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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.

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In other words, the thesis is : $g$ is not of the form $a(z-z_0)^m$ for $m > 0$. So a stronger claim, that maybe is simpler to treat, should be : If $p\in \mathbb{C}[z]$ is nonconstant polynomial and not a multiple of $z$, and if $u '(z)=z^m/p(z)$ then $u\notin \mathbb{C}(z)$. –  Pietro Majer Mar 28 '12 at 6:45
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@Pietro: That is wrong. Take $u(x)=-1/3{\frac {3{x}^{2}+3x+1}{ \left( x+1 \right)^{3}}}$. Its derivative is $\frac{x^2}{(x^2+2*x+1)^2}$. –  Mark Sapir Mar 28 '12 at 8:01
    
good point ! –  Pietro Majer Mar 28 '12 at 8:12
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1 Answer 1

This is not a solution, but could be partially helpful in one.

Let $n$ be the number of zeroes of $p$. Then a counterexample occurs if and only $f^{(k)}$ for some $k$ has a zero of order $k(n+1)$.

Proof: Let $f(z)=1/p(z)$. Since each pole of $f$ increases in order by $1$ each time it is differentiated, the total order of the poles of $f^{(k)}$ is equal to $\deg p+ nk$. So the total order of the zeroes (on the projective line) is equal to $\deg p+ nk$.

Either $k<\deg p$ or $k\geq \deg p$.

If $k<\deg p$, then, since the order of the zero of $f(z)$ at $\infty$ is $\deg p$, the order of the zero of $f^{(k)} (z)$ at $\infty$ is $\deg p-k$, so the total order of the zeroes of $f^{(k)}$ on $\mathbb C$ is $kn+k$. So $f^{(k)}$ has a unique zero if and only if it has a zero of order $kn+k$.

If $k\geq \deg p$, then there is always more than one zero of $f^{(k)}$ in $\mathbb C$. This is because every zero has order at most $\deg p-1$, since the coefficients of the power series of a rational function of degree $d$ satisfy a recurrence relation of length $d$ and so can have at most $d-1$ consecutive zeroes. (unless the function is a polynomial, which $f$ isn't.) So the number of zeroes, including $\infty$, is at least $k(n+1)/(\deg p -1)$, which is greater than $2$. QED

This has two relevant consequences. The first is that, to find a counterexample, we must solve $k(n+1)$ equations in $n$ variables (the roots of $p$), and $k>2$. If these equations are even a little bit transverse, we would be done, but I don't know how to prove that.

The second is that $\deg p> k (n+1)$, so there must be quite a lot of multiple roots of $p$. Since we know $k>2$ and $n>1$, $\deg p$ must be at least $10$.

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