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Let $u$ be an harmonic function in the open simply connected set $U$. Then $u$ has a harmonic conjugate $v$ in $U$ (i.e. $f=u+iv$ is analytic on $U$). Suppose $u$ bounded in $U$. Can I say that $v$ is also bounded?

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No. Consider $-\text{arg}(z)$ whose conjugate is $\log(|z|)$ on $\mathbb{C}-\mathbb{R}^{-}$.

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  • $\begingroup$ why minus arg(z)? $\endgroup$ – Federica Maggioni Dec 14 '12 at 11:12
  • $\begingroup$ You wrote the problem so that you're real part was bounded. If one takes the principal branch of $log$ one gets $\log(|z)|+\arg(z)$, which has bounded imaginary part. I just multiplied by $i$ to get bounded real part. $\endgroup$ – Alex Youcis Dec 14 '12 at 11:51

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