If an analytic function $f(z)$ has two zeros which are close together, there is a zero of $f^\prime(z)$ which is nearby. This is clear if $f(z)$ is a quadratic polynomial, and intuitive in general from the Hadamard product: $f(z)$ is a quadratic polynomial times a function which is nearly constant in the neighborhood of the zeros.

How rigorous can we make this statement?

Without loss of generality, we can translate so the origin is the midpoint of the two zeros, rotate so they are both real, and dilate so the zeros are $\pm1$, so $f(z)=(1-z^2)g(z)$, with (suppose) $g(z)\ne0$ on $|z|<1$. Thus $\log(g(z))$ is well defined on the disk, and we can write $g(z)=\exp(h(z))$ with analytic $h(z)$.

Thus we seek $z$ (hopefully in $|z|<1$) satisfying $-2z+(1-z^2)h^\prime(z)=0$.

Edit: Based on robjohn's comment, we need to quantify the heuristic above that the 'remainder' of the Hadamard product is roughly constant near the zeros. Thus assume also that $g^\prime(0)$ is 'small', where the proposed solution below makes this precise.

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    $\begingroup$ The function $f(z)=e^{2\pi iz}-1$ is $0$ at each integer, yet $f'(z)=2\pi i\;e^{2\pi iz}$ is never $0$. $\endgroup$
    – robjohn
    Nov 2 '11 at 18:02
  • $\begingroup$ @robjohn: Nicely done, and so much for folklore. The problem is not the 'wlog' either, as one can take $\exp(2\pi iz/\epsilon)-1$ for arbitrary $\epsilon>0$. $\endgroup$
    – stopple
    Nov 2 '11 at 18:11
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    $\begingroup$ You might be interested in Sendov's Conjecture: If $p(z)=(z-z_1)(z-z_2)\cdots(z-z_n)$ is a polynomial where the zeros $z_k$ are in the closed unit disc, then each of the discs $D_n=\{w:|w-z_k|\le1\}$ contains a zero for $p'$. $\endgroup$
    – AD.
    Nov 2 '11 at 19:25
  • $\begingroup$ Forgot to mention the constrain that $p$ in the above comment is of degree $\ge2$. $\endgroup$
    – AD.
    Nov 3 '11 at 4:47

Edit: Here's a cleaner solution based on the assumption that $\log(g(z))$ is roughly constant near the zeros.

Write $\log(g(z))=h(z)=h(0)+h^\prime(0)z+h^{\prime\prime}(0)z^2/2+O(z^3)$, and consider the series expansion of $$ -2z+(1-z^2)h^\prime(z). $$ Up to terms which are $O(z^2)$, this is $$ h^\prime(0)+\left(h^{\prime\prime}(0)-2\right)z. $$ Thus neglecting the $O(z^2)$, the expression is $0$ for $$ z=\frac{h'(0)}{2-h''(0)}. $$ This is in the unit disk for $h^\prime(0)$ sufficiently small, as long as $h^{\prime\prime}(0)$ is not too close to $2$.

Question: What is the 'meaning' of the second derivative of $\log(g(z))$ at $0$ being near $2$?

  • $\begingroup$ The $2$ appears because of the normalization of the zeros to be $\pm 1$. In general if the zeros are $\pm\lambda$, one sees $2/\lambda^2$. $\endgroup$
    – stopple
    Nov 3 '11 at 22:02

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