I have read an interesting question recently. This is about a series having infinitely many zeros inside the unit disc $U$. The series is
\begin{equation*}
f(z)=\sum_{k=1}^\infty 5^k z^{n_k},
\end{equation*} where $\{n_k\}$ is a sequence of integers such that $n_1 \ge 2$ and $n_{k+1}>2kn_k$. I have shown that $f \in H(U)$ and it does not have radial limit at any point of the unit circle $\partial U$. However, I can't show that it has infinitely many zeros inside $U$. The tools that I can apply are graduate level complex analysis such as Rouche's theorem, the open mapping theorem, the maximum modulus theorem, Schwarz lemma, Runge's theorem, the Mittag-Leffler theorem and etc. Anyone can help?
Proof of a series having infinitely many zeros in the unit disc
complex-analysis
Best Answer
This function has the property that on the circle $|z|=1-\frac{1}{n_k}, k \ge 10$, $|h(z)|>C5^k$, where we can take for example $C=\frac{1}{100}$
(note that $(1-\frac{1}{n_k})^{n_k} \ge \frac{1}{3}$ as that sequence converges increasingly to $\frac{1}{e}$ with the inequality already happening for $k \ge 10$ and $n_k>k$, while for $m>k,(1-\frac{1}{n_k})^{n_m} \le (\frac{1}{e})^{2^{m-k}(m-1)...k} < 6^{-m}$ so the corresponding terms form a geometric sum that converges to a small finite number, while the smaller terms sum in absolute value to obviously less than $\frac{1}{4}5^{k}$ by the trivial estimate and the corresponding geometric series)
Assume now that $h$ has finitely many zeroes only (could be none here of course). Let $B$ a finite Blaschke product with the same zeroes (including multiplicities etc, where we take $B=1$ if $h$ has no zeroes). Then $g=\frac{h}{B}$ has no zeroes in the unit disc and is analytic and still satisfying $|g(z)| > C5^k, |z|=1-\frac{1}{n_k}$, since $|B(z)| <1, |z| <1$.
But then $\frac{1}{g}$ is analytic in the unit disc and by maximum modulus, $|\frac{1}{g}| < \frac{1}{C5^k}, |z| \le 1-\frac{1}{n_k}$. If we let $k \to \infty$ we get $\frac{1}{g}=0$ in the unit disc and that is a contradiction.
Note that $h(z)-w$ has the same properties as $h$ (taking all $k$ large enough so $5^k > 200|w|$ say, so we can use $C=\frac{1}{200}$ say, so the same proof applies to show that $h$ takes every complex value infinitely many times in the unit disc!