Sharp radius for univalent convex functions Context
Koebe 1/4 Theorem states the following:

Theorem.- Let $f \in \mathcal{S}$, that is, the set of univalent (analytic and injective) with $f(0)=0$ and $f’(0)=1$ functions from $\mathbb{D}=\lbrace z \in \mathbb{C}: \vert z \vert < 1 \rbrace$ to $\mathbb{C}$, then $D(0,1/4)=\lbrace z \in \mathbb{C}: \vert z \vert < 1/4 \rbrace \subset f(\mathbb{D})$. Furthermore, if there exists a $w \in \mathbb{C}$ with $\vert w \vert = 1/4$ and $w \notin f(\mathbb{D})$, then $f$ is a rotation of the Koebe function, that is to say, of the form
$$
f(z) = \frac{z}{(1-\lambda z)^2} \quad \text{for some } \lambda \text{ with } \vert \lambda \vert = 1.
$$

This result can be improved if we restrict ourselves to the subclass $\mathcal{K} = \lbrace f \in \mathcal{S} : f(\mathbb{D}) \text{ is a convex domain} \rbrace$, where we have $D(0,1/2) \subset f(\mathbb{D})$. To show this, take $w \notin f(\mathbb{D})$ and consider the function
$$ 
h(z) = \frac{w^2-(f(z)-w)^2}{2w}
$$
which belongs to $\mathcal{S}$ and satisfies $w/2 \notin h(\mathbb{D})$, so by Koebe-1/4 we have  $\vert w \vert \geq 1/2$. This radius is also sharp because we have the function
$$
l(z) = \frac{z}{1-z}.
$$
My question
I was wondering if it is possible to characterize functions $f \in \mathcal{K}$ with optimal radius, that is, there exists $w \notin f(\mathbb{D})$ with $\vert w \vert = 1/2$ as in Koebe-1/4 theorem, showing that the only possibilities are rotations of $l$, that is, of the form
$$ 
f(z) = \frac{z}{1-\lambda z} \quad \text{for some } \lambda \text{ with } \vert \lambda \vert = 1.
$$
My attempt
Write
$$f(z)=z+a_2z^2+a_3z^3+\dots$$
if there exists such an $w$, then
$$
h(z) = \frac{w^2-(f(z)-w)^2}{2w} = \frac{f(z)^2}{2w} + f(z) = z + z^2 \left( a_2 + \frac{1}{2w} \right) + \dots 
$$
must be a rotation of the Koebe function, so
$$ 
\left| a_2 + \frac{1}{2w} \right| = 2 \implies \vert a_2 \vert = 1.
$$
How can I conclude? Thanks in advance.
 A: There are many ways to show that if $f \in K$, and any $|a_n|=1$ for some $n \ge 2$ then $f$ is a rotation (in $K$) of the basic $\frac{z}{1-z}$ - so in other words precisely $\frac{z}{1-\lambda z}$.
So if $|a_n|=1$ for $n \ge 2$, then $g(z)=zf'(z)=z+2a_2z+...+na_nz^n+...$ is starlike, with $n|a_n|=n$, hence $zf'(z)$ is a rotation of the Koebe function by the usual results about starlike functions (which follow from the Herglotz representation of positive real part functions on the unit disc) and now one can integrate
I would do the original problem directly without going through $|a_2|=1$: namely by a rotation of the function ($\lambda f(\bar \lambda z), |\lambda|=1$) we can assume  $- \frac{1}{2}$ doesn't belong to the image of $f$ and it is then enough to prove that $f(z)=\frac{z}{1-z}$.
Now by the convexity of the image domain, there is a support line through $-\frac{1}{2}$ that avoids the image domain and by the minimal property of distance $\frac{1}{2}$, that support line must be the vertical line (otherwise there is a point with lower absolute value outside the image $f(D)$). So $f(D)$ is a subset of the half-plane $\Re z >- \frac{1}{2}$
But now this means that $f(z)$ is subordinate to $h(z)=\frac{z}{1-z}$ so exists $w(0)=0$ Schwarz function with $h(w(z))=f(z)$ ($f(D) \subset h(D)$ as $h(D)$ is the full half-plane $\Re z > -\frac{1}{2}$, hence $w(z)=h^{-1}(f(z))$ is well defined, $h(0)=0$, $h$ has image contained in the unit disc)
But obviously $f'(0)=1=h'(0)$, so $w'(0)=1$ hence by Schwarz lemma, $w(z)=z, f=h$ so we are done!
