Show that if $f$ is holomorphic in the unit disk, is bounded, and converges uniformly to zero in the sector $\theta < \arg z < \phi $ as $|z| \rightarrow 1$, then $f=0.$
I'm having trouble proving this fact, I don't even know where to begin.
Any hints are appreciated.
Choose $n$ so $n\phi/2>2\pi$. Define $$g(z)=\prod_{j=0}^nf(e^{ij\phi/2}z).$$
Since $f$ is bounded it follows that $g(re^{it})\to0$ uniformly as $r\to1$. Hence $g=0$.
If $Z$ is the zero set of $f$ this shows that the disk is the union of finitely many rotations of $Z$. So $Z$ is uncountable, hence $f=0$.