This is a homework question so please don't provide the full proof.
Hi everyone,
Let $f$ be holomorphic on the open square $U = \{z\in \mathbb{C}: |Re \,z| <1, |Im \,z| < 1\}$ and continuous on its closure $\bar{U}$. Suppose $f=0$ on $\{z\in \bar{U}: Re \, z =1\}$, prove $f = 0$ on $U$.
My attempt
Method 1: By Max Modulus Theorem, $\max_{z\in U} |f(z)| = \max_{z\in \partial U} |f(z)|$, so it suffices to show that $f$ is identically $0$ on the four edges of the rectangle. To this end, I am trying to use the Identity Theorem, but I don't see how it can be applied.
Another method: Define $g(z) = f(iz + i)$, then by assumption, $g$ is identically $0$ on $(-1, 1)$, holomorphic on $\{z\in \mathbb{C}: -1< Re \,z <1, -2<Im \,z < 0\}$. Then I am trying to use the Schwarz reflection principle to extend this function to a holomorphic function on the entire plane and then apply the Identity Theorem. But since this function is only holomorphic on a subset of the lower half plane, I don't see how Schwarz reflection principle can be applied.
Some hints would be greatly appreciated. Thanks a lot.
Best Answer
Hint: consider $f(z)f(iz)$. Show that if a product of two analytic functions is $0$ then one of them must be $0$.