TITLE: Show that $f(z)=0$ for all $z$, where $f$ is an analytic function on the closed unit disc with additional conditions.
QUESTION [6 upvotes]: Let $D$ denote the open ball of unit radius about origin in the complex plane $\Bbb C$.
Let $f$ be a continuous complex-valued function on its closure $D$ which is analytic on $D$. If $f(e^{it}) = 0$ for $0 < t  <\frac{\pi}{2}$ , show that $f(z) = 0 $ for all $z$.

Here is what I tried:
$f$ is analytic on $D$. If I can find a sequence $(z_n)_n$ in $D$ such that $f(z_n)=0\forall n$ and additionally $(z_n)_n$ has a limit point in $D$ then we are done.
I think $f(e^{it}) = 0$  may help finding one sequence but I am not sure.
Will you kindly help?

REPLY [11 votes]: Let $\omega \ne 1$ be a 5-th root of unity ($\omega^5 = 1$) and define
$$
  g(z) = f(z)f(\omega z)f(\omega^2 z)f(\omega^3 z)f(\omega^4 z) \, .
$$
$g$ is analytic in the unit disk $\Bbb D$ and continuous on it closure because $f$ is.
For every $z \in \partial \Bbb D$, at last one of the points
$z, \omega, z, \omega^2 z, \omega^3 z, \omega^4 z$ has an argument 
between $0$ and $\pi/2$, so that $g(z) = 0$ on the boundary of 
the unit disk.
Now apply the maximum principle.
In the same way one can show that $f$ is identically zero if $f(z)=0$
on any segment of the unit circle. 
Another option would be to use the Schwarz reflection principle (which is
a more advanced topic however). It implies that $f$ can be continued
analytically to a larger domain containing the arc $\{ e^{it} \mid
0 < t < \pi /2 \}$ in its interior. Then the identity theorem can
be applied.