Following @mercio's comments, I've rewritten my question in terms of zeros instead of saddles. Also, after more careful consideration, I've decided that perhaps the path I seek might not depend on the presence of poles or choice of end points, but I'm not sure. In any case, please help me prove or disprove the following statement:

In any connected region on which a meromorphic function is defined, there exists a path along which the function is smooth and analytic, and the zeros of the function and all its higher derivatives (inside the region) lie.

I was also thinking about whether such a path would be unique, but I don't need uniqueness for my problem. However, I need such a path to exist in the limit where the size of the region becomes infinite. Actually, I also need the path to be regular and differentiable, but I'll leave that question for later.

The reason I would like such a path is to consider a restriction of the function or its derivatives to the path. If the path is $z(s)=x(s)+\mathit{i} y(s)$, where s is arc length, and the function or one of its derivatives $w(z) = u(z)+\mathit{i}v(z)$, then

$$ \frac{du}{ds}=\frac{\partial u}{\partial x}x'(s)+\frac{\partial u}{\partial y}y'(s) $$ $$ \frac{dv}{ds}=\frac{\partial v}{\partial x}x'(s)+\frac{\partial v}{\partial y}y'(s) $$

are 0 iff $w'(z)=0$. Also, since $w(z)$ is continuous and finite everywhere on the path, at least 1 extremum with $\frac{du}{ds}=\frac{dv}{ds}=0$ must exist between every pair of consecutive zeros of $w(z)$. Accordingly, the extrema must coincide with where $w'(z)=0$. I could therefore conclude $w'(z)$ has at least 1 zero for each pair of consecutive $w(z)$ zeros. Moreover, the $w'(z)$ zeros would be between the corresponding $w(z)$ zeros along the path.

I'm now wondering whether this kind of reasoning is even necessary, or there is some easier way to show the result I want. I suppose it's adequate if the path just has zeros of the function and its derivative, then I can use different paths for higher derivatives. However, it would be nice if I could use the same path for all derivatives.

As an example, I've tried to search for such a path for the digamma function $\psi(z)$, which I'm studying for Convexity of reciprocal polygamma. Using Mathematica, I computed zeros of $\psi^{(m)}(z)$ for $0\leq m\leq 8$, $-1<x<0$, and $y>0$. I then looked at the path obtained from connecting zeros from $\psi^{(m)}(z)$ with increasing $m$:

polygamma function zeros

The contours in the figure are labeled by the base-10 log of their values. Since the zeros can be connected in different ways, the path I obtained isn't unique.

Next, I parametrized the path connecting all zeros by arc length, $s$, then plotted $\psi^{(m)}(z)$ vs $s$:

Polygamma parametrized by arc length

$\psi^{(2)}(z)$ has zeros near $s=0$ and $s=0.5$. As expected, $\psi^{(3)}(z)$ has a corresponding zero near $s=0.05$ between those zeros. However, $\psi^{(2)}(z)$ also has extrema near $s=0.01$ and $s=0.35$, for which $\psi^{(3)}(z)$ has no corresponding zeros. Instead, these correspond to points where both $x'(s)=y'(s)=0$, which occurs because the path is irregular. For this path, therefore, my conclusion is invalid. It is then necessary to ask if a regular path exists, perhaps if I included more zeros, or connected the zeros in a different way?

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    $\begingroup$ what's a saddle point ? $\endgroup$ – mercio Nov 11 '16 at 8:02
  • $\begingroup$ en.wikipedia.org/wiki/Saddle_point $\endgroup$ – obsolesced Nov 11 '16 at 10:04
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    $\begingroup$ see, if you were talking about a function $\Bbb R^2 \to \Bbb R$ I would understand, but meromorphic functions are functions $\Bbb C \to \Bbb C$, so I would like you to give an example of a saddle point of a meromorphic function. $\endgroup$ – mercio Nov 11 '16 at 10:15
  • $\begingroup$ I thought the log-gamma was an example, but I realized it's not meromorphic because of the branch cut along the negative real axis. Consider instead the digamma function, which has saddles at zeros of the trigamma function. They are points where the derivative is 0. I can upload a picture if you like. $\endgroup$ – obsolesced Nov 11 '16 at 10:28
  • $\begingroup$ @mercio, I wasn't aware "complex saddles" isn't an accepted term in the community, as I come from a physics background, where it's commonly used. Would there be a more appropriate term? Critical/inflection points? I suppose I could avoid the term completely, and rephrase the question in terms of zeros of all derivatives, but I was thinking of saddles at the time of writing. $\endgroup$ – obsolesced Nov 11 '16 at 12:22

Not an answer, but sufficient for my needs. I realized my statement that the $u(z)$ and $v(z)$ extrema along the path must coincide with where $w'(z)=0$ is false. For example, it is possible to have $\frac{du}{ds}=0$ if $\frac{\partial u}{\partial x}=0$ and $y'(s)=0$, without $\frac{\partial u}{\partial y}=0$. So even if such a path exists, and is regular and differentiable, I cannot conclude $w'(z)$ has a zero for each pair of $w(z)$ zeros in the region.

For example, consider $w(z)=z^2-1$ in some annulus around $z=0$. Then only $w(z)$ has zeros in the annulus; no derivatives do. The path $e^{\mathit{i}\theta}$ for $0\leq\theta\leq\pi$ connects the $w(z)$ zeros, and possesses extrema that do not coincide with $w'(z)=0$.

Perhaps a better question to ask is whether, in general, in the complex plane, the number of $w'(z)$ zeros is at least that of $w(z)$ less 1, but the question doesn't really serve my needs either.


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