$f$ constant on boundary implies $f$ is constant? I have the following question:
Prove of disprove: Let $f$ be a bounded holomorphic function on $\mathbb{D}$ and continous on $\overline{\mathbb{D}}$. Suppose that for a constant $0<\delta < 1/10, f(e^{i\theta}) = 1 + 2i$ for all $-\delta \pi < \theta < \delta \pi$. Then $f \equiv 1 + 2i$ on $\overline{\mathbb{D}}$
I am not really sure what to do. I first thought of the maximum/minimum modulus principle and can prove the claim in the special case that $f(0) = 1 + 2i$. However, I don't know how to prove (or disprove) in the general case. Any help would be appreciated.

A really non-elementary answer: It is known that functions $f\in H^\infty(\mathbb{D})$ (i.e. bounded holomorphic functions) have non-tangential limits almost everywhere (Fatou's theorem) and moreover, if the non-tangential limit is $0$ in a set of positive measure, then $f$ is identically zero.
Applying this to $f-(1+2i)$ yields the result immediately.
I am not saying this as a way of proving the result, I'm just saying this to point out that the claim is indeed true.
