I posted [this question](https://math.stackexchange.com/questions/860973/) on MSE a few time ago, but it did not receive much attention. I thought there might be an elementary answer so didn't want to post it directly on MO. My apologies if this question is not a good fit here.


Let $f:\bar{\mathbb C}\to \bar{\mathbb C}$ be a rational function, and take a circle $C$ not crossing the zero- and polar-locus of $f$. The argument principle tells us the total variation $\Delta$ of $\arg f$ along $C$, obtained by performing the analytic continuation of $g : t\in[0,1]\mapsto \arg f(z(t))$ for some parameterization $z(\bullet)$ of $C$:$$ \Delta = g(1)-g(0)=2\pi(Z-P)$$ where $Z$ is the count of roots of $f$ encircled by $C$ and $P$ that of its poles (of course this value can also be obtained by integrating $\frac{f'}{if}$ along $C$). 


What it does not tell is the size of $g([0,1])$. **Question:** how big can this interval get? Is there a known bound, *e.g.* in terms of the degree of $f$ and/or the zeroes and poles encircled by $C$? 


Also, although the answer might not depend on the special shape of $C$, I don't want to assume $C$ to be «small enough» or to match too restrictive hypothesis of the like.



Variation of the argument of a rational function along a circle
There are some easy bounds. Let $f(z) = \dfrac{g\_1(z) g\_2(z)}{h\_1(z) h\_2(z)}$
where $g\_1, g\_2, h\_1, h\_2$ are polynomials not identically $0$, $g\_1$ and $h\_1$ having their roots
inside the circle $C$ and $g\_2$ and $h\_2$ having their roots outside.
Then the length of $g([0,1])$ is at most $2 \pi (\deg(g\_1) + \deg(h\_1)) + \pi (\deg(g\_2) + \deg(h\_2))$. This is best possible in the case where only one of 
$g\_1, g\_2, h\_1, h\_2$ is non-constant.


