0
$\begingroup$

Is there an example of an analytic function in the unit disc whose zeros are only the points $z_n=1-1/n$?

  • 0
    If there is such a function it should be $\prod_{n=1}^\infty (z-z_n)$.2012-04-16
  • 0
    @azarel: that doesn't converge.2012-04-16
  • 0
    @GEdgar Thanks for pointing that out.2012-04-16

3 Answers 3

5

Another one: $$ f(z) = \frac{1}{\Gamma\left(\frac{1}{z-1}\right)} $$ This is analytic in the plane, except one point $z=1$, and has zeros exactly $1-1/n$, $n=1,2,3\dots$. Unlike Henning's, which also has zeros ${} \gt 1$.

  • 0
    Thanks for the examples given, but how did you come to it? Maybe Henning's could have come to mind, but how did you think about this in general?2012-04-16
  • 0
    I think it's just a matter of already knowing $1/\Gamma(z)$ as a standard example of a holomorphic function whose zeroes form a singly infinite arithmetic sequence.2012-04-16
3

How about $f(z) = \sin(\frac{\pi}{1-z})$?

3

For the general question of functions with prescribed zeros, consider Weierstraß' factorization theorem.

  • 0
    Weierstrass factorization theorem is usually stated for *entire* functions, though.2012-04-16
  • 0
    True - I refered to this sentence from the Wikipedia page: "The theorem generalizes to the following: sequences in open subsets (and hence regions) of the Riemann sphere have associated functions that are holomorphic in those subsets and have zeroes at the points of the sequence."2012-04-16
  • 1
    Just use a fractional linear transformation to go from a question about a sequence of prescribed zeros with a finite limit $\alpha$ to a sequence of prescribed zeros with a limit of $\infty$: get an entire function, and transform it back (obtaining a function analytic on ${\mathbb C} \backslash \{ \alpha \}$).2012-04-16