Meromorphic function with zeroes on unit circle 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. 
 A: 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.
A: if $f$ is holomorphic on $\mathbb{C}$, then for $|\zeta| \lt 1 $
$$
f(\zeta) = \frac1{2\pi i}\oint_{|z|=1} \frac{f(z)dz}{z-\zeta} = 0
$$
