The Routh-Hurwitz Stability Criterion is essentially an algorithm to determine how many roots a polynomial has in the right-hand-side of the complex plane (that is, how many of its roots have positive real part) without having to actually calculate those roots (which is, in principle, impossible for high-degree polynomials). However, sometimes this algorithm runs into the division by zero error, and you have to calculate it using limits. However, this video by Neso Academy claims that there is an easier way: simply replace s in the polynomial with 1/z. Does that really always work? If so, how?
Replacing s in the polynomial with 1/z, as far as I understand it, effectively "reverses" it (that's how I implemented it in my implementation of the Routh-Hurwitz Stability Criterion). For example, the polynomial a*s^2 + b*s + c becomes c*z^2 + b*z + a. And the polynomial a*s^3 + b*s^2 + c*s + d becomes d*s^3 + c*s^2 + b*s + a. So, does "reversing" the polynomial preserve the number of the roots in the right-hand-side of the complex plane (that is, the number of roots with a positive real part)? If so, why?
As you've said, reversing the polynomial $p(x)$ with complex roots $\{r_1,r_2,\dots,r_n\}$ gives a polynomial $q(x)$ with complex roots $\left\{\frac{1}{r_1},\frac{1}{r_2},\dots,\frac{1}{r_n}\right\}$.
Furthermore, for some nonzero root $r=a+bi$ of $p(x)$, the real part of $\frac{1}{r}$ must have the same sign as the real part of $r$, as
$$ \frac{1}{r}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i $$
and $a^2+b^2>0$.
So $p(x)$ and $q(x)$ must have the same number of roots with positive (or negative, or zero) real part.