0
$\begingroup$

Is there an Analytic function $f(z) \neq 0$ throughout the open unit disc such that $f(\frac{ni^n}{n+1}) = 0$

I tried using identity theorem. But the issue is that the sequence $\frac{ni^n}{n+1}$is not convergent. Whereas the identity theorem requires that there exists a sequence in the open disc whose limit point is in the open disc. Then how can I prove/ disprove this statement

$\endgroup$
1
  • 2
    $\begingroup$ $f(z) \neq 0$ throughout the open unit disc such that $f(\frac{ni^n}{n+1}) = 0$ is self contradictory. I Believe you mean $f$ not identically $0$ with $f(\frac{ni^n}{n+1}) = 0$ for each positive integer $n$. $\endgroup$ Oct 21, 2023 at 5:17

1 Answer 1

2
$\begingroup$

There is such a function. For proving this it is enough to show that the given points have no limit points in the open unit disk. [This is Theorem 15.11 in my copy of Rudin's RCA, but the theorem number may be different in your copy]. Note that $|\frac {ni^{n}} {n+1}|=\frac {{n}} {n+1} \to 1$. This proves that there are no limit points in the open unit disk. By the quoted theorem, we can find an analytic function with simple zeros at the given points and no other zeros in the open unit disk.

EDIT: I just found an explicit example. Take $f_n(z)= \sin (\frac {\pi} {1-i^{-n}z})$. Note that there are only a finite number of distinct $f_n$'s (since $i^{4}=1$). Take $f$ to be their product.

$\endgroup$
2
  • $\begingroup$ Your proposed $f$ depends not only on $z$ but also on $n$, which doesn't make sense since $n$ is not fixed $\endgroup$ Oct 21, 2023 at 5:35
  • $\begingroup$ @BrevanEllefsen Thank you. I have corrected the answer. $\endgroup$ Oct 21, 2023 at 5:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .