# Expansion of the zeros of a polynomial with smooth coefficients

I have given a polynomial of the form $p(z;\varepsilon) = \sum^N_{n=0} f_n(\varepsilon) z^n$ where $z \in \mathbb C$ and $\varepsilon \in (0,\infty)$. I want to expand the zeros $\zeta_j(\varepsilon)$ in a series of the form $\zeta_j(\varepsilon) = \varepsilon^{\sigma_j} \sum^\infty_{k=0} a_{j,k} \varepsilon^k$ where the $\sigma_j \in \mathbb Z$ and the coefficients $a_{j,k}$ are complex. My general questions are: is this called Puiseux expansion? How do i find these expansions? For example, take $p(z;\varepsilon) = e^\varepsilon +(\frac{1}{\varepsilon} + 1) z + \varepsilon^2 z^2 - \varepsilon z^3$ then i dont even know how to find these functions $\zeta_j$. I guess there is a formal way to write down these expansions and then afterwards one checks where it does make sense analytically. Any help (suggestions) appreciated!

First of all, even solving $x - f(\varepsilon) = 0$ in terms of a power series in epsilon isn't possible for arbitrary smooth $f$. You at least want to assume $f$ is analytic at some point ($0$ without loss of generality), and even then you can't expect the power series you get out to be convergent for all values of $\varepsilon$, so you will have to settle for some neighborhood of your starting point (which again I will take to be $\varepsilon = 0$).
Even this isn't enough though: Consider the equation $x^2-\varepsilon=0$. This has perfectly well behaved coefficients at $0$, but there is no analytic in $\varepsilon$ expansion of $\sqrt{\varepsilon}$ at $0$. We can fix this however by formally introducing a symbol $\varepsilon^{1/2}$ and expressing our solutions in terms of that (in this case $\pm\varepsilon^{1/2}$), more analytically this is just choosing a branch of the square root function.
For simplicity I'd like to also assume that the lead coefficient is not $0$ when $\varepsilon = 0$, in this case we may divide through and assume we have something monic. I will leave how to handle the case when the lead coefficient vanishes as something for those interested to work out. Now we have the following version of the Newton-Puiseux theorem:
Let $p(z,\varepsilon) = x^k + \sum_{n=0}^{k-1} f_n(\varepsilon)z^n$ with each $f_n$ analytic at $0$. If $\zeta$ is a $d$-tuple root of the polynomial $p(z,0)$, then there exist $d$ solutions (counted with appropriate multiplicities) to $p(z,\varepsilon)= 0$ in the ring $\mathbb{C}[[\varepsilon^{1/d}]]$ (power series in $\varepsilon^{1/d}$) with leading term $\zeta$, moreover these converge for values of $\varepsilon^{1/d}$ in some neighborhood of $0$.