Suppose $f(z)=\sum_{n=0}a_nz^n$ such that the radius of convergence is 1 and $a_0\neq0$.Let $s_n(z)=a_0+a_1z+\cdots+a_nz^n$ be the partial sums.Let $1>\delta>0$ such that $f\neq 0$ on $|z|=1-\delta$.Prove that there is $K>0$ and integer N such that for all $n\ge N$ the modulus of the product of zeros of $s_n(z)$ inside $|z|=1-\delta$ is at least $K$.

We know that $f(0)\neq0$,what bothers me is that since for sufficiently large $n$,the zeros of $s_n(z)$ should be near $f$,what if $f$ has no zeros inside $|z|=1-\delta$.The reason to choose a smaller circle than the unit one seems to avoid the zeros of $s_n$ goes to the boundary.I sense that I shall invoke the radius of convergence to derive some constraints on the coefficients so as to ensure polynomial $s_n$ will have zeros inside the circle,which is near zero.

  • $\begingroup$ You have a typo, we know in fact that $f(0) \neq 0$. If $f$ has no zeros in $\{z : \lvert z\rvert < 1-\delta\}$, no problem, the value of the empty product is $1$. The reason to choose a smaller circle is that we then have uniform convergence on the closed disk, and all functions are defined (and continuous) on the closed disk. Now, you already mentioned Hurwitz's theorem, so what does that tell us about the product of the zeros of $s_n$ in the smaller disk? $\endgroup$ May 14, 2014 at 20:52
  • $\begingroup$ @DanielFischer I get what you are saying,and in this way this problem seems very simple I don't know what's the nontrivial point here. $\endgroup$
    – Daniel S.
    May 15, 2014 at 7:38

1 Answer 1


A Hurwitz-based proof may be intuitively clear, but writing it out in every detail takes a bit of work. I'd say: $f$ has distinct zeros $a_1,\dots,a_m$ with multiplicities $k_1,\dots,k_m$. Let $r>0$ be small enough so that the $r$-neighborhoods of the zeros are disjoint and stay inside of $|z|=1-\delta$. Also make sure that $r<\frac12 \min|a_j|$. By Hurwitz, for large enough $n$ the function $s_n$ has $k_j$ zeros inside of the $r$-neighborhood of $a_j$; and it has no other zeroes inside of $|z|=1-\delta$. It follows that the product of its zeroes is at least $$\prod_{j=1}^m (|a_j|-r)^{k_j} \ge \prod_{j=1}^m (|a_j|/2)^{k_j} $$ which is the desired lower bound.

An alternative (in my opinion, better) approach is to use Jensen's formula.

Remark. The statement could be made cleaner and stronger: for every $\delta\in (0,1)$ there exist $K>0$ and $N$ such that for all $n\ge N$ the modulus of the product of zeros of $s_n$ inside $|z|=1-\delta$ is at least $K$.

Indeed, if $\delta$ happens to be such that $f$ has a zero on the circle $|z|=\delta$, you can use some larger value of $\delta$ instead. There are at most countably many values of $\delta$ to avoid, since $f$ has at most countably many zeros in the unit disk.


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