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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?

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

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  • $\begingroup$ Good counter-example! Thanks :-) $\endgroup$ – CS1994 Sep 3 '19 at 15:00

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