Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $0$. – robjohn Nov 2 '11 at 18:02
@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 $\epsilon>0$. – stopple Nov 2 '11 at 18:11
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 $p'$. – AD. Nov 2 '11 at 19:25
Forgot to mention the constrain that $p$ in the above comment is of degree $\ge2$. – 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 $2$?

share|cite|improve this answer
The $2$ appears because of the normalization of the zeros to be $\pm 1$. In general if the zeros are $\pm\lambda$, one sees $2/\lambda^2$. – stopple Nov 3 '11 at 22:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.