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Let's consider $h(z)$ analytic in $ B(0,1)$ and continuous in $\overline{B(0,1)}$ , such that $ Re(h(z))=0 $ in $\partial D(0,1)$. Prove that $h(z)$ is constant.

Well... Since $h$ is continuous on a compact set, then attains it's maximum and it's minimum on it. That points are in $\partial D(0,1)$ since otherwise , the function is constant ( maximum modulus principle). More than that I don't know what can I do )=

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If $\text{Re}(h) = 0$ on $B(0,1)$ then Cauchy-Riemann gives you that the function is constant on the ball. Otherwise, there is some $z\in B(0,1)$ for which $|Re(z)| > 0$. From here you should be able to argue that the function attains a maximum on the open ball whence is constant by MMP, right? –  John Martin Oct 8 '12 at 2:20
    
@JohnMartin the hypothesis says that $Re(h)=0$ but on $\partial D(0,1)$ ( i.e $z$ such that $|z|=1$). –  Daniel Oct 8 '12 at 3:57
    
I know that, but nothing is said about the real part on the (open) disc - it is possible that the real part is zero on the disc in which case CR implies that the function is constant there, and otherwise you could find a point making the modulus on the interior greater than on the boundary which would of course make the function constant. In any case, it seems that an acceptable answer has been given. –  John Martin Oct 8 '12 at 14:02

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Since $h$ and in particular $\textrm{Re}(h)$ is continuous on the closed unit disc it follows that $\lim_{r \uparrow 1} \max_{|z|=r} |\textrm{Re}\;h(z)| = 0$. Now the maximum of $|\textrm{Re}\;h(z)|$ for $|z| \leq r$ is attained on the circle $|z|=r$. This follows from the maximum/minimum principle for $e^h$. This implies that this maximum is non-decreasing in $r$. Since its limit is $0$ it must be identically zero. Therefore $\textrm{Re}\;h=0$ on the open unit disc. Then again it follows from the maximum/minimum principle for $e^h$ that $h$ must be constant.

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Thanks WimC!! =D! –  Daniel Oct 8 '12 at 5:53

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