Hyperbolic Translations / Blaschke Factors

For \(a\in {\mathbb{D}}\), the maps \begin{align*} \psi_a \coloneqq{a-z\over 1-\overline{a}z} \end{align*} are hyperbolic translations because they preserve the hyperbolic metric on the Poincaré disc. They’re also commonly called Blaschke factors, and also sometimes taken to be \begin{align*} \phi_a \coloneqq{z-a \over 1-\overline{a} z} = - \psi_a .\end{align*} A rational map of the form \begin{align*} \Psi_{\mathbf{a}}(z) = \lambda \prod_{1\leq k\leq n} \psi_a(z) = \lambda \prod_{1\leq k \leq n} {a_i - z\over 1 - \overline{a_i} z},\qquad \mathbf{a}\coloneqq{\left[ {a_1,\cdots, a_n} \right]} \end{align*} with zeros \(a_i \in {\mathbb{D}}\) is called a Blaschke product and is a map \({\mathbb{D}}\to {\mathbb{D}}\) that preserves \(S^1\).

Some useful properties:

  • \(\psi_a \in \mathop{\mathrm{Aut}}({\mathbb{D}})\)
  • \(\psi_a(S^1) = S^1\)
  • \(\psi_a(0) = a\) and \(\psi(a) = 0\)
  • With this choice of sign, \(\psi_{a} ^{-1}= \psi_{a}\), so \(\psi_a^2 = \operatorname{id}\).
  • \(\psi_a'(z) = {{\left\lvert {a} \right\rvert}^2 - 1 \over \qty{1-\overline{a} z}^2 }\)
  • \(\psi_a(\lambda z) = \lambda\psi_{\overline{\lambda }a}(z)\)

Inverting: set \(f(z) = w\) and solve for \(z\): \begin{align*} {a-z \over 1 - \overline{a}z} &= w \\ \implies a-z - w(1-\overline{a} z) &= 0 \\ \implies z&= {w-a \over \overline{a} w - 1} = {a-w\over 1-\overline{a} w} .\end{align*}

Differentiating: the quotient rule \begin{align*} \psi'_a(z) = {-(1-\overline{a} z) + \overline{a}(a-z) \over \qty{1-\overline{a} z}^2} = {-1 + {\left\lvert {a} \right\rvert}^2 \over \qty{1-\overline{a} z}^2} .\end{align*}

Scaling: use a fun trick, insert \(1=\overline{\lambda }\lambda\) like so \begin{align*} \psi_a(\lambda z) &= {a - \lambda z \over 1 - \overline{a} \lambda z}\\ &= {\lambda\overline{\lambda }a - \lambda z \over 1 - \overline{a} \lambda z} \\ &= \lambda {\overline{\lambda }a - z \over 1 - \overline{\overline{\lambda }a} z} \\ &= \lambda \psi_{\overline{\lambda }a}(z) .\end{align*}

Being an involution: check \(\psi_a(\psi_a(z))\) satisfies the Schwarz lemma and has two fixed points, forcing it to be the identity.

Every map \(g\in \mathop{\mathrm{Aut}}({\mathbb{D}})\) is of the form

\begin{align*} \mathop{\mathrm{Aut}}({\mathbb{D}}) = \left\{{ \lambda \psi_a(z) {~\mathrel{\Big\vert}~}a\in {\mathbb{D}}, \lambda \in S^1 }\right\} ,\end{align*} i.e. they are all Blaschke factors and rotations.

  • That these maps are biholomorphisms: they’re compositions of \(z\mapsto \lambda z\) and \(z\mapsto {z-a\over 1-\overline{a} z}\), which are biholomorphisms.

  • Let \(f \in \mathop{\mathrm{BiHol}}(\Delta)\) be arbitrary, fix \(a\in \Delta\) with \(f(a) = 0\)

  • Write \(M(z) = {z-a\over 1-\overline{a} z}\), then note that \(M(a) = 0\) and this is a biholomorphism.

  • \(g\coloneqq f\circ M^{-1}\in \mathop{\mathrm{BiHol}}(\Delta)\) sends \(0\to0\) and is thus a rotation, so \(g(z) = \lambda z\).

  • Write \(g\circ M = f \circ M \circ M^{-1}= f\), which exhibits \(f\) in the desired form.

  • Claim: this representation is unique. Consider \(f'(z)\), this determines \(\operatorname{Arg}(\lambda)\).

The Schwarz Lemma

If \(f: {\mathbb{D}}\to {\mathbb{D}}\) is holomorphic with \(f(0) = 0\), then

  • \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) for all \(z\in {\mathbb{D}}\)
  • \({\left\lvert {f'(0)} \right\rvert} \leq 1\).

Moreover, if

  • \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\) for any \(z_0\in {\mathbb{D}}\setminus\left\{{0}\right\}\), or
  • \({\left\lvert {f'(0)} \right\rvert} = 1\),

then \(f\) is a rotation, i.e. \(f(z) = \lambda z\) for some \({\left\lvert {\lambda} \right\rvert} = 1\).

Let \(f:{\mathbb{D}}\to {\mathbb{D}}\) be holomorphic with \(f(0) = 0\). Then either

  • \(f(z) = e^{i\theta}z\) is a rotation, or
  • \({\left\lvert {f'(0)} \right\rvert} < 1\) and \({\left\lvert {f(z)} \right\rvert} < {\left\lvert {z} \right\rvert}\) for all \(z\in {\mathbb{D}}\), noting the strict inequalities.

  • Idea: apply the maximum modulus principle to \(g(z) \coloneqq f(z)/z\).
  • \({\left\lvert {g(z)} \right\rvert} \leq 1\):
    • Expand \(f\) at \(z=0\) as \(\sum_{k\geq 0} c_k z^k\). Since \(f(0) = c_0\), we have \(c_0 = 0\).
    • So \(g(z) \coloneqq f(z)/z\) is holomorphic on \({\mathbb{D}}\), since the singularity at \(z=0\) is removable.
    • Set \({\left\lvert {z} \right\rvert} = r < 1\), then \({\left\lvert {g(z)} \right\rvert} = {\left\lvert {f(z)} \right\rvert}/r \leq 1/r\) since \({\left\lvert {f(z)} \right\rvert} \leq 1\).
    • By MMP, \({\left\lvert {g(z)} \right\rvert} \leq 1/r\) holds in the entire disc \({\left\lvert {z} \right\rvert} \leq r\), so take \(r\to 1\) to get \({\left\lvert {g(z)} \right\rvert} \leq 1\)
  • \({\left\lvert {f'(0)} \right\rvert} \leq 1\) with equality iff \(f\) is a rotation:
    • Note that \(f(0) = 0\), so we can write \(g(0) = \lim_{z\to 0} {f(z) - f(0) \over z-0} \coloneqq f'(0)\).
    • So \(1 = {\left\lvert {f'(0)} \right\rvert} = {\left\lvert {g(0)} \right\rvert}\).
    • But \({\left\lvert {g(z)} \right\rvert} \leq 1\) on \({\mathbb{D}}\) and \(g(z) = 1\) in the interior, so by MMP this makes \(g\) constant.
    • So again \(f(z) = cz\) with \({\left\lvert {c} \right\rvert} = 1\).
  • \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\implies f\) is a rotation:
    • Again \({\left\lvert {g(z)} \right\rvert} \leq 1\), but \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert} \implies {\left\lvert {g(z_0)} \right\rvert} = 1\), so \(g\) attains a maximum on \({\left\lvert {z} \right\rvert}\leq 1\), making it constant, so \(f(z) = cz\).
    • Then \({\left\lvert {z_0} \right\rvert} = {\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {cz_0} \right\rvert}\) since \(f(z_0) = z_0\), so \({\left\lvert {c} \right\rvert} = 1\) and \(c = e^{i\theta}\) for some \(\theta\).




Blaschke Factors

Let \(f: {\mathbb{D}}\to {\mathbb{D}}\) with \(\left\{{a_k}\right\}_{k\leq n}\) the zeros of \(f\) in \({\mathbb{D}}\). Show that \begin{align*} {\left\lvert {f(z)} \right\rvert}\leq \prod_{k\leq n} {\left\lvert { \psi_{a_k}(z) } \right\rvert} .\end{align*}


Define \(\Psi(z) \coloneqq\prod_{k\leq n} \psi_{a_k}(z)\) and \(g(z) \coloneqq f(z)/\Psi(z)\). The claim is that \({\left\lvert {g(z)} \right\rvert} \leq 1\), which implies the result directly. Note that \({\left\lvert {\Psi(z)} \right\rvert} = 1\) for \({\left\lvert {z} \right\rvert} = 1\), so \(\lim_{r\to 1^-} {\left\lvert {\Psi(re^{it})} \right\rvert} = 1\) along any ray. Now for \(z_r \coloneqq re^{it}\) for \(r< 1\), \begin{align*} {\left\lvert {g(z_r)} \right\rvert} = { {\left\lvert {f(z_r)} \right\rvert} \over {\left\lvert {\Psi(z_r)} \right\rvert} } \leq {1\over {\left\lvert {\Psi(z_r)} \right\rvert} } \leq {1\over \displaystyle\sup_{t\in [0, 2\pi]} {\left\lvert {\Psi(z_r) } \right\rvert} } \overset{r\to 1^-}\longrightarrow 1 .\end{align*}


If \(f: {\mathbb{D}}_R(a)\to{\mathbb{D}}_M(0)\) with \(f(a) = 0\), then \begin{align*} {\left\lvert {f(z)} \right\rvert}\leq {M\over R}{\left\lvert {z-a} \right\rvert} .\end{align*}


Set \(g(z) \coloneqq{f(Rz + a) \over M}\), then \(g: {\mathbb{D}}\to {\mathbb{D}}\) with \(g(0) = f(a)/M = 0\), so unwinding Schwarz yields \begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {f(Rz+a)\over M} \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert {f(Rz+a)} \right\rvert} &\leq M{\left\lvert {z} \right\rvert} \\ \implies {\left\lvert {f(w)} \right\rvert} &\leq M{\left\lvert {w-a\over R} \right\rvert} \qquad w = Rz+a\implies z={w-a\over R} .\end{align*}

Let \(\psi_a\) be a Blaschke factor and use the Schwarz lemma to prove that \(\psi_a \circ \psi_a = \operatorname{id}_{\mathbb{D}}\).


First, \(\psi_a\) maps \({\mathbb{D}}\to {\mathbb{D}}\). Fix \(z\in S^1\), then \begin{align*} {\left\lvert {\psi_a(z)} \right\rvert} \coloneqq {\left\lvert {a-z\over 1-\overline{a} z} \right\rvert} &= {\left\lvert {a-z\over 1-\overline{a} z} \right\rvert} \cdot {\left\lvert {\overline{z}} \right\rvert}^{-1}\\ &= {\left\lvert {a-z\over \overline{a} - \overline{z}} \right\rvert} \\ &= {\left\lvert {a-z \over \overline{a-z} } \right\rvert} \\ &\coloneqq{\left\lvert {w \over \overline{w} } \right\rvert} \\ &= {\left\lvert {w \over {\left\lvert {w} \right\rvert}^2/w } \right\rvert}\\ &= {\left\lvert {w^2 \over {\left\lvert {w} \right\rvert}^2} \right\rvert} \\ &= 1 .\end{align*} \(\psi_a\) is holomorphic on \({\mathbb{D}}\) since it has a simple pole at \(z=1/\overline{a}\), but \({\left\lvert {a} \right\rvert}<1\) implies \({\left\lvert {1/\overline{a}} \right\rvert} > 1\). Let \(g\coloneqq\psi_a\circ \psi_a\), then \(g(0) = 0\) so Schwarz applies. Since \(g(a) = a\) with \(a\neq 0\), the 2nd part of Schwarz also applies since \({\left\lvert {g(a)} \right\rvert} = {\left\lvert {a} \right\rvert}\) and \(g(z) = \lambda z\) is a rotation. Since \(a = g(a) = \lambda a\), this forces \(\lambda = 1\), so \(g\) is the identity.

Prove the following: suppose \({\left\lvert {f(z)} \right\rvert}\leq 1\), then for all \(z, w\in {\mathbb{D}}\), \begin{align*} \left|\frac{f(z)-f(w)}{1-\overline{f(w)} f(z)}\right| \leq\left|\frac{z-w}{1-\overline{w} z}\right| \quad\text{ and } \left|f^{\prime}(z)\right| \leq \frac{1-|f(z)|^{2}}{1-|z|^{2}} .\end{align*} If equality holds for some \(z\neq w\) in either expression, then \(f= \lambda F\) where \(F\) is a linear fractional transformation and \({\left\lvert {\lambda} \right\rvert} = 1\), so \(f\in \mathop{\mathrm{Aut}}({\mathbb{D}})\).

Note that this does not require \(f(0) = 0\).




Show that if \(f: \Delta\to\Delta\) is a biholomorphism with \(f(0) = 0\) then \(f\) is a rotation.


By Schwarz, \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\), and if we write \(h \coloneqq f^{-1}\) then \(h(0) = 0\) and \({\left\lvert {h(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) as well. Schwarz says \(f\) will be a rotation if there is any \(z_0\) such that \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\). Write \(f(z) = w\), we’ll then show that in fact \({\left\lvert {f(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) for all \(z\in {\mathbb{D}}\). \begin{align*} {\left\lvert {z} \right\rvert} = {\left\lvert {(h\circ f)(z)} \right\rvert} = {\left\lvert {h(w)} \right\rvert} \leq {\left\lvert {w} \right\rvert} = {\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert} .\end{align*}

By Schwarz, \({\left\lvert {f'(0)} \right\rvert} \leq 1\) – the claim is that we have equality, so that by Schwarz \(f(z) = \lambda z\) for some \({\left\lvert {\lambda} \right\rvert} = 1\). Use that \(f'(0) \neq 0\) since \(f\) is a bijection near zero and if \(g\coloneqq f^{-1}\) then \(g'(0) = 1/f'(0)\). Moreover Schwarz applies to \(g\), so \(1 \geq {\left\lvert {g'(0)} \right\rvert} \geq {1\over {\left\lvert {f'(0)} \right\rvert} }\), forcing \({\left\lvert { f'(0) } \right\rvert} = 1\). By the equality clause in the Schwarz lemma for \(f\), \(f\) is rotation.

Show that if \(f:{\mathbb{H}}\to {\mathbb{D}}\) is holomorphic and \(f(i) = 0\) then \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z-i\over z+i} \right\rvert}\).


Note that \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {g(z)} \right\rvert} \impliedby {\left\lvert {(f\circ g^{-1})(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} ,\end{align*} so one can use the Schwarz lemma on \(F \coloneqq f\circ g^{-1}\). Noting that \(g(z) \coloneqq{z-i\over z+i}: {\mathbb{H}}\to {\mathbb{D}}\) is the Cayley map, the inverse is \(g^{-1}(z) = i{1-z\over 1+z}: {\mathbb{D}}\to {\mathbb{H}}\). Then \(F(0) = f(g^{-1}(0)) = f(i) = 0\) by assumption, so Schwarz yield \({\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\).

Suppose \(f:{\mathbb{D}}\to {\mathbb{D}}\) with \(f(0) = 0\) and that there exists an \(r\in (0, 1)\) with \(f(r) = f(-r) = 0\). Show that \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z^2-r^2 \over 1-r^2 z^2} \right\rvert} .\end{align*}


The key observation is that this factors: \begin{align*} {z^2 - r^2 \over 1-r^2 z^2} = {r-z\over 1+rz}{-r-z\over 1-rz} = \psi_{r}(z) \psi_{-r}(z) ,\end{align*} so this inequality will follow from Schwarz on \(g(z) \coloneqq f(z)/\psi_a(z)\psi_{-a}(z)\). Schwarz does apply since \({\left\lvert {f} \right\rvert}\leq 1\) in \({\mathbb{D}}\) and \({\left\lvert {\psi_a(z)} \right\rvert} = 1\) on \(S^1\), so \({\left\lvert {g(z)} \right\rvert} \leq 1\) on \(S^1\) and by the MMP this inequality holds in all of \({\mathbb{D}}\). So \({\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\) and unwinding gives the desired inequality.

Use a version of the Schwarz lemma to prove Liouville’s theorem.


Suppose \(f\) is entire and bounded, we’ll show \(f\) is constant. If \(f\) is bounded by \(M\), then \(f({\mathbf{C}}) \subseteq {\mathbb{D}}_M(0)\). Without loss of generality, replace \(f\) with \(g(z) \coloneqq f(z) - f(0)\), so \(g(0) = 0\) and is still bounded by \(M' \coloneqq M + {\left\lvert {f(0)} \right\rvert}\) by the triangle inequality. This is still finite since \(0\) is not a singularity since \(f\) is entire.

By the radius \(R\) variant of the Schwarz lemma, for every \({\mathbb{D}}_R(0)\), \begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {M\over R}{\left\lvert {z} \right\rvert} \qquad \text{for } z\in {\mathbb{D}}_R(0) .\end{align*} Using that \(g(0) = 0\), \begin{align*} {\left\lvert {g(z) - g(0) \over z} \right\rvert} \leq {M\over R}\overset{R\to\infty}\longrightarrow 0 ,\end{align*} where dividing by \(z\) is not an issue since \(z=0\) is a zero of \(g\) of at least order one. This forces \(g(z) = g(0) = 0\) for all \(z\), so \(f(z) = f(z_0)\) is constant.

Let \(f:{\mathbb{D}}\to {\mathbb{D}}\) with \(f(0) = f'(0) = 0\). Show that \({\left\lvert {f''(0)} \right\rvert} \leq 2\) and describe all \(f\) for which this is an equality.


By Schwarz, \({\left\lvert {f(z)} \right\rvert}\leq z\). Write \(g(z) \coloneqq f(z)/z\), then \({\left\lvert {g(z)} \right\rvert}\leq 1\) and \(g:{\mathbb{D}}\to {\mathbb{D}}\) is holomorphic since \(f\) has a zero of order at least one at \(0\). Note that \(f(z) = c_2z^2 + { \mathsf{O}}(z^3)\), where \(c_0 = 0\) since \(f(0) = 0\) and \(c_1 = 0\) since \(f'(0) = 0\). Thus \(g(z) = c_2z + { \mathsf{O}}(z^2)\), and \(g(0) = 0\), so Schwarz applies and

  • \({\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\)
  • \({\left\lvert {g'(0)} \right\rvert} \leq 1\)

We have \(g'(z) = c_2 + { \mathsf{O}}(z)\) so \(g'(0) = c_2 = {f^{(2)}(0) \over 2!}\). \begin{align*} 1 \geq {\left\lvert {g'(0)} \right\rvert} = {\left\lvert {c_2} \right\rvert} = {\left\lvert {f^{(2)}(0) \over 2} \right\rvert} \implies {\left\lvert { f^{(2)}(0) } \right\rvert} \leq 2! .\end{align*}

Suppose this is an equality – then Schwarz on \(g\) shows \(g\) is a rotation, so \begin{align*} {f(z)\over z} = g(z) = \lambda z \implies f(z) = \lambda z^2 \qquad \lambda \in S^1 .\end{align*}

Note: if \({\left\lvert {z} \right\rvert}\leq 1\), we have \(g({\mathbb{D}}) \subseteq \overline{{\mathbb{D}}}\), but the open mapping theorem always guarantees \(g({\mathbb{D}}) = {\mathbb{D}}\).

Suppose \(f:{\mathbb{D}}\to {\mathbb{D}}\) with \(f(a) = a\) a fixed point where \({\left\lvert {f'(a)} \right\rvert} < 1\). Show that for any initial point \(z_0\), the sequence \(z_k \coloneqq f(z_{k-1})\) converges to \(a\).


First suppose \(a=0\) – then Schwarz applies, and since \({\left\lvert {f'(a)} \right\rvert} < 1\) is strict, \(f\) is not a rotation.

For any choice of \(z_0\in {\mathbb{D}}\), there is an \(r\) with \(0< {\left\lvert {z_0} \right\rvert} < r < 1\) and a constant \(C<1\) such that \({\left\lvert {f(z)} \right\rvert} \leq C{\left\lvert {z} \right\rvert}\) for \({\left\lvert {z} \right\rvert} < r\).

With such an \(r\) and \(C<1\) in hand,

\begin{align*} {\left\lvert {z_k} \right\rvert} = {\left\lvert {f(z_{k-1})} \right\rvert} \leq C{\left\lvert {z_{k-1}} \right\rvert} = C{\left\lvert {f(z_{k-2})} \right\rvert} \leq C^2 {\left\lvert {z_{k-2}} \right\rvert} \cdots \implies {\left\lvert {z_k} \right\rvert} \leq C^k{\left\lvert {z_0} \right\rvert} \overset{k\to\infty}\longrightarrow 0 ,\end{align*} which proves the \(a=0\) case.

The claim is that for any given \(r\), the constant \(C\coloneqq M/r\) works, where \(M\coloneqq\max_{{\left\lvert {z} \right\rvert} = r} {\left\lvert {f(z)} \right\rvert}>0\). The scaled Schwarz lemma gives \({\left\lvert {f(z)} \right\rvert}\leq {M\over r}{\left\lvert {z} \right\rvert} = C{\left\lvert {z} \right\rvert}\), and \({\left\lvert {C} \right\rvert} \leq 1\) since \({\left\lvert {M} \right\rvert} \leq r\), which follows because \({\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\) on \({\mathbb{D}}\) itself.

For \(a\neq 0\), take a Blaschke factor \(\psi_a(z)\) and consider \(F \coloneqq\psi_a^{-1}\circ f\circ \psi_a\). The claim is that this reduces to the case \(a=0\).

Note \(F(0) = 0\), so \(0\) is a fixed point of \(F\). Moreover, a clever calculation shows \begin{align*} F'(0) &= (\psi_a^{-1})'(f(\psi_a(0))) \cdot f'(\psi_a(0)) \cdot \psi_a'(0) \\ \\ &= (\psi_a^{-1})'(f( a )) \cdot f'(a ) \cdot \psi_a'(0) \\ \\ &= (\psi_a^{-1})'(a) \cdot f'(a ) \cdot \psi_a'(0) \qquad \text{since } f(a) = a \\ \\ &= (\psi_a)'(a) \cdot \psi_a'(0) \cdot f'(a) \qquad \text{since } \psi_a^{-1}= \psi_a \\ \\ &= (\psi_a)'( \psi_a(0) ) \cdot \psi_a'(0) \cdot f'(a) \\ \\ &= (\psi_a \circ \psi_a)'(0) \cdot f'(a) \\ &= 1 \cdot f'(a) ,\end{align*} so \({\left\lvert {F'(0)} \right\rvert} = {\left\lvert {f'(a)} \right\rvert} < 1\). Now setting \(w_k \coloneqq\psi_a(z_n)\) and writing \(f = \psi_a \circ F \circ \psi_a^{-1}\), by continuity we have \begin{align*} f(z_k) = \psi_a(F(w_k)) \overset{k\to\infty}\longrightarrow \psi_a(0) = a .\end{align*}

Show that the only holomorphic map \(f:{\mathbb{D}}\to {\mathbb{D}}\) two distinct fixed points \(a\neq b\) is the identity.


Note that without loss of generality we can assume \(a=0\) so \(f(0) = 0\) and \(b\neq 0\). If not, if \(a,b\neq 0\) then let \(F\coloneqq\psi_a \circ f \circ \psi_a\), then \(F(0) = 0\) and \(F(b') = b'\) for \(b\coloneqq\psi_a(b)\).

Since \(f(0) = 0\), Schwarz applies, so \({\left\lvert {f(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) with equality attained because \({\left\lvert {f(b)} \right\rvert} = {\left\lvert {b} \right\rvert}\), and \(f(z) = \lambda z\) must be a rotation. Since \(b = f(b) = \lambda b\), we have \(\lambda = 1\).


Let \(f\in \mathop{\mathrm{Hol}}({\mathbb{D}})\). Show that if \(f\) has a fixed point \(a\) then \({\left\lvert {f'(a)} \right\rvert} \leq 1\), and that \begin{align*} {\left\lvert {f(0)} \right\rvert}^2 + {\left\lvert {f'(0)} \right\rvert}^2 \leq 1 .\end{align*}


Set \(f(a) = a\) in Schwarz-Pick: \begin{align*} \left|f^{\prime}(a)\right| \leq \frac{1-|f(a)|^{2}}{1-|a|^{2}} \implies {\left\lvert {f'(a)} \right\rvert} \leq {1 - {\left\lvert {a} \right\rvert}^2 \over 1 - {\left\lvert {a} \right\rvert}^2} \leq 1 .\end{align*} Set \(a=0\): \begin{align*} \left|f^{\prime}(0)\right| \leq \frac{1-|f(0)|^{2}}{1-|0|^{2}} \implies {\left\lvert {f'(0)} \right\rvert}^2 \leq 1 - {\left\lvert {f(0)} \right\rvert}^2 .\end{align*}

Does there exist a map \(f: {\mathbb{D}}\to {\mathbb{D}}\) with

  • \(f\qty{1\over 2} = {3\over 4}\)
  • \(f'\qty{1\over 2} = {2\over 3}\)


Apply Schwarz-Pick: \begin{align*} {\left\lvert {f'\qty{1\over 2} } \right\rvert} \leq {1 - {\left\lvert {f\qty{1\over 2}} \right\rvert}^2 \over 1 - {\left\lvert {1\over 2} \right\rvert}^2 } = {7\over 2}< {2\over 3} ,\end{align*} so this is not possible.

Suppose \(f: {\mathbb{D}}\to {\mathbb{D}}\) with \(f(0) = 0\) and \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {e^z} \right\rvert}\) when \({\left\lvert {z} \right\rvert} = 1\). Find an upper bound for \(f\qty{1+i\over 2}\).


Consider \(g(z) \coloneqq f(z)/e^z\) – since \(g(0) = 0\) and \(g: {\mathbb{D}}\to {\mathbb{D}}\), Schwarz applies and \begin{align*} {\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {ze^z} \right\rvert} \leq e{\left\lvert {z} \right\rvert} .\end{align*} So \begin{align*} {\left\lvert {f\qty{1+i\over 2}} \right\rvert} \leq e{\left\lvert {1+i\over 2} \right\rvert}= {e\sqrt{2} \over 2} .\end{align*}

Suppose \(f:{\mathbb{H}}\to {\mathbf{C}}\) with \({\left\lvert {f(z)} \right\rvert}< 1\) and \(f(i) = 0\). Find an upper bound for \(f(2i)\).


Compose with the inverse Cayley map \(g(z) \coloneqq i{1+z\over 1-z}\) so \(g: {\mathbb{D}}\to {\mathbb{H}}\) to get \(F\coloneqq f\circ g:{\mathbb{D}}\to {\mathbb{D}}\), where \(F(0) =f(g(0))=f(i) = 0\). So Schwarz applies and \({\left\lvert {F(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\). Now a small trick: \begin{align*} {\left\lvert {f(2i)} \right\rvert} = {\left\lvert {(f\circ g\circ g^{-1})(2i)} \right\rvert} = {\left\lvert { F \qty{z+i\over z-i}\Big|_{z=2i} } \right\rvert} = {\left\lvert {F \qty{1\over 3} } \right\rvert} \leq {1\over 3} .\end{align*}

Use Rouché’s theorem and the Schwarz lemma to prove the following: if \(f:{\mathbb{D}}\to \overline{ {\mathbb{D}}}\) is holomorphic with \(f(0) = 0\), then there are exactly \(m\) solutions (counted with multiplicity) to \(f(z) = (2z)^m\) in the disc \({\left\lvert {z} \right\rvert} < 1/2\).


First note that the image of \(f\) is in fact \({\mathbb{D}}\) rather than \(\overline{{\mathbb{D}}}\), using the open mapping theorem and that the domain \({\mathbb{D}}\) is open. So Schwarz applies to \(f\). Write \(g(z) \coloneqq f(z) - (2z)^m\), the claim is that \(g\) has \(m\) zeros. Toward applying Rouché, identify

  • The big part of \(g\): \(M(z) \coloneqq-(2z)^m\)
  • The small part of \(g\): \(m(z) \coloneqq g(z) - M(z) = f(z)\).

Now \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} = {1\over 2} < {\left\lvert {m(z)} \right\rvert} = {\left\lvert {2z} \right\rvert}^m = 1 \qquad \text{on } {\left\lvert {z} \right\rvert} = {1\over 2} ,\end{align*} so Rouché applies: \({\sharp}Z_m = {\sharp}Z_M\) on \({\left\lvert {z} \right\rvert} < {1\over 2}\), where \(m(z) = f(z)\) and \(M(z) = -(2z)^m\) which has exactly \(m\) zeros.

Note that this works with \(g(z) \coloneqq(cz)^m\) and \(R = {1\over c}\).

Suppose \(f\) is holomorphic and \({\left\lvert {f(z)} \right\rvert}\leq 1\) for \(\Re(z) > 0\) with \(f(1) = 0\). Find an upper bound for \(f(2)\).


Use the conformal map \(g: z\mapsto -1{z+1\over z-1}\) to map \(\Re(z)>0\) to \({\mathbb{D}}\). Composing \(F: {\mathbb{D}}\xrightarrow{g} -i{\mathbb{H}}\xrightarrow{f} {\mathbb{D}}\) yields a map \(F = f\circ g:{\mathbb{D}}\to {\mathbb{D}}\). Since \(F(0) = f(g(0)) = f(1) = 0\), Schwarz applies and \({\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\). Using the standard trick, \begin{align*} {\left\lvert {f(2)} \right\rvert} = {\left\lvert {(f\circ \circ g^{-1})(2)} \right\rvert} = {\left\lvert {F(g^{-1}(2))} \right\rvert} = {\left\lvert {F\qty{z-1\over z+1}\Big|_{z=2}} \right\rvert} = {\left\lvert {F\qty{1\over 3}} \right\rvert} \leq {\left\lvert {1\over 3} \right\rvert} .\end{align*}

Suppose \(f:{\mathbb{D}}\to {\mathbb{D}}\) and \(f(0) = 0\). Show that \begin{align*} {\left\lvert {f(z) + f(-z)} \right\rvert} \leq 2{\left\lvert {z} \right\rvert}^2 ,\end{align*} with equality only if \(f(z) = \lambda z^2\) for some \({\left\lvert {\lambda} \right\rvert} = 1\).


Define \(F(z) \coloneqq{ f(z) + f(-z) \over 2 z^2}\), then \(F(0) = 0\) and thus Schwarz applies and yields the desired inequality. If \({\left\lvert {F(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) for any \(z\), then \(F(z) = \lambda z^2\) is a rotation and \(f(z) + f(-z) = \lambda z\). Noting that \(F\) is an even function, \(f(z) = \lambda z^2 + f_o(z)\) for \(f_o\) some odd function.

\(f_0\) is identically zero.

Given this, the result follows immediately since \(f(z) = \lambda z^2\).

Note that on the LHS, \({\left\lvert { f(re^{it})} \right\rvert} \to 1\) as \(r\to 1\) for any \(t\). So this must be true on the RHS, and the first term \(\lambda {\left\lvert { re^{it}} \right\rvert}^2 \to 1\), forcing \({\left\lvert {h(re^{it})} \right\rvert} \to 0\). So \(h\equiv 0\) on \({\left\lvert {z} \right\rvert} = 1\), and by the MMP, \({\left\lvert {h} \right\rvert} = 0\) in \({\mathbb{D}}\), making \(h\equiv 0\) on \({\mathbb{D}}\).

#complex/exercise/completed #complex/exercises/completed