Hadamard gap theorem

Definition. A sequence $\{n_k\}$ of nonnegative integers is said to be lacunary if there is some $\delta > 0$ such that $n_{k + 1}/n_k \geq 1 + \delta$ for all $k \geq 1$.

A typical example of a lacunary sequence is $n_k = 2^k$.

Theorem (Hadamard Gap Theorem).  Suppose $\{n_k\}$ is a lacunary sequence and the power series $f(z) = \sum_{n = 0}^\infty a_n z^n$ with $a_n$ nonzero precisely when $n = n_k$ for some $k$ and has radius of convergence 1. Then $f(z)$ cannot be analytically extended to any larger region containing a point on the unit circle.

Proof: Suppose $f(z)$ extends to an analytic function on a region containing the open unit disk $B$ and a point $w_0$ on the unit circle. WLOG, we may assume $w_0 = 1$. Then $f(z)$ is analytic on a region $\Omega$ containing $B \cup \{1\}$. Define $g(w) = f(w^m(1 + w)/2)$ where $m$ is a positive integer. Since $1^m(1 + 1)/2 = 1$ and $|w^m(1+w)/2| = |w|^m|1+w|/2 < |w|^m \leq 1$ when $|w| \leq 1$ and $w \neq 1$ because $|1 + w| < 2$, so $g(w)$ is well-defined when $|w| \leq 1$, hence when $|w| < 1 + \epsilon$ for some $\epsilon > 0$. Write $g(w) = \sum_{m = 0}^\infty b_m w^m$ for $|w| < 1 + \epsilon$. Note that the powers of $w$ in the terms of $[w^m(1 + w)/2]^{n_k}$ range from $n_k m$ to $n_k (m + 1)$. Since $\{n_k\}$ is lacunary, we can choose $m$ to be so large that $n_{k + 1} m > n_k (m + 1)$ for all $k$, then the powers of $w$ in all of $[w^m(1 + w)/2]^{n_k}$ are distinct. Now
\[ \sum_{n = 0}^{n_k} a_n z^n = \sum_{m = 0}^{n_k (m + 1)} b_m w^m. \]
When $|w| < 1 + \epsilon$, RHS converges as $k \rightarrow \infty$. But LHS is the partial sums of $\sum_{k = 1}^\infty a_{n_k} z^{n_k}$ as $k$ varies, so the power series converges for all $z = w^m(1 + w)/2$ where $|w| < 1 + \epsilon$, which includes the case $z = 1$. The image of $B(0, 1 + \epsilon)$ under the map $w^m(1 + w)/2$ is an region $W$ containing 1, so the power series of $f(z)$ at 0 converges on $W$, contradicting its radius of convergence being 1. QED.

Reference:
Gamelin - Complex Analysis
Rudin - Real and Complex Analysis (3rd Ed.)

Conformal equivalence of annuli

For $0 < r < R$, let $A(r, R) = \{z \in \mathbb{C}: r < |z| < R\}$, the annulus centered at the origin with inner radius $r$ and outer radius $R$. We are going to show that the conformal equivalence classes of such anuuli in the complex plane are parametrized by the ratio $R/r$.

Theorem. $A(r_1, R_1)$ is conformally equivalent to $A(r_2, R_2)$ if and only if $R_1/r_1 = R_2/r_2$.
Proof: Suppose $R_1/r_1 = R_2/r_2$. Then there exists $k > 0$ such that $R_2 = k R_1$ and $r_2 = k r_1$. The map $f(z) = kz$ gives a conformal equivalence from $A(r_1, R_1)$ onto $A(r_2, R_2)$.
Conversely, let $f$ be a biholomorphic map from $A(r_1, R_1)$ onto $A(r_2, R_2)$. By scaling if necessary,  we may assume $r_1 = r_2 = 1$. Let $A_1 = A(1, R_1)$ and $A_2 = A(1, R_2)$. Fix some $1 < r < R_2$.
Let $C = \{z \in \mathbb{C}: |z| = r\}$.
Since $f^{-1}$ is continuous, $f^{-1}(C)$ is compact, so we can find some $\delta > 0$ such that $A(1, 1 + \delta) \cap f^{-1}(C) = \emptyset$. Let $V = f(A(1, 1 + \delta))$. Since $f$ is continuous, $V$ is connected, so either $V \subseteq A(1, r)$ or $V \subseteq A(r, R_2)$. By replacing $f$ with $R_2/f$, we may assume the first case holds.
Claim: $|f(z_n)| \rightarrow 1$ whenever $|z_n| \rightarrow 1$.
Proof of claim: Let $\{z_n\} \subseteq A(1, 1 + \delta)$. Note that $\{f(z_n)\}$ does not have a limit point in $A_2$ since otherwise $\{z_n\}$ would have a limit point in $A_1$ by continuity of $f^{-1}$, contradicting $|z_n| \rightarrow 1$. Hence $|f(z_n)| \rightarrow 1$ or $|f(z_n)| \rightarrow R_2$ (it must converge by continuity). The latter case is ruled out as $f(z_n) \in V \subseteq A(1, r)$.
Similarly, we also have:
Claim:  $|f(z_n)| \rightarrow R_2$ whenever $|z_n| \rightarrow  R_1$.
Now set $\alpha = \log R_2/ \log R_1$. Define $g: A_1 \rightarrow \mathbb{R}$ by \[ g(z)  = \log |f(z)|^2 - \alpha \log |z|^2 = 2(\log |f(z)| - \alpha \log |z|). \] We know that $\log |h|$ is harmonic whenever $h$ is holomorphic and nonzero, so $g$ is a harmonic function. By the two claims, $g$ extends to a continuous function on $\overline{A_1}$ vanishing on $\partial A_1$. This forces $g$ to vanish identically on $A_1$. In particular, \[ 0 = \frac{\partial g}{\partial z} = \frac{f'(z)}{f(z)} - \frac{\alpha}{z}. \] Take some $1 < c < R_1$ and let $\gamma(t) = ce^{it}, t \in [0, 2\pi]$. We have \[ \alpha = \frac{1}{2 \pi i}\int_\gamma \frac{\alpha}{z} dz = \frac{1}{2 \pi i}\int_\gamma \frac{f'(z)}{f(z)} dz = \mathrm{Ind}_{f \circ \gamma}(0), \] so $\alpha > 0$ is an integer. Observe that \[ \frac{d}{dz}(z^{-\alpha}f(z)) = z^{-\alpha - 1}(-\alpha f(z) + zf'(z)) = 0, \] thus $f(z) = Kz^{\alpha}$ for some nonzero constant $K$. Since $f$ is injective, $\alpha = 1$. Therefore $R_2 = R_1$.

Reference:
Rudin - Real and Complex Analysis (3rd Ed.)

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