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

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 $\omega^2$.