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I'm currently stuck on the following problem.

Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$\bar f(z)$ $\forall z\in V$ then f is constant on V.

I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.

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    $\begingroup$ Maybe use $f^3 = f \bar f$ is a pure real analytic function? $\endgroup$ – JonathanZ Jan 4 at 2:29
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    $\begingroup$ The hypothesis implies that $\overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$. $\endgroup$ – Kavi Rama Murthy Jan 4 at 5:40
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No need. First solve $$ Z^2=\bar{Z}\qquad (1) $$ Eq.(1) implies $Z^3=|Z|^2$ which has $S=\{1,j,j^2,0\}$ as set of solutions, with $$ j=e^{\frac{2i\pi}{3}} $$ then even a continuous function $f:V\to S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.

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$$\|f(z)\|^2 = \|f(z)^2\|=\|\overline{f}(z)\|=\|f(z)\| $$ so $\|f(z)\|$ is either $0$ or $1$, constantly, since $\|f(z)\|$ is continuous. In the former case $f(z)\equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $\xi^2=\overline{\xi}$, so our $f$ is constantly $0$, $1$, $\omega$ or $\omega^2$.

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