Show that the image, $f(\mathbb{D}_1(0))$, for $f(z) = z + \sum_{n=2}^\infty a_n z^n$ an injective analytic function with $f(0) = 0$ and $f'(0) = 1$, is symmetric with respect to the real axis, and that $f$ maps the upper half-disk into either the upper or lower half of $\mathbb{C}$.
$a_n$ is a real quantity for all $n$
Geometrically, $w$ and $\bar{w}$ are symmetric with respect to the real axis, so I thought it would suffice to show that if $w \in f(\mathbb{D}_1(0))$, then $\bar{w}\in f(\mathbb{D}_1(0))$.
First note that if $z \in \mathbb{D}_1(0)$, then $\bar{z} \in \mathbb{D}_1(0)$, as $|z| < 1 \implies |\bar{z}| < 1$.
Now, suppose $w \in f(\mathbb{D}_1(0))$. This implies that there exists a $z \in \mathbb{D}_1(0)$ such that $f(z) = w$.
Now, take the complex conjugate of this equation to get:
$$\bar{z} + \sum_{n=2}^\infty a_n \bar{z}^n = \bar{w}$$
Since $\bar{z} \in \mathbb{D}_1(0)$, we get that there exists an element of the unit disk which $f$ maps to $\bar{w}$, thus $\bar{w} \in f(\mathbb{D}_1(0))$. This implies $f(\mathbb{D}_1(0))$ is symmetric with respect to the real axis.
Now I must show that the $f$ maps the upper half-disk onto one side of the real line.
My thought was that I could let $z_1, z_2$ belong to the upper half-disk with Im($z_1$), Im($z_2) \ne 0$ (It is clear from the formula that $f$ maps all real numbers to real numbers), and show that, if they map to opposite sides of the real-axis, then any smooth curve $C$ joining $z_1$ and $z_2$ must contain an element that $f$ maps to the real line (I believe this is true because $f$ is analytic, and thus continuous). Then, if I can show that $f(z) \in \mathbb{R}$ iff $z \in \mathbb{R}$, I'd reach a contradiction, as $C$ can be taken to not intersect the real axis.
I'm not 100% convinced of the validity of this approach, but it feels like it would be true to me (if someone could either justify this or show me why I'm wrong that would be awesome).
I tried to do this letting $z$ be in the upper half-disk with Im($z) \ne 0$. Then let $z = re^{i \theta}$, with $r \ne 0$, and $\theta \ne n \pi$ for integer $n$.
$$Im(f(z)) = r\sin(\theta) + \sum_{n=2}^\infty a_n r^n \sin(n \theta)$$
remembering that $\{\sin(n \theta): n \in \mathbb{N}\}$ is a linearly-independent set when $\theta \ne k\pi$, it follows that $Im(f(z)) \ne 0$.
We could also show this is impossible if we could prove that $f([-1,1]) = \mathbb{R}$, as $f$ is injective, and so no $z \in \mathbb{D}_1(0)$ with non-zero imaginary part could map to the real line. However, this approach seems much less tractable to me.
Is any of this valid? I feel like my idea is fairly nice, but I'm not sure how justified it really is.
Hint:
Use this fact that every univalent (injective) function with real coefficients, send points on real line to real line, since ${\bf Im}\,f(z)=0$ implies $f(z)=\overline{f(z)}=f(\overline{z})$ and this implies $z=\overline{z}$ or ${\bf Im}\,z=0$.