The derivative of a polynomial $f$ over $\mathbb{C}$ has one less zero than $f$. Including various entire transcendental $f$, we can show that the pair of counts of zeros $(\#Z(f), \# Z(f'))$ ranges over all $(m,n)$ where $\infty = n \ge m-1$, or $m=\infty$ and $n < \infty$ (Hadamard factorization / great Picard possibly being used to justify some examples, for instance here or here). I wonder if there is an example where $n \le m-2 < \infty$, or how to show there is not. (The title references $m=2$, $n=0$.) The mean value theorem shows that $n \ge m-1$ for real analytic $f : \mathbb{R} \to \mathbb{R}$, but I don't see how to leverage this for holomorphic functions.
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$e^z-1$ jas a zero whenever $z=n2\pi i$, and its derivative is $e^z$ which is nonvanishing right? – Chris Jan 29 '25 at 08:19
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1@Chris So it doesn't have two zeros, but an infinity of those. – mathcounterexamples.net Jan 29 '25 at 08:21
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Yes, I mean exactly two zeros. Let me update the title to reflect this, thanks for the catch – llf Jan 29 '25 at 08:21
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@mathcounterexamples.net good point, missed that. – Chris Jan 29 '25 at 08:22
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1A possible starting point: if $f$ as in the title exists, we have $f = pe^g$ and $f' = e^h$ for entire functions $g,h$ and some polynomial $p$ of degree $2$. Differentiating the first equation and equating to the other gives a relation between $g$ and $h$. – llf Jan 29 '25 at 08:36
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Yes, such a function exists. Define $$h(z)=\frac{e^{i\pi z}+2z-1}{z(1-z)}.$$ The numerator evaluates to zero at $0$ and $1$, and thus this function is entire. In particular, it has an antiderivative $g(z)$. Let $f(z)=z(1-z)e^{g(z)}$. We compute $$f'(z)=e^{g(z)}\big(z(1-z)g'(z)+1-2z\big)=e^{g(z)+i\pi z},$$ so $f'$ is everywhere nonvanishing.
Carl Schildkraut
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Excellent example (easily generalized to any degree polynomial btw so one gets $f$ with arbitrary finitely many zeroes st $f'$ is non vanishing – Conrad Jan 30 '25 at 23:13