I'm a bit stumped on this question. I've been thinking about it, and wanted to perhaps apply the Schwarz Lemma, and or, the argument principle, but I really have no idea how to begin. The question is:

Does there exist a function meromorphic on the complex plane whose set of zeros is precisely the unit circle ?

Any hints or feedback is appreciated : )

Originally I had holomorphic, which was an error. Though I don't know if meromorphic will make a difference.

  • $\begingroup$ I edited my answer. $\endgroup$ – M.G Dec 14 '15 at 3:05

No, such a meromorphic function cannot exist.

Identity Principle : Let $D$ be an open connected subset of $\mathbb{C}$ and let $f, g \in \text{Hol}(\mathbb{D})$. If the following set $$ \big\{ z \in \mathbb{C} :~~ f(z) = g(z) \big\}$$ has a non-isolated point then $f \equiv g$ on $D$.

Answer to your question : Suppose $f(z)$ is a meromorphic function such that $f(z) = 0$ for all $z \in \mathbb{C}$ satisfying $|z| = 1$. Let $E$ be the set of poles of $f$. Then $f \in \text{Hol}(\mathbb{C}\backslash E)$. Apply the Identity Principle to $f(z)$ and $0(z)$ i.e the function defined to be zero on $\mathbb{C}\backslash E$. Since the unit circle clearly has non-isolated points, we conclude that $f \equiv 0$ on $\mathbb{C}\backslash E$.

Moral : The zeroes of non-constant meromorphic functions are isolated and the Identity Principle is a corollary of this fact.

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  • $\begingroup$ I'm not sure how to see the immediately down the identity principle. $\endgroup$ – mathgenesis22813 Dec 14 '15 at 1:34
  • $\begingroup$ Why the down vote ? $\endgroup$ – M.G Dec 14 '15 at 1:42
  • $\begingroup$ It was accidental. $\endgroup$ – mathgenesis22813 Dec 14 '15 at 1:43
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    $\begingroup$ thx for your useful comment on my answer. i've inserted the missing constant $\endgroup$ – David Holden Dec 14 '15 at 2:03

if $f$ is holomorphic on $\mathbb{C}$, then for $|\zeta| \lt 1 $ $$ f(\zeta) = \frac1{2\pi i}\oint_{|z|=1} \frac{f(z)dz}{z-\zeta} = 0 $$

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