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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.

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