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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.