# If $f$ is analytic and $f(z)^2$ = $\bar f(z)$ then $f$ is constant

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.

• Maybe use $f^3 = f \bar f$ is a pure real analytic function? – JonathanZ supports MonicaC Jan 4 '19 at 2:29
• The hypothesis implies that $\overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$. – Kavi Rama Murthy Jan 4 '19 at 5:40

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.
$$\|f(z)\|^2 = \|f(z)^2\|=\|\overline{f}(z)\|=\|f(z)\|$$ 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 $$\omega^2$$.