Let $f$ be a complex-valued continuous function defined on the closure of the open unit ball of $\mathbb{C}$ being analytic in the open unit ball. Then, if $f(e^{it})=0\,\,\forall t\in(0,\frac{\pi}{2})$, can we say that $f(z)=0\,\,\forall z\in\mathbb{C}$?
The problem is that though the function is continuous and zero on a connected set, it is not given analytic at the points where it is zero. Can we apply isolated-ness of zeroes of analytic functions to this example? Any ideas. Thanks beforehand.
Hint. Map the upper half plane conformally to the open disk by $\psi$. Then define $g=f\circ \psi$. For example $$ g(z)=f\left(\frac{z-i}{z+i}\right). $$
Observe that $g$ extends continuously to the closure of the upper half plane (i.e., upper half plane union with the real line), and $g$ vanishes at an interval of the real line, namely $(-\infty,-1]$. Then use the Schwarz reflection principle, and observe that $g$, with its reflection vanishes in a set with limits points inside its domain, and hence $g\equiv 0$.