D is the closed disk of unit radius centred at the origin, in the complex plane. F is a continuous complex function on D. F is analytic on the interior of D. F(cost,sint)=0 for all t less than 90 degrees and greater than 0 degree. Prove the F(z)=0 for all z in D.
Using continuity, I know that F(1) and F(i) are also 0. How do I proceed beyond this?
Put $g(z)=f(z)f(iz)f(-z)f(-iz)$. Then $g$ has the same properties as $f$, but we have $g(\exp(i\theta))=0$ for all $\theta$. Let $g(z)=\sum a_n z^n$, show that for $0<r<1$, one has $$\int_0^{2\pi}|g(r\exp(i\theta))|^2 d\theta= \sum |a_n|^2r^{2n}$$ Now $g$ is bounded, hence by the Dominated convergence theorem, $\int_0^{2\pi}|g(r\exp(i\theta))|^2 d\theta\to 0$ if $r\to 1$, hence $\sum |a_n|^2=0$, it is easy to finish.