# Can an algebraic function be zero both at $z=0$ and at its leading singularity?

Apologies for asking possibly strange questions, but I am just a poor computer scientist trying to understand a mathematical paper on singularity analysis of algebraic functions that is apparently not meant to be read by me...

When trying to figure out one of the multiple "it is obvious that..." in the paper, it seems to me that the authors might implicitly assume that some $\mathbb{N}$-algebraic functions are non-zero at their leading singularity $\rho > 0$ (and as a consequence, their Puiseux expansions at $\rho$ have a constant term). The idea I have got is that this could be somehow related to the fact that algebraic functions in question are all zero at $z=0$.

So my question is, if in general $f(0) = 0$ implies $f(\rho) \neq 0$ for $\mathbb{N}$-algebraic functions $f$ with leading singularity $\rho$. (Let us suppose that the leading singularity is unique.)

Some simple examples seem to support this: for instance, if an algebraic function containing $\sqrt{1-z}$ has to be zero at $z = 0$, it can be achieved by adding some constant terms, e.g. $f(z) = 1 - \sqrt{1-z}$. However, the constant term $1$ is what makes the function non-zero for $z=\rho=1$.

Does something like this hold in general or are my thoughts just completely wrong? It may be that I am just missing something really trivial, as my knowledge of analysis is poor and dusty.

• That no one has commented in the nine hours this question has been up emboldens me to admit to my ignorance here. I know what an algebraic function is, but I don't know what an N-algebraic function is. I also don't know what a leading singularity is. And I surely don't know whether $z\sqrt{1-z}$ would be an example of an N-algebraic function zero at both $z=0$ and at its leading singularity, but I'd like to find out. Jun 2, 2016 at 23:07
• @GerryMyerson Thanks for reply. $\mathbb{N}$-algebraic systems are defined as solutions of "well-posed" systems of algebraic equations, where each polynomial in the system has nonnegative integer coefficients. Moreover, there is a requirement that the solution has to be a power series with nonnegative coefficients as well - only such functions can be called $\mathbb{N}$-algebraic. The definition can be found, e.g., here. As their power series contain negative coefficients, I think that $z\sqrt{1-z}$ and $\sqrt{1-z}$ are not $\mathbb{N}$-algebraic. Jun 3, 2016 at 8:28