finite number of zeros of a complex function Suppose that $f$ is analytic on a region $D = \{|z|>R\}$, and the limit of $\vert f \vert$ at infinity exists. Show that $f$ is either the constant function $0$, or has only finitely many zeros.
I thought since the limit at infinity exists then Laurent seires of $f$ only contains the principal part plus a constant, but how do I proceed to show that this will only have finite zeros ?
 A: This is false. Let $g$ be analytic in $\{z:|z|<\frac 1 R\}$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(\frac 1 z)$. Then the hypothesis  is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=\sin( \frac {\pi} {\frac 1 R -z})$. It vanishes at the points $z=\frac 1 R -\frac 1 n$.
However, if $f$ is assumed to be holomorphic in $\{|z| \geq R\}$ (and $\lim_{z \to \infty} f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $\{z:|z| \geq R\}$.  Once again define $g(z)=f(\frac 1 z)$ This is holomorphic in some open set containing $\{z:|z| \leq \frac 1 R\}$ . The hypothesis tells you that $g$ has  a removable singularity at $0$. The extended function is holomorphic in an open set containing $\{|z| \leq \frac 1 R\}$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$. 
