Boundary behavior of an analytic function

Question

Let $f$ be a function holomorphic in a simply-connected domain $D$; for simplicity, assume that the boundary $\partial D$ of $D$ is piece-wise analytic with positive inner angles. Let $0\in \partial D$.

If we know that $$ \lim_{z\to 0} \frac{f(z)}{z}=0, $$ (by that, we have in mind a non-tangential limit from $D$, say, within a sector with vertex at $0$), is it true that $$ \lim_{z\to 0} f'(z)=0, $$ again, in the non-tangential sense?

Notice that the only assumption is the analyticity of $f$ in $D$. If the answer is no, what is a possible counterexample, and what are the minimal sufficient conditions?

Answer 1

The answer is yes. Apply Cauchy's formula $$f'(z)=\frac{1}{2\pi i}\int_{\gamma_z}\frac{f(\zeta)d\zeta}{(\zeta-z)^2},$$ where $\gamma_z$ is the circle $\{ \zeta:|z-\zeta|=\epsilon|z|\}$, and epsilon is small, so that the disk is within a non-tangential sector. Then, if $|f(\zeta)|\leq\delta|z|$ on $\gamma_z$, the straightforward estimate of the integral gives you $|f'(z)|\leq \delta/\epsilon$, and you obtain your statement.