Let $f:D(0,1) \to \mathbb{C}$ holomorphic injective such that $f(0) = 1 = f'(0)$ and $f^{(k)}(0) \in \mathbb{R} \; \forall k$.
Show that if $\Omega^+ = \{z \in D(0,1):im(z)\geq 0\}$ and $\Omega^- = \{z \in D(0,1):im(z)\leq0\}$, then either $f(\Omega^+)\subset \Omega^+$ or $f(\Omega^+)\subset \Omega^-$
Obs: $D(0,1)$ denotes the disc of center $0$ and radius $1$
f is holomorphic, then it is analytic and $f(re^{i\theta})=\sum \frac{f^{(k)}(0)}{k!}(re^{i\theta})^k = \sum \frac{f^{(k)}(0)}{k!}r^k(\cos(k\theta)+i\sin(k\theta))$.
I know that, by the principle of reflection, as f is continuous and holomorphic on $\Omega^+$, there is a unique extension to $D(0,1)$ to which $f(z) = \overline{f(\bar{z})}$.
Now, how do I show that the image of the upper semiplane is either on the upper plane or in the lower plane? I wanted to find a contradiction by assuming that exists $z_1,z_2 \in \Omega^+$ such that $f(z_1) \in \Omega^+$ and $f(z_2)\in \Omega^-$ but no idea came through.
Note that $f(\bar{z}) = \overline{f(z)}$ on $D(0,1)$. If $z_1, z_2$ are two distinct points on upper half plane such that $\Im f(z_1) > 0, \Im f(z_2) < 0$, then there exists $z_3$ on the straight line joining them such that $f(z_3) \in \mathbb{R}$, then $f(\overline{z_3}) = f(z_3)$, but $f$ is injective.