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Given $f:\mathbb{D}\to\mathbb{D}$ be analytic function on unit diskm $f(0)=0,f'(0)={1 \over 2}$, I was asked to show that $f(\mathbb{D}) \supset B(0,7-4\sqrt{3})$.

[Observation] Write $$f(z)={1 \over 2}z+\sum_{n\geq 2}a_n z^n$$ By Cauchy's formula, $a_n={1 \over {2\pi i}}\int_{|z|=r<1}{f(z)\over z^{n+1}}\ dz$ hence $|a_n|\leq {1 \over r^n},\forall n\geq 2$. Let $r \to 1-$ we get $|a_n|\leq 1$.Now we have \begin{equation} \begin{split} f(z) &\geq {1\over 2}|z|-\sum_{n\geq 2}|a_n||z|^n\\ &\geq {1\over 2}|z|-|z|^2\cdot{1 \over {1-|z|}} \end{split} \end{equation} where the right side of the inequality takes maximum at $|z|={{3-\sqrt{6}}\over 3}$, and this means $f$ maps $\{|z|={{3-\sqrt{6}}\over 3}\}$ outside of the circle $B(0,{{5-2\sqrt{6}}\over 2})$. Now it is a simple argument by Rouche's theorem to deduce that $f(\mathbb{D})\supset B(0,{{5-2\sqrt{6}}\over 2})$. But this is the best estimate I get which still does not meet the requirement, so how can I refine it?

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Let $\omega\in\mathbb{D}$. We'll use the standard unit disk automorphism $$\varphi_\omega(z)=\frac{\omega-z}{1-\overline{\omega}z}$$

Suppose that $\omega$ is not in the image of $f$. Then $\varphi_\omega\circ f\not=0$ on $\mathbb{D}$, and since $\mathbb{D}$ is simply connected we can define a branch of log on $f(\mathbb{D})$. This gives us a composition $\sqrt{\varphi_\omega\circ f}$. Note that $f(0)=0$ tells us $(\sqrt{\varphi_\omega\circ f})(0)=\sqrt{\omega}$. Now we post-compose with $\varphi_{\sqrt{\omega}}$. This gives us a function from $\mathbb{D}$ to $\mathbb{D}$ that sends $0$ to $0$. Then Schwarz's Lemma applies, so the modulus of the derivative at $0$ is bounded by $1$: $$\left|\varphi_{\sqrt{\omega}}'(\sqrt{\omega})\right|\left|\frac{1}{2\sqrt{\omega}}\right|\left|\varphi_\omega'(0)\right|\left|f'(0)\right|\le 1$$

Now we calculate that $\varphi_{\sqrt{\omega}}'(\sqrt{\omega})=(\left|\sqrt{\omega}\right|^2-1)^{-1}$ and $\varphi_\omega '(0)=\left|\omega\right|^2-1$. Taking into account that $|f'(0)|=\frac{1}{2}$: $$\frac{|\omega|^2-1}{4\left|\sqrt{\omega}\right|(|\omega|-1)}\le1$$ $$\frac{|\omega|+1}{4\left|\sqrt{\omega}\right|}\le 1$$ $$|\omega|-4\left|\sqrt{\omega}\right|+1\le 0$$ Solving for $\left|\sqrt{\omega}\right|$ in the case of equality gives us $2\pm\sqrt{3}$, and squaring this gets $7\pm4\sqrt{3}$. The derivative with respect to $\sqrt{\omega}$ is decreasing at $2-\sqrt{3}$, so if $|\omega|$ is less than $7-4\sqrt{3}$ we get a contradiction.

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