Number of roots of a family of polynomials

Question

For all $\varepsilon \in \mathbb{C} \setminus \{0\}$, consider the polynomial $$ f_\varepsilon(z) = (z^3 + \varepsilon)^3 + (z^5 + \varepsilon)^2. $$ I want to prove (or disprove) that, for all sufficiently small $\varepsilon$, the polynomial $f_\varepsilon$ has exactly $9$ distinct roots of multiplicity $1$ near the origin. (More precisely, given $0 < \delta < 1$, for all sufficiently small $\varepsilon$, the polynomial $f_\varepsilon$ has exactly $9$ distinct simple roots in the disk $\{|z| < \delta\}$.)

Answer 1

There are several ideas outlined in the comments above, but I want to give a proof which uses only ideas from differential topology. In particular, the following proof avoids using almost any general facts about polynomials.

The relevant quantity will be the intersection number $I(f_0|_D, 0)$, where $D$ is the closed disk of radius $1 / 2$ centered at the origin. By the transversality theorem $0$ will be a regular value of $f_\varepsilon$ for generic $\varepsilon$, in which case $f_\varepsilon^{-1}(0)$ will contain exactly $I(f_0|_D, 0)$ points. Thus, we want to show that $I(f_0|_D, 0) = 9$.

To compute $I(f_0|_D, 0)$, it suffices to find some $g : D \to \mathbb{C}$ such that

1. $0$ is a regular value of $g$;
2. $g$ is orientation preserving at each point of $g^{-1}(0)$;
3. there exists a homotopy $H : D \times I \to \mathbb{C}$ from $f_0$ to $g$ such that $0 \not\in H(\partial D \times I)$.

Then intersection theory (cf. Guillemin and Pollack) tells us that $I(f_0|_D, 0) = I(g, 0) = \# g^{-1}(0)$.

Consider $g(z) = z^9 + \delta$ for any $\delta < 2^{-10}$, and let $H(z, t) = z^9 + (1 - t)z^{10} + t\delta$. For all $z \in \partial D$, we have $$ |H(z, t)| \geq |z^9| - (1 - t)|z^{10}| - \delta > 2^{-9} - (1 - t)2^{-10} - 2^{-10} \geq 0, $$ so (3) is satisfied. (2) is satisfied since $g$ is holomorphic, and (1) is immediate. Clearly $g$ has nine roots, which proves $I(g, 0) = 9$, completing the proof.