I'm trying to solve the following exercise
Be $f:\{B\{0,r\} - \{0\}\} \longrightarrow \mathbb{C},\ r>0 $ holomorphic such that in ${0}$ has an essential singularity. Show that if $$f(\{B\{0,r\} - \{0\}\}\cap \mathbb{R})\subset \mathbb{R},\ \mathbb{R}\subset f(\{B\{0,r\} - \{0\}\}),$$ then $f(\{B\{0,r\} - \{0\}\})=\mathbb{C}.$
By the Great Picard Theorem we know that $ f $ attains all complex, with the exception of at most one point, an infinite number of times. My problem is how to show that in this case this exception does not occur.
Thank You.
The premises say that $f$ omits no real value:
$$\mathbb{R}\subset f(B\{0,r\}\setminus \{0\}).$$
So if $f$ omits any value $w$, that must be non-real.
Now consider the function
$$g(z) = \overline{f(\overline{z})}.$$
$g$ is holomorphic on $B\{0,r\}\setminus \{0\}$, and if $f$ omits $w$, then $g$ omits $\overline{w}$.
But since $f$ is real on $(B\{0,r\}\setminus \{0\}) \cap \mathbb{R}$, we have $g(t) = f(t)$ for $t \in (B\{0,r\}\setminus \{0\}) \cap \mathbb{R}$, and by the identity theorem, $g \equiv f$. So if $f$ omits any non-real value, it omits two values, which cannot be by Picard's theorem.