The range of a univalent functions Let $\Sigma=\{g:\mathbb{C\backslash\overline{\mathbb{D}}}\to\mathbb{C}$ be univalent$\big{|}g(z)=z+b_0+\frac{b_1}{z}+...\}$
Suppose $f:\mathbb{D}\to\mathbb{C}$ is univalent and has a Taylor expansion $f(z)=a_1z+a_2z^2+...$. If there exists $g\in\Sigma$ satisfying $f(\mathbb{D})\cap g(\mathbb{C\backslash\overline{\mathbb{D}}})=\emptyset,$ then $|a_1|\leq 1,$ and the equality holds if and only if $f(z)=e^{i\theta}z$,$g(z)=z$.
My idea: It's not hard to prove that $Aera(f(\mathbb{D}))=\pi \sum\limits_{n=1}^\infty n|a_n|^2$. With some effort, we can prove that $\mathbb{C}\backslash g(\mathbb{C}\backslash\overline{\mathbb{D}})\subseteq \{\omega\big{|}|\omega-b_0|\leq 2\}$. Thus $f(\mathbb{D})\subseteq \mathbb{C}\backslash g(\mathbb{C}\backslash\overline{\mathbb{D}})\subseteq \{\omega\big{|}|\omega-b_0|\leq 2\}$ and hence $Aera(f(\mathbb{D}))=\pi \sum\limits_{n=1}^\infty n|a_n|^2\leq 4\pi$. However, this estimate is not accurate and can not meet the needs of the problem. 
I really appreciate any help I can get! 
 A: According to the Gronwall area theorem, the area of the complement of $g(\Bbb C \setminus \overline {\Bbb D})$ is
$$
\pi \left( 1 - \sum_{n=1}^\infty n |b_n|^2 \right)
$$
and the area of $f(\Bbb D)$ is $\pi \sum_{n=1}^\infty n|a_n|^2$.  Therefore, if $f(\Bbb D)$ and $g(\Bbb C \setminus \overline {\Bbb D})$ are disjoint, we have
$$
\pi \left( 1 - \sum_{n=1}^\infty n |b_n|^2 \right) \ge \pi \sum_{n=1}^\infty n|a_n|^2
$$
which implies
$$
|a_1| \le \sum_{n=1}^\infty n|a_n|^2 + \sum_{n=1}^\infty n |b_n|^2  \le 1 \, .
$$
Equality holds only if $a_n = 0$ for $n \ge 2$, and $b_n = 0$ for $n \ge 1$, which implies $f(z)=e^{i\theta}z$ and $g(z) = z + b_0$.
A: By the area theorem for $g$, we have that the area of $\mathbb C-g(\mathbb{C\backslash\overline{\mathbb{D}}})$ is $\pi(1-\sum_{k \ge 1}k|b_k|^2) \le \pi$ with equality iff $b_k=0, k \ge 1$. But then area $f(\mathbb{D}) \le \pi$ so as noted in the OP $|a_1| \le 1$. If we have equality, we first get $a_k=0, k \ge 2$ and $b_k =0, k \ge 1$ so $f(z)=e^{i\theta}z, g(z)=z+b_0$ and then the condition on disjointness of images forces $b_0=0$ so done!
