I met an interesting phenomenon. Suppose $f(z)=\frac{1}{p(z)}$ where p(z) is a polynomial in $\mathbb{C}[z] $. If there exists a $ k \in \mathbb{N} $ and $ k>1 $ such that after you take $k$-th derivative for $f(z)$ (i.e $f^{(k)}(z)=\frac{g(z)}{h(z)}$), $g(z)$ has zero points, then it must have at least two $\it{distinct} $ zero points.

I can use some elementary approach to prove special case: (i) when $k=2$, according to the explicit formula for $f^{(2)}(z)$, I can prove this claim directly. (ii) when the $\deg p(z) = 2$, according to the partial fraction, I can also prove this directly. In general, the first approach seems very hard to apply and the second approach can give some information (actually, it will give a series of nonlinear relations on the roots of $p(z)$). What I am thinking next is that if we regard this relations as hypersurface, i want to show that the intersection of all these surface will only give no solution. But for the lack of the knowledge on this nonlinear part, I can not complete the proof.

(P.S. Since the number of the relations of the roots is much more than the number of the roots, which seems forcing the roots to be non existed; and this is exactly the reason why I believe this result to some extent.)

I am wondering whether someone can give a better approach which can be easily generalized to the general case.