If $f$ is analytic on $B(R,z_0)$ with $z_0 $ on the real axis and if $ \forall \zeta \in \mathbb{R} \cap B(R,z_0), \Im f(\zeta) = 0 $ then $ \partial _z f (\zeta) = 0 $.
I can write $f = u + iv$ with $u,v : \mathbb{C} \rightarrow \mathbb{R} \leadsto \partial _z f = \frac{1}{2} (\partial_x - i \partial_y) (u + iv) \stackrel{\text{Cauchy-Riemann}}{=} (\partial_y + i \partial_x) v $. But since $\Im f(\zeta) = 0 $ then $ v(\zeta) = 0 $ and so $\partial _z f (\zeta) = 0$.
Is this true? I don't think so but can't figure why...