Some calculations with polynomials in the proof of the Routh Hurwitz Test In the article with the proof of the Routh Hurwitz test I couldn't see why the following result is true:

Let $$p(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_{1}s+a_0$$ and $$ \eta_{*} = \frac{a_n}{a_{n-1}}$$ $$g_{\eta}(s):=p(s)-\eta (a_{n-1}s^n+a_{n-3}s^{n-2}+\cdots)$$  Then one and only one zero of $g_{\eta}(s)$ tends to $\frac{-a_{n-1}}{a_n-\eta a_{n-1}}$ as $\eta \in \mathbb{R}$ approaches $\eta_{*}$ from the origin.

Notice that when $\eta=0$ then $g_{\eta}(s)=p(s)$ and when $\eta=\eta^*=\frac{ a_n}{a_{n-1}}$ then $$g(s)=p(s)-\frac{a_n}{a_{n-1}}(a_{n-1}s^n+a_{n-3}s^{n-2}+a_{n-5}s^{n-4}+\cdots)$$ a polynomial of degree $n-1$.
The name of the article is "Elementary proof of the Routh-Hurwitz test", and in the part quoted above, the author has already demonstrated that $p(s)$ and $g(s)$ have the same number of purely imaginary roots. I'd appreciate any help. Thanks in advance!
 A: The claim is close to the following: For $\varepsilon\approx 0$ and some polynomial $r(z)$ of degree $n-1$, the polynomial
$$
p(z)= εz^n+r(z)
$$
has exactly one root $\sim -\frac{r_0}{ε}$. The basic idea for a proof is that $p$ still has $n$ roots, $n-1$ accounted for as close to those of $r$, so the remaining one must be $\sim ε^{-1}$ for reasons of the Viete equations.
If $\zeta$ is any root of $r$, then it is an approximate root of $p$ with an exact root close-by if $ε$ is small enough, one can use the Rouché theorem.
For the last root rescale the equation using $w=εz$,
$$
ε^{n-1}p(z)=w^n+r_0w^{n-1}+εr_1w^{n-2}+...
$$
Now the later terms are the perturbation. The $n-1$ roots at zero of $0=w^n+r_0w^{n-1}$ correspond to the previously identified roots close to those of $r$. The only non-trivial root of the leading terms is $\omega=-r_0$, which again is close to a root of the full polynomial in $w$.
Thus there is exactly one root going to infinity like $-\frac{r_0}{ε}$.
The actual situation in question has $p(z)=εq(z)+r(z)$ with $\deg(p)=n$, so what counts as perturbation term gets a little more involved in the second step, but nothing much changes.
