Suppose I have a polynomial satisfying the condition
$$\prod_{i = 1}^{r}(z - a_i) = n$$
where all $r, a_1, \dots, a_r, n \in \mathbb{N}$ are non-zero and $r$ is assumed greater than $2$.
Letting $z \in \mathbb{C}$ allows for the definition of a polynomial $P(z)$ with integer coefficients where
$$P(z) = z^r + \cdots + c_1z + c_0$$
and
$$\frac{dP(z)}{dz} = rz^{r - 1} + \cdots + c_1.$$
The constants may be given in terms of $a_1, \dots, a_r, n \in \mathbb{N}$ above. I have two questions about how to interpret this and what I am allowed to do from a mathematical operations standpoint.
The first question refers to the fact that in a sense this polynomial is already factored, so am I allowed to take the derivative of the L.H.S.? this would give
$$\frac{d}{dz}\prod_{i = 1}^{r}(z - a_i) = \frac{d}{dz}(n).$$
Since $n$ is assumed to be a constant it would then follow that
$$\frac{d}{dz}\prod_{i = 1}^{r}(z - a_i) = 0.$$
Expanding out the polynomial shows
$$\frac{d}{dz}\prod_{i = 1}^{r}(z - a_i) = \frac{d}{dz}(z^r + \cdots + c_1z + (-1)^r\prod_{i = 1}^{r}a_i) = 0$$
where it follows that
$$\frac{d}{dz}\prod_{i = 1}^{r}(z - a_i)= rz^{r - 1} + \cdots + c_1 = 0.$$
Sinse
$$\frac{dP(z)}{dz} = rz^{r - 1} + \cdots + c_1,$$ it follow that
$$\frac{dP(z)}{dz} = 0.$$
However, this implies that the derivative of $P(z)$ is equal to $0$ for all $z$. How are we to interpret this and use it to find general solutions, or at least a relationship between roots? One thought I had was to take $\ln$ of both sides since
$$\frac{P'(z)}{P(z)} = \frac{d}{dz}\ln{P(z)}$$
and use the fact that
$$\frac{d}{dz}\ln(\prod_{i = 1}^{r}(z - a_i)) = \frac{d}{dz}\ln(n).$$
This allows for
$$\frac{1}{z - a_1} + \frac{1}{z - a_2} + \cdots + \frac{1}{z - a_r} = 0.$$
However, I do not see a way that this would be useful.
Indeed, if the polynomial $P$ is constant, we would actually expect the derivative to be zero.
Your last equation just rearranges this fact. Multiply it through by $P(z)$, then you have the differential of $P(z)$ on the LHS (via the product rule).