Complex Analysis Proof Workshop

A Proof Strategy: Move the Point to the Origin

The Residue Theorem turns local information about singularities into a global contour integral. This workshop uses a different complex-analysis strategy: transform a problem until a powerful theorem applies directly. The Schwarz lemma controls holomorphic maps of the unit disk that fix the origin. Disk automorphisms let us convert a general point and its image into that setting, leading to the Schwarz–Pick inequality.

Write $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. We will prove the needed transformations map $\mathbb{D}$ onto itself, and then track their derivatives carefully. The image and inverse-image checks matter: solving an algebraic equation for an inverse is not, by itself, enough to show that the inverse is a map from the disk to the disk.

Automorphisms of the Unit Disk

For $a\in\mathbb{D}$, define a fractional linear map that sends $a$ to $0$. Its denominator is nonzero throughout the disk because $|\overline{a}z|=|a||z|<1$ for $z\in\mathbb{D}$.

Definition: For $a\in\mathbb{D}$, the disk map $\phi_a:\mathbb{D}\to\mathbb{C}$ is given by $$ \phi_a(z)=\frac{z-a}{1-\overline{a}z}. $$

The following identity verifies that $\phi_a$ maps the disk into itself:

|1-\overline{a}z|^2-|z-a|^2 =(1-|a|^2)(1-|z|^2).

Indeed, expanding the left side gives $1-2\operatorname{Re}(\overline{a}z)+|a|^2|z|^2-|z|^2+2\operatorname{Re}(z\overline{a})-|a|^2$; the real-part terms cancel, leaving the right side. Since $a,z\in\mathbb{D}$, the right side is positive, so $|z-a|<|1-\overline{a}z|$, and hence $|\phi_a(z)|<1$.

Theorem (Disk Automorphism): For each $a\in\mathbb{D}$, $\phi_a$ is a holomorphic bijection from $\mathbb{D}$ onto itself, with holomorphic inverse $$ \phi_a^{-1}(w)=\frac{w+a}{1+\overline{a}w}. $$

Proof. The denominator $1+\overline{a}w$ is nonzero for $w\in\mathbb{D}$, since $|\overline{a}w|<1$. Solving $w=(z-a)/(1-\overline{a}z)$ for $z$ gives $z=(w+a)/(1+\overline{a}w)$. But we must also verify that this formula maps $\mathbb{D}$ into $\mathbb{D}$. For $w\in\mathbb{D}$,

1-\left|\frac{w+a}{1+\overline{a}w}\right|^2 =\frac{|1+\overline{a}w|^2-|w+a|^2}{|1+\overline{a}w|^2} =\frac{(1-|a|^2)(1-|w|^2)}{|1+\overline{a}w|^2}>0.

The numerator identity follows by expanding both squared moduli: their terms involving $\operatorname{Re}(\overline{a}w)$ cancel in the difference. Thus the inverse formula does map $\mathbb{D}$ into $\mathbb{D}$. Substitution into the two formulas gives $\phi_a(\phi_a^{-1}(w))=w$ and $\phi_a^{-1}(\phi_a(z))=z$. Both maps are holomorphic on $\mathbb{D}$, since their denominators do not vanish there. This proves the bijection and the asserted inverse. Also $\phi_a(a)=0$. $\square$

Worked Example: Checking an Automorphism and Its Inverse

Take $a=1/2$. Then $\phi_{1/2}(z)=(z-1/2)/(1-z/2)$, and the inverse formula gives $\phi_{1/2}^{-1}(w)=(w+1/2)/(1+w/2)$. For $z=0$, $\phi_{1/2}(0)=-1/2$, which lies in $\mathbb{D}$, and $\phi_{1/2}^{-1}(-1/2)=0$. For $w=1/3$, the inverse gives

\phi_{1/2}^{-1}(1/3) =\frac{1/3+1/2}{1+(1/2)(1/3)} =\frac{5/6}{7/6} =\frac{5}{7},

which is in $\mathbb{D}$. The general identity in the automorphism proof, not just these sample values, ensures that every point of the disk is mapped into the disk in both directions.

The Schwarz Lemma

The Schwarz lemma gives a size bound for holomorphic maps that fix $0$. Its central idea is to divide out the zero at the origin. The resulting holomorphic function can then be bounded using the Maximum Modulus Principle.

Theorem (Schwarz Lemma): Suppose $f:\mathbb{D}\to\mathbb{D}$ is holomorphic and $f(0)=0$. Then $|f(z)|\leq |z|$ for every $z\in\mathbb{D}$, and $|f'(0)|\leq1$. If equality holds in either inequality at a nonzero point or in the derivative bound, then $f(z)=cz$ for a constant $c$ with $|c|=1$.

Proof. Since $f(0)=0$, its Taylor expansion about $0$ has zero constant term. Thus $g(z)=f(z)/z$ for $z\neq0$ extends holomorphically to $\mathbb{D}$, with $g(0)=f'(0)$. Fix $r$ with $0<r<1$. On $|z|=r$, the fact that $f$ maps into $\mathbb{D}$ gives $|g(z)|=|f(z)|/r\leq1/r$. The Maximum Modulus Principle applied to the closed disk of radius $r$ therefore gives $|g(z)|\leq1/r$ whenever $|z|\leq r$. For any fixed $z\in\mathbb{D}$, take $r>|z|$ and let $r$ increase to $1$. It follows that $|g(z)|\leq1$, so $|f(z)|=|z||g(z)|\leq|z|$. At $z=0$, the derivative bound follows from $|g(0)|\leq1$, that is, $|f'(0)|\leq1$.

If equality holds at some $z\neq0$, then $|g(z)|=1$, and $g$ attains its maximum modulus at an interior point. The Maximum Modulus Principle implies that $g$ is constant, say $g(z)=c$, with $|c|=1$. Hence $f(z)=cz$. If instead $|f'(0)|=1$, then $g$ attains its maximum modulus at $0$, so the same principle gives the same conclusion. Conversely, $f(z)=cz$ with $|c|=1$ has equality in both bounds. $\square$

Worked Example: Applying the Schwarz Lemma to a Power

Let $f(z)=z^2$. This function maps $\mathbb{D}$ into $\mathbb{D}$ and satisfies $f(0)=0$, so the Schwarz lemma gives $|z^2|\leq|z|$. Directly, this is $|z|^2\leq|z|$, which holds because $0\leq|z|<1$. At $z=i/2$, the values are $f(i/2)=-1/4$, so $|f(i/2)|=1/4$, while $|i/2|=1/2$. The inequality is strict at this point, consistent with the equality characterization: $z^2$ is not of the form $cz$ with $|c|=1$.

From the Schwarz Lemma to Schwarz–Pick

A general holomorphic map need not fix the origin, and the point where we want an estimate need not be $0$. Apply $\phi_a$ to move the input point $a$ to $0$, and apply $\phi_{f(a)}$ to move its image to $0$. The resulting composition is a holomorphic disk map fixing $0$, so the Schwarz lemma applies.

Theorem (Schwarz–Pick Inequality): If $f:\mathbb{D}\to\mathbb{D}$ is holomorphic, then for all $a,z\in\mathbb{D}$, $$ \left|\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right| \leq \left|\frac{z-a}{1-\overline{a}z}\right|. $$ In particular, $$ |f'(a)|\frac{1-|a|^2}{1-|f(a)|^2}\leq1. $$

Proof. Fix $a\in\mathbb{D}$ and define $F=\phi_{f(a)}\circ f\circ\phi_a^{-1}$. Each factor maps $\mathbb{D}$ into itself, as established for the disk automorphisms above, so $F$ is a holomorphic map from $\mathbb{D}$ into itself. Also, $\phi_a^{-1}(0)=a$, and therefore $F(0)=\phi_{f(a)}(f(a))=0$. The Schwarz lemma gives $|F(w)|\leq|w|$ for every $w\in\mathbb{D}$. Taking $w=\phi_a(z)$, and using $\phi_a^{-1}(\phi_a(z))=z$, yields $|\phi_{f(a)}(f(z))|\leq|\phi_a(z)|$. Substituting the definition of each $\phi$ gives the stated inequality. Both denominators are nonzero: for example, $|\overline{f(a)}f(z)|<1$ because both values lie in $\mathbb{D}$.

For the derivative form, the chain rule gives $F'(0)=\phi_{f(a)}'(f(a))f'(a)(\phi_a^{-1})'(0)$. Differentiating the displayed formulas for the automorphisms gives $\phi_b'(b)=1/(1-|b|^2)$ for $b\in\mathbb{D}$, and $(\phi_a^{-1})'(0)=1-|a|^2$. Consequently,

F'(0)=f'(a)\frac{1-|a|^2}{1-|f(a)|^2}.

The Schwarz lemma gives $|F'(0)|\leq1$, which proves the derivative inequality. $\square$

Worked Example: A Derivative Bound for the Square Map

For $f(z)=z^2$ and $a=1/2$, we have $f(a)=1/4$ and $f'(a)=2a=1$. The derivative form of Schwarz–Pick gives

|f'(a)|\frac{1-|a|^2}{1-|f(a)|^2} =1\cdot\frac{1-1/4}{1-1/16} =\frac{3/4}{15/16} =\frac{4}{5}\leq1.

The direct pointwise inequality at $z=0$ gives the same type of check: \[ \left|\frac{f(0)-f(1/2)}{1-\overline{f(1/2)}f(0)}\right| =\frac{1}{4} \leq \frac{1}{2} =\left|\frac{0-1/2}{1-(1/2)0}\right|. \] Here $f(0)=0$, $f(1/2)=1/4$, and the denominators are $1$. This illustrates how Schwarz–Pick compares the disk distances between two inputs and their images.

Worked Example: Equality for a Disk Automorphism

Let $f(z)=\phi_{1/2}(z)=(z-1/2)/(1-z/2)$, and choose $a=1/2$, $z=0$. Since $f(1/2)=0$ and $f(0)=-1/2$, the left side of the Schwarz–Pick inequality is

\left|\frac{f(0)-f(1/2)}{1-\overline{f(1/2)}f(0)}\right| =\left|\frac{-1/2-0}{1-0}\right| =\frac{1}{2}.

The right side is \[ \left|\frac{0-1/2}{1-(1/2)0}\right|=\frac{1}{2}. \] Thus equality holds. This is consistent with the fact that disk automorphisms preserve the disk distance measured by the two fractions in the theorem. The calculations use the already verified automorphism property, so both the input and output points lie in $\mathbb{D}$.

Why the Normalization Works—and a Common Pitfall

The proof has a reusable structure: identify the point where the hypotheses of a known theorem are missing, transform the problem so those hypotheses hold, apply the theorem, and translate the result back. Here the Schwarz lemma only handles maps fixing $0$, so the two disk automorphisms normalize the chosen input and its image. Their derivatives then record exactly how the estimate changes when we undo that normalization.

A common pitfall is to solve for an inverse formula and immediately claim a bijection of the disk. The algebra establishes an inverse expression, but a bijection from $\mathbb{D}$ onto itself also requires that the expression map every $w\in\mathbb{D}$ back into $\mathbb{D}$. The strict positivity calculation in the automorphism proof verifies this missing point. Likewise, in Schwarz–Pick, the compositions are disk maps only because each factor has been shown to preserve the disk.

The derivative inequality can be rearranged as \[ |f'(a)|\leq\frac{1-|f(a)|^2}{1-|a|^2}. \] It bounds the derivative in terms of both the input location and the image location. A plain bound such as $|f'(a)|\leq1$ is generally not what Schwarz–Pick asserts away from the origin. The factors involving $a$ and $f(a)$ are essential consequences of the normalization.

Check Your Understanding

Use the definitions and proofs in this workshop to answer the following questions.

What identity shows that $\phi_a(z)$ lies in $\mathbb{D}$ when $a,z\in\mathbb{D}$?
Why does finding an algebraic inverse formula alone not prove that $\phi_a$ is a bijection from $\mathbb{D}$ onto itself?
In the Schwarz lemma proof, why does $g(z)=f(z)/z$ extend holomorphically to $z=0$?
Which two disk automorphisms are composed with $f$ to apply the Schwarz lemma at an arbitrary point $a$?
What additional factor appears in the derivative form of Schwarz–Pick, and why can it not generally be omitted?