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I posted this question 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.

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  • $\begingroup$ very nice question. I think the answer depend on the shape of $C$.For example put $f(z)=z$. Now let $C:[0,1]\to \mathbb{C}$ is a simple closed curve surrounding the origin which image of its restriction to $[0,1/2]$ is identical to the image of the spiral curve $\lambda(s)=e^{-s+is}$, for $s$ sufficiently large. for such curve the image of $g$ is large. So is it a good Idea that we expect to bound the length of $g$ in term of length of $C$? $\endgroup$ – Ali Taghavi Jul 11 '14 at 9:03
  • $\begingroup$ @AliTaghavi I'm not so sure it is such a great question actually. I was aware that if $C$ is spiralling enough then you can get large variations (that's why the question is asked for circles). But you're surely right that there should exists a bound in terms of the length (or more likely «curvature deviation») of $C$. $\endgroup$ – Loïc Teyssier Jul 11 '14 at 9:19
  • $\begingroup$ I think the question is interesting because it assigne a new quantity to a simple closed curve and a vec. field along it.the quantity is the length of interval image of $g$. As a possible another quantity could be "the total variation" of $g$. $\endgroup$ – Ali Taghavi Jul 11 '14 at 10:02
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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.

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