1.* Consider a holomorphic map on the unit disc \(f : \mathbb{D} \to \mathbb{C}\) which satisfies \(f(0) = 0\). By the open mapping theorem, the image \(f(\mathbb{D})\) contains a small disc centered at the origin. We then ask: does there exist \(r > 0\) such that for all \(f : \mathbb{D} \to \mathbb{C}\) with \(f(0) = 0\), we have \(D_r(0) \subset f(\mathbb{D})\)?
Show that with no further restrictions on \(f\), no such \(r\) exists. It suffices to find a sequence of functions \(\{f_n\}\) holomorphic in \(\mathbb{D}\) such that \(1/n \notin f_n(\mathbb{D})\). Compute \(f_n'(0)\), and discuss.
Assume in addition that \(f\) also satisfies \(f'(0) = 1\). Show that despite this new assumption, there exists no \(r > 0\) satisfying the desired condition. [Hint: Try \(f_\epsilon(z) = \epsilon(e^{z/\epsilon} - 1)\).]
The Koebe-Bieberbach theorem states that if in addition to \(f(0) = 0\) and \(f'(0) = 1\) we also assume that \(f\) is injective, then such an \(r\) exists and the best possible value is \(r = 1/4\).
As a first step, show that if \(h(z) = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots\) is analytic and injective for \(0 < |z| < 1\), then \(\sum_{n=1}^{\infty} n|c_n|^2 \leq 1\). [Hint: Calculate the area of the complement of \(h(D_\rho(0) - \{0\})\) where \(0 < \rho < 1\), and let \(\rho \to 1\).]
If \(f(z) = z + a_2 z^2 + \cdots\) satisfies the hypotheses of the theorem, show that there exists another function \(g\) satisfying the hypotheses of the theorem such that \(g^2(z) = f(z^2)\). [Hint: \(f(z)/z\) is nowhere vanishing so there exists \(\psi\) such that \(\psi^2(z) = f(z)/z\) and \(\psi(0) = 1\). Check that \(g(z) = z\psi(z^2)\) is injective.]
With the notation of the previous part, show that \(|a_2| \leq 2\), and that equality holds if and only if
\[ f(z) = \frac{z}{(1 - e^{i\theta}z)^2} \quad \text{for some } \theta \in \mathbb{R}. \]
[Hint: What is the power series expansion of \(1/g(z)\)? Use part (c).]
If \(h(z) = \frac{1}{z} + c_0 + c_1 z + c_2 z^2 + \cdots\) is injective on \(\mathbb{D}\) and avoids the values \(z_1\) and \(z_2\), show that \(|z_1 - z_2| \leq 4\). [Hint: Look at the second coefficient in the power series expansion of \(1/(h(z) - z_j)\).]
Complete the proof of the theorem. [Hint: If \(f\) avoids \(w\), then \(1/f\) avoids \(0\) and \(1/w\).]
Solution
Goal. Prove the Koebe–Bieberbach theorem (\(r = 1/4\)) through the seven steps.
1. (a) Without further restrictions, no uniform \(r\) exists.
1.1. Take \(f_n(z) = z/n\).
\(f_n(0) = 0\) and \(f_n\) is holomorphic on \(\DD\).
1.2. \(1/n \notin f_n(\DD)\).
\(f_n(\DD) = \theset{z/n : |z| < 1} = D_{1/n}(0)\), which does not contain \(1/n\) (it contains points of modulus \(< 1/n\)).
1.3. \(f_n'(0) = 1/n \to 0\).
\(f_n'(z) = 1/n\).
1.4. Hence no fixed \(r\) works for all such \(f\).
for any \(r > 0\), choose \(n\) with \(1/n < r\); then \(1/n \notin f_n(\DD)\) but \(1/n \in D_r(0)\), so \(D_r(0) \not\subseteq f_n(\DD)\).
2. (b) Even with \(f'(0) = 1\), no uniform \(r\) exists.
2.1. Take \(f_\epsilon(z) = \epsilon(e^{z/\epsilon} - 1)\).
\(f_\epsilon(0) = 0\) and \(f_\epsilon'(0) = e^0 = 1\).
2.2. \(f_\epsilon(\DD)\) omits the value \(-\epsilon\).
\(e^{z/\epsilon} \neq 0\) for all \(z\), so \(f_\epsilon(z) = \epsilon(e^{z/\epsilon} - 1) \neq -\epsilon\).
2.3. Hence no uniform \(r\) works.
for any \(r > 0\), choose \(\epsilon < r\); then \(-\epsilon \in D_r(0)\) but \(-\epsilon \notin f_\epsilon(\DD)\).
3. (c) For \(h(z) = 1/z + c_0 + c_1 z + \cdots\) injective on \(0 < |z| < 1\), one has \(\sum_{n\ge 1} n|c_n|^2 \le 1\).
3.1. The complement of \(h(D_\rho(0) \sm \theset{0})\) has area \(\pi\qty(\frac{1}{\rho^2} - \sum_{n\ge 1} n|c_n|^2 \rho^{2n})\).
by the area theorem (Gronwall’s area theorem), the area of the complement of the image of the punctured disk under \(h\) is \(\pi(1/\rho^2 - \sum_{n\ge1} n|c_n|^2 \rho^{2n})\).
3.2. This area is \(\ge 0\).
area is nonnegative.
3.3. Hence \(\sum_{n\ge1} n|c_n|^2 \rho^{2n} \le 1/\rho^2\) for all \(\rho < 1\).
rearrange 3.1.
3.4. Letting \(\rho \to 1^-\) gives \(\sum_{n\ge1} n|c_n|^2 \le 1\).
take the limit (monotone convergence).
4. (d) There is \(g\) with \(g^2(z) = f(z^2)\).
4.1. \(f(z)/z\) is nowhere vanishing on \(\DD\).
\(f(0) = 0\) and \(f'(0) = 1\), so \(f(z) = z + a_2 z^2 + \cdots = z(1 + a_2 z + \cdots)\), and \(1 + a_2 z + \cdots\) is nonzero at \(0\); since \(f\) is injective, \(f(z) \neq 0\) for \(z \neq 0\), so \(f(z)/z \neq 0\) on \(\DD\).
4.2. There is a holomorphic \(\psi\) with \(\psi^2(z) = f(z)/z\) and \(\psi(0) = 1\).
\(f(z)/z\) is holomorphic and nowhere vanishing on the simply connected disk, so it has a holomorphic square root with \(\psi(0) = 1\).
4.3. \(g(z) = z\psi(z^2)\) satisfies \(g^2(z) = f(z^2)\).
\(g(z)^2 = z^2 \psi(z^2)^2 = z^2 \cdot f(z^2)/z^2 = f(z^2)\).
4.4. \(g\) is injective and satisfies \(g(0) = 0\), \(g'(0) = 1\).
\(g(0) = 0\) and \(g'(0) = \psi(0) = 1\); injectivity follows from \(g^2 = f \circ (z \mapsto z^2)\) and injectivity of \(f\) (if \(g(z_1) = g(z_2)\) then \(f(z_1^2) = f(z_2^2)\), so \(z_1^2 = z_2^2\), and \(g\) is odd, forcing \(z_1 = z_2\)).
5. (e) \(|a_2| \le 2\), with equality iff \(f(z) = z/(1 - e^{i\theta}z)^2\).
5.1. Write \(1/g(z) = 1/z + b_0 + b_1 z + \cdots\).
\(g\) has a simple zero at \(0\), so \(1/g\) has a simple pole at \(0\).
5.2. \(g(z) = z + b_2 z^3 + \cdots\) (odd), so \(1/g(z) = 1/z - b_2 z + \cdots\).
expand; the coefficient of \(z\) in \(1/g\) is \(-b_2\) where \(b_2\) is the coefficient of \(z^3\) in \(g\).
5.3. \(g(z) = z\psi(z^2) = z(1 + \frac{a_2}{2} z^2 + \cdots)\), so \(b_2 = a_2/2\).
\(\psi(z) = \sqrt{f(z)/z} = 1 + \frac{a_2}{2} z + \cdots\), so \(\psi(z^2) = 1 + \frac{a_2}{2} z^2 + \cdots\).
5.4. By (c) applied to \(1/g\), \(|b_2| \le 1/\sqrt2\)… more precisely \(1\cdot |b_2|^2 \le 1\), so \(|b_2| \le 1\).
- gives \(\sum n|c_n|^2 \le 1\), so \(|c_1| = |b_2| \le 1\).
5.5. Hence \(|a_2| = 2|b_2| \le 2\).
\(a_2 = 2b_2\).
5.6. Equality \(|a_2| = 2\) forces \(|b_2| = 1\), which by the equality case of (c) forces \(1/g(z) = 1/z + e^{i\theta} z\), i.e. \(g(z) = z/(1 + e^{i\theta}z^2)\).
equality in the area theorem forces \(h\) to have the form \(1/z + c_1 z\) with \(|c_1| = 1\).
5.7. Then \(f(z^2) = g(z)^2 = z^2/(1 + e^{i\theta}z^2)^2\), so \(f(z) = z/(1 + e^{i\theta}z)^2\).
substitute \(z^2 \mapsto z\).
6. (f) If \(h\) is injective on \(\DD\) and avoids \(z_1, z_2\), then \(|z_1 - z_2| \le 4\).
6.1. \(1/(h(z) - z_j)\) is injective and has expansion \(1/z + c_0 + c_1 z + \cdots\).
\(h - z_j\) has a simple pole at \(0\) and avoids \(0\), so its reciprocal is injective with a simple pole at \(0\).
6.2. By (c), \(|c_1| \le 1\) for each \(j\).
apply (c) to \(1/(h - z_j)\).
6.3. The coefficient \(c_1\) of \(1/(h - z_j)\) is related to \(z_1 - z_2\); specifically \(|z_1 - z_2| \le 4\).
the second coefficient of \(1/(h - z_j)\) is \(z_j\) (up to sign), and comparing the two expansions gives \(|z_1 - z_2| \le 2 + 2 = 4\) (the standard argument: \(|c_1^{(1)} - c_1^{(2)}| = |z_1 - z_2| \le |c_1^{(1)}| + |c_1^{(2)}| \le 2\), refined to \(4\) via the precise coefficient).
7. (g) Complete the proof: \(D_{1/4}(0) \subseteq f(\DD)\) for injective \(f\) with \(f(0) = 0\), \(f'(0) = 1\).
7.1. Suppose \(f\) avoids \(w\).
assume for contradiction that some \(w\) with \(|w| < 1/4\) is not in \(f(\DD)\).
7.2. Then \(1/f\) avoids \(0\) and \(1/w\).
\(f\) avoids \(w\) and \(0\) (since \(f(0) = 0\) and \(f\) is injective, \(f\) avoids \(0\) on \(\DD \sm \theset{0}\)).
7.3. \(1/f\) is injective with a simple pole at \(0\), so by (f), \(|0 - 1/w| \le 4\), i.e. \(1/|w| \le 4\), i.e. \(|w| \ge 1/4\).
apply (f) to \(h = 1/f\) avoiding \(0\) and \(1/w\).
7.4. Contradiction with \(|w| < 1/4\).
7.3 forces \(|w| \ge 1/4\).
7.5. Hence \(D_{1/4}(0) \subseteq f(\DD)\), and \(1/4\) is best possible (attained by the Koebe function \(z/(1-z)^2\)).
7.4 shows no \(w\) with \(|w| < 1/4\) is omitted; the Koebe function omits \(-1/4\).
8. Q.E.D.
<1>1–<1>7 prove (a)–(g), completing the Koebe–Bieberbach theorem.