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We know that the zeros of an analytic non-constant function are always isolated. A proof is here. Let $L(v)$ be an analytic function in $v$, where $v\in\mathbb{R}$.

Let us write $L(v) \equiv L(v,p)$ where $p \equiv p(v)$ is an analytic function. Moreover, it is injective (one-one) too.

Can we say that the zeros of $L(v,p)$ are also isolated ? I mean "isolated" on the $v$-$p$ plane, i.e, If $L(v_{0},p_{0}) = 0$, then $\exists$ $\epsilon > 0$ such that there is no other zero of $L(v,p)$ in a disk of radius $\epsilon$ centered at $(v_{0},p_{0})$.

I also have a followup question here.

Update: It is ok to assume that $L(v)$ is such that there exists an analytic function $p \equiv p(v)$ such that $L(v,p)$ is a bivariate polynomial.

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    You contradicted yourself again. $L(v,p(v))$ is not a bivariate polynomial: it only has one variable $v$.2012-12-11

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