# Functions on the unit circle having finitely many zeroes

I actually consider functions on the strip $$S(0,\pi) = \{ z | 0 < Im(z)<\pi \}$$ in the complex plane, but by a biholomorphism, I can map to the unit disc. In particular, the functions I consider are bounded and continuous on the closed strip, and analytic in its interior. Moreover, they are also "regular", which means for me that there exists a $$\kappa >0$$ such that they continue to a bounded analytic function on the strip $$S(-\kappa, \pi + \kappa).$$ They also converge uniformly in the limits $$z\to \pm \infty, z\to i\pi \pm \infty$$. On the unit disc, this all translates to them being bounded, analytic functions on the (open) unit disc, and this extends to a disc slightly bigger than the unit disc, except for around the points mapped to from $$\pm \infty$$ (I usually consider these to be $$\pm 1$$). My question is: even with this lack of knowledge about what happens at the points $$\pm 1$$, can we still conclude that the functions must have finitely many zeroes in the closed unit disc?

No. The function $$\sin(z)/(z-2i)$$ is "regular" in your sense.