Koebe's 1/4-Theorem

Notation: $B(a, r) = \{z \in \mathbb{C}: |z - a| < r \}, B = B(0, 1), \mathbb{S}$ is the Riemann sphere.

Let $\mathcal{S}$ be the collection of all injective analytic function $f$ on the unit disk with  $f(0) = 0$ and $f'(0) = 1$. We know from the Open Mapping Theorem that $f(B)$ must contain a disk $B(0, r_f)$. For the class $\mathcal{S}$, there is a universal $r > 0$ such that $f(B)$ contains $B(0, r)$ for all $f \in S$.

Theorem (Koebe's 1/4-Theorem). If $f \in \mathcal{S}$ then $B(0, 1/4) \subseteq f(B)$.

We need to the following result, which is of independent interest.

Theorem (The Area Theorem). Let $g:  B \backslash \{0\} \rightarrow  \mathbb{C}$ be an injective analytic function with Laurent series expansion at 0 $$g(z) = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots.$$ Then $$\sum_{n = 1}^\infty n|c_n|^2 \leq 1.$$
Proof: Note that $g$ is a conformal equivalence between $B$ and $g(B) \subseteq \mathbb{S}$ and $\infty \in g(B)$. Fix any $0 < r < 1$. Consider $U_r = \mathbb{S} \backslash \overline{B(0, r)}$. It is an open set in $\mathbb{C}$ with boundary $\gamma$ parametrized by $\theta \mapsto h(re^{i\theta})$ where $\theta$ runs from $2\pi$ to 0. The area of $U_r$ is given by $\begin{align*} \mathrm{Area}(U_r) &= \frac{1}{2i}\int_\gamma \bar{w} dw = -\frac{1}{2i} \int_0^{2\pi} \overline{h(re^{i\theta})}h'(re^{i\theta})ire^{i\theta}d\theta \\ &= - \frac{1}{2i} \int_0^{2\pi}  [ \left(-\frac{1}{r^2} + |c_1|^2 + 2|c_2|^2 + 3|c_3|^2 + \cdots \right)r^2 \\ &{\,\,} + \textrm{ terms with nonzero integral powers of } e^{i\theta} ]  \\ &= \pi \left(\frac{1}{r^2} - \sum_{n = 1}^\infty n|c_n|^2 r^2 \right). \end{align*}$
Since area is always non-negative, let $r \rightarrow 1$ and we are done.

Proof of Koebe's Theorem: Let $f \in \mathcal{S}$. Then $f$ omits some $w_0 \in \mathbb{C}$ (or otherwise $f^{-1}$ is a bounded entire function, which is then constant by Liouville's Theorem, contradiction), so $h = 1/f$ omits 0 and $1/w_0$. Observe that the Taylor series of $f$ at 0 takes the form $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$, so $$h(z) = \frac{1}{z + a_2 z^2 + \cdots} = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots.$$  Let $z_0 = 0 \textrm{ or } 1/w_0$. We have $$g(z) = \frac{1}{h(z) - z_0} =  \frac{z}{1 + (c_0 - z_0)z + c_1 z^2 + \cdots} = z + (z_0 - c_0)z^2 + \cdots.$$ In particular, $g(0) = 0$ and $g'(0) = 1$. Also note that $g$ is injective. Thus $g \in \mathcal{S}$.
We claim that there exists some $u \in \mathcal{S}$ such that $u(z)^2 = g(z^2)$. Since $g$ is injective and $g(0) = 0$, the analytic function $g(z)/z$ omits 0 on $B$, we can find some analytic function $\varphi$ on $B$ such that $\varphi(z)^2 = g(z)/z$ and $\varphi(0) = 1$. Let $u(z) = z\varphi(z^2)$. Then $g(z^2) = z^2 \cdot g(z^2)/z^2 = z^2\varphi(z^2)^2 = u(z)^2$. Since $\varphi(0) = 1$ and $\varphi'(0) = \frac{1}{2}(g(z)/z)^{-1/2}\frac{d}{dz}(g(z)/z)|_{z = 0} = \frac{1}{2} (z_0 - c_0)$, $$u(z) = z\varphi(z^2) = z + \frac{1}{2}(z_0 - c_0)z^3 + \cdots.$$ Hence $u(0) = 0$ and $u'(0) = 1$. To show that $u \in \mathcal{S}$ it remains to prove that $u$ is injective. Suppose $u(z) = u(w)$. Then $g(z^2) = g(w^2)$. By injectivity of $g$, $z^2 = w^2$ or $z = \pm w$. If $z = -w$, then $u(w) = u(z) = u(-w) = -u(w)$, so $g(w^2) = u(w)^2 = 0$ and $w^2 = 0$ by injectivity of $g$ again, i.e. $w = z = 0$. In any case we have $z = w$.
Now consider the Laurent expansion of $1/u$ at 0, we have $$\frac{1}{u(z)} = \frac{1}{z + \frac{1}{2}(z_0 - c_0)z^3 + \cdots} = \frac{1}{z}(1 - \frac{1}{2}(z_0 - c_0)z^2 + \cdots) = \frac{1}{z} - \frac{1}{2}(z_0 - c_0)z + \cdots.$$ By the Area Theorem, $|-\frac{1}{2}(z_0 - c_0)| \leq 1$, or $|z_0 - c_0| \leq 2$. Therefore $|1/w_0| \leq |1/w_0 - c_0| + |0 - c_0| \leq 2 + 2 = 4$, or $|w_0| \geq 1/4$. It follows that $\mathbb{C} \backslash f(B) \subseteq \{w \in \mathbb{C}: |w| \geq 1/4\}$, so $B(0, 1/4) \subseteq f(B)$. QED.


Reference:
Andersson - Topics in Complex Analysis