Zelaron Gaming Forum - Number of zeros of a certain type of rational function

Perhaps you can assist me with some intuition. I'm looking for the number of zeros of $\reverse f_{m,n}(z) : = \frac{\mathrm{d}^m}{\mathrm{d}z^m}\left[\left(\frac{P(z)}{Q(z)}\right)^n\right]$ , where $\reverse P(z)$ and $\reverse Q(z)$ are nonconstant polynomials with no common zeros, and $\reverse m,\,n$ are nonnegative integers. As an example, the case $\reverse P(z) = 2(z-1)^4(z-3)(z-7),\:Q(z) = (z+1)(z-2),\:m=0,\:n=5$ yields $\reverse f_{0,5}(z) = \frac{32(z-1)^{20}(z-3)^5(z-7)^5}{(z+1)^5(z-2)^5}$ , which has thirty zeros (the zeros of the polynomial in the numerator).

For some convenient notation, let $\reverse p : = \deg{P(z)},\: q : = \deg{Q(z)}$ and $\reverse Q(z) = A(z-z_1)(z-z_2)\dots (z-z_q)$ , where $\reverse A$ is a complex number. I tried to decompose the rational function $\reverse f_{m,n}(z)$ in terms of its poles as follows:

$\reverse f_{m,n}(z) = H_{m,n}(z) + \frac{C_{z_1,1}}{(z-z_1)^{m+1}} + \frac{C_{z_1,2}}{(z-z_1)^{m+2}} + \dots + \frac{C_{z_1,n}}{(z-z_1)^{m+n}} + \frac{C_{z_2,1}}{(z-z_2)^{m+1}} + \frac{C_{z_2,2}}{(z-z_2)^{m+2}} + \dots + \frac{C_{z_q,n}}{(z-z_q)^{m+n}}.\:\: ( 1 )$

Here, the $\reverse C_{z_i,j}$ are complex numbers, and $\reverse H_{m,n}(z)$ is a polynomial of degree $\reverse \max\{(p-q)n-m,\,0\}$ . Specifically, if $\reverse p\ge q$ and $\reverse 0\le m\le (p-q)n$ (which is precisely when $\reverse H_{m,n}(z)$ does not vanish), we can write the terms in the right-hand side of equation $\reverse ( 1 )$ as a fraction with a common denominator, such that the polynomial $\reverse H_{m,n}(z)[(z-z_1)(z-z_2)\dots(z-z_q)]^{m+n}$ dominates the degree in its numerator. Hence, $\reverse f_{m,n}(z)$ has $\reverse (p-q)n-m+q(m+n) = pn + (q-1)m$ zeros in this case.

There seem to be two more distinct cases, but I'm not sure how to prove what the number of zeros is in them. The answer in those (remaining) cases should (probably, based on my numerical experiments) be

$\reverse pn + (q-1)m$ if $\reverse p < q,$

and

$\reverse q(m+n)-(m+1),$ if $\reverse p \ge q$ and $\reverse m > (p-q)n.$

Any ideas?