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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...

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  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Jun 8 '12 at 12:55
  • $\begingroup$ Is it clearer now? $\endgroup$ – User11111 Jun 8 '12 at 14:22
  • $\begingroup$ Yes, perfectly clear, and I have answered your question. $\endgroup$ – Georges Elencwajg Jun 8 '12 at 16:41
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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 $[(\partial_y + i \partial_x) v](\zeta)=0$.
This is like falsely arguing that

"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$"

The gist of the matter is that in any context you care to consider, there is no relation between the value of a function at a point and that of some derivative of that function at that same point.

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$f(z)=z$ 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.

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