# $f$ analytic on $B(R,z_0)$ with $z_0$ on the real axis and $\forall\zeta\in\mathbb{R}\cap B(R,z_0),\Im f(\zeta)=0$ then $\partial _z f (\zeta) = 0$

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...

• Please, write quantifiers correctly: your first line doesn't make sense, especially that implication sign. Under my interpretation of your question, what you sk about is indeed not true. – Georges Elencwajg Jun 8 '12 at 12:55
• Is it clearer now? – User11111 Jun 8 '12 at 14:22
• Yes, perfectly clear, and I have answered your question. – Georges Elencwajg Jun 8 '12 at 16:41

No, this is false: the function $f(z)=z$ gives a counterexample, since for all $\zeta \in \mathbb R$ you have $Im(f)(\zeta)=0$ but nevertheless $(\partial_zf)(\zeta)=1$ for those $\zeta$.
Your argumentation in the general case is correct up to the equality $\partial _z f = (\partial_y + i \partial_x) v$.
But you may not argue that since $v(\zeta)=0$ then $[(\partial_y + i \partial_x) v](\zeta)=0$.
"Since $\frac {\operatorname d}{\operatorname dx}sin =cos$ and since $\operatorname {sin}(0)=0$, then $\frac {\operatorname d}{\operatorname dx}sin(0)=cos(0)=0$"
$f(z)=z$ is a counterexample. The mistake is in concluding from $v(\zeta)=0$ that $\partial _z v(\zeta)=0$. A function of two variables may well vanish on a line but have a nonzero gradient there: $v(x,y)=y$ is an example of such function.