Explicitly write down an example of a non-zero analytic function in \(|z|<1\) which has infinitely zeros in \(|z|<1\).
Why does not the phenomenon in (1) contradict the uniqueness theorem?
Solution
Take \[ f(z)=\sin\!\left(\frac{1}{1-z}\right). \] The function \(z\mapsto(1-z)^{-1}\) is holomorphic on the unit disk, since its only pole is at the boundary point \(z=1\). Hence \(f\) is holomorphic on \(|z|<1\), and it is not identically zero.
For every sufficiently large positive integer \(n\), \[ z_n=1-\frac{1}{n\pi} \] lies in the unit disk and satisfies \[ f(z_n)=\sin(n\pi)=0. \] Thus \(f\) has infinitely many zeros in \(\mathbb D\).
This does not contradict the identity theorem. The zeros \(z_n\) converge to \(1\), but \(1\notin\mathbb D\). Hence the zero set has no accumulation point in the domain of holomorphy. The identity theorem forces a holomorphic function to vanish identically only when its zeros have an accumulation point inside the domain.