{\beta}{\sqrt{z}}$ and my numerical tests are wrong.

• Have you heard of Stokes' Phenomenon? For example, see en.wikipedia.org/wiki/Stokes_line – Ron Gordon Dec 21 '13 at 23:44
• @RonGordon I hadn't heard of that, but that sounds like something to look at. It doesn't really say how to find a Stokes line but I'm sure it's mentioned somewhere. – Mr. G Dec 22 '13 at 0:04
• Unfortunately, the behaviour of a function of a complex variable is rather difficult to judge from just its behaviour for real values. For example, if you just look at real values, the function $f(x) = \sin x$ is nicely bounded and just oscillates between $-1$ and $+1$, but when you consider $f(z) = \sin z$ as a function of a complex variable it is unbounded. I would suspect that your numerical tests are not wrong, but the complex function might just be more ... complex :) – Old John Dec 22 '13 at 0:19
• Well said, Old John. I believe there is an entire body of theorems addressing the behavior of holomorphic functions, which we could skip if we could just pretend R is the story. Actually, the story is somewhat the other way around; if we know what's going on with the complex function, we might learn something about it as a real function. – Betty Mock Dec 22 '13 at 0:26
• "The shortest path between two truths in the real domain passes through the complex domain" - Jacques Hadamard (1865-1963) – Old John Dec 22 '13 at 0:28

This function extends to any simply connected domain in $\mathbb C$, which misses the origin. For example, the most typical such domain is $\mathbb C\smallsetminus(-\infty,0]$. You can not thus talk about the behaviour of this function, as $z\to\infty$, because this function can not be defined in any domain which contain a whole neighborhood of $\infty$!
• Is the part about missing the origin only there because there's a $1/\sqrt{x}$ term? That term 'doesn't exist' for small $x$. Like, near the origin, $f(x)$ behaves linearly. – Mr. G Dec 22 '13 at 0:41
• The origin can not be an element of the domain where the extension of $f$ is defined because of the appearance of the $\sqrt{x}$. Even if $f$ extend to a Riemann surface (roots extend to Riemann surfaces), still zero has to be missing. – Yiorgos S. Smyrlis Dec 22 '13 at 0:45
• I see. So if the asymptotic behavior of $f(x)$ was something simpler like $x$, then there wouldn't be a problem? – Mr. G Dec 22 '13 at 0:51
• It should be made clear what "this function" means here: it is the function $\alpha\sqrt{x}+\beta/\sqrt{x}$ itself, not the (classified) function $f$ in the question. To @Mr.G If the asymptotic behavior was something like $x$, there would be the same problem: nobody can tell you what the asymptotic of $f$ is in the complex plane just by looking at its asymptotics on the real line. Examples were already given in comments, here is one more: add $e^{-x^2}$ to the function: it is very tiny on the real line, and huge in the complex plane. – Post No Bulls Dec 28 '13 at 17:06