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 […]

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## Sharp radius for univalent convex functions

- Post author By Q+A Expert
- Post date March 26, 2020
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- Tags 1/2) \subset f(\mathbb{D})$. To show this, 1/4)=\lbrace z \in \mathbb{C}: \vert z \vert < 1/4 \rbrace \subset f(\mathbb{D})$. Furthermore, Context Koebe 1/4 Theorem states the following: Theorem.- Let $f \in \mathcal{S}$, if there exists a $w \in \mathbb{C}$ with $\vert w \vert = 1/4$ and $w \notin f(\mathbb{D})$, 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, of the form $$ f(z) = \frac{z}{1-\lambda z} \quad \text{for some } \lambda \text{ with } \vert \lambda \vert = 1. $$ My attempt Write $$f(z, showing that the only possibilities are rotations of $l$, so $$ \left| a_2 + \frac{1}{2w} \right| = 2 \implies \vert a_2 \vert = 1. $$ How can I conclude? Thanks in advance., so by Koebe-1/4 we have $\vert w \vert \geq 1/2$. This radius is also sharp because we have the function $$ l(z) = \frac{z}{1-z}. $$ My ques, take $w \notin f(\mathbb{D})$ and consider the function $$ h(z) = \frac{w^2-(f(z)-w)^2}{2w} $$ which belongs to $\mathcal{S}$ and satisfies $, that is, that is to say, the set of univalent (analytic and injective) with $f(0)=0$ and $f’(0)=1$ functions from $\mathbb{D}=\lbrace z \in \mathbb{C}: \vert z \vert, then $$ 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, then $D(0, then $f$ is a rotation of the Koebe function, there exists $w \notin f(\mathbb{D})$ with $\vert w \vert = 1/2$ as in Koebe-1/4 theorem, where we have $D(0