To show $f(z)=0$ on unit closed disk , where $f$ is analytic on unit disk

Question

let $D$ be the open unit ball around origin.let $f$ be a continuous complex valued function on its closure which is analytic on $D$. if $f(e^{it})=0$ ,for $0<t<\pi/2.$ show that $f(z)=0$ for all $z$

if i some how show $f$ is zero on boundary of unit ball , then by maximum modlus principle $f$ will be zero. but how can i show that $f$ is also zero on rest of unit circle?

Answer 1

Using the Schwarz reflection principle, you can define $f$ so that it is analytic in a neighbourhood of $e^{it}$ for $0 < t < \pi/2$. But then the zeros of $f$ are not discrete, which implies $f$ is identically $0$.

Answer 2

Hint : consider $g(z)=f(z)f(iz)f(-z)f(-iz)$.

Answer 3

Let $g(z)=f(z)f(e^{\pi/2}z)f(e^{\pi}z)f(e^{3\pi/2}z)$. Then $g$ is continuous in the closed unit disk, and analytic in the open disk, and $g=0$ in the boundary of the disk. Hence, by virtue of the maximum principle, $g\equiv 0$, and consequently $f\equiv 0$.