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I can propose an algorithm for doing this for ``Belyi curves'', the curves whose coefficients are algebraic numbers. According to Belyi's theorem, such a curve of genus $g$ is defined by a combinatori …

These invariants are traditionally called moduli. Riemann found that there are $3g-3$ complex parameters defining a curve of genus $g>1$. The modern statement is that the space of curves of genus $g$ …

I think that the answer is no. Here is a somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are not related by any equation of the form $f(x)=g(y …

$2g=da+ma_1-2a-d-m+2$, where $a=a_1m,$ and $m$ is the g.c.d ($d,a$). Edit. Let $\chi(S)=2-2g$ be the Euler characteristic. Hurwitz formula gives $$\chi(S)=2a-r,$$ where $r$ is the ramification: a bra …

Such $f$ exists on every open Riemann surface: R.C. Gunning and R. Narasimhan. Immersion of open Riemann surfaces. Math. Ann., 174:103–108, 1967.

Let $C$ be the Riemann sphere, $p=0$. Then $$\omega(z)=\left(\sum_{-\infty}^\infty c_nz^n\right)dz.$$ Here the part with negative powers converges for $|z|>0$, while the part with positive powers con …

EDIT. My previous answer was incorrect. So I replace it. The answer is no. A counterexample is $$y^{2m}+(z^{m-1}y-x^m)^2.$$ This is of degree $2m$ but $\delta$ is like $\epsilon^{2m^2}$ near the point …

An algorithm exists in principle, at least when the genus is $0$. But it is very difficult unless the degree is small. For example, if the function is supposed to be a polynomial, as in your example, …

In your example, everything is easy to calculate. Use the fact that your $f$ is a composition of $f(z)=z+1/z$ and $h(z)=z^2$. The plane is partitioned into $8$ regions by the coordinate axes and the …

There are many simpler sequences which are not orbits, for example, it is impossible to have $a_n=a_{n+1}\neq a_{n+k}$ for some $k>1$. This can be generalized to $a_n=a_{n+m}\neq a_{n+km},\; k>1$. In …

One large area of research which addresses questions of this type is related to the Lee-Yang theorem, which gives a sufficient condition in terms of coefficients for zeros of a polynomial to lie in th …

Algebraically - no. Your problem is reduced to computing the area of intersection of an ellipse with a disk. There can be different configurations, but one of them is when the centers coincide, and th …

Several proofs are available. If you are interested in a short algebraic proof, I can recommend S. Lang, Introduction to algebraic and Abelian functions, Addison-Wesley, Reading, MA, 1972. First 25 pa …

The answer is yes, in principle, and this is explained in the reference that you cite. But a "simple" criterion probably does not exist. The state of the art is described in this book Sottile, Frank …

I've heard from a reliable source that Grothendieck decided on this name because he wanted to include the first person who discovered the thing and the last person who made an important contribution:- …