Conformal maps, Möbius transformations, Schwarz lemma

In one sentence

Conformal means angle-preserving, which for holomorphic maps is exactly $f' \ne 0$; Möbius transformations are the maps of the sphere to itself and send circles-or-lines to circles-or-lines; and Schwarz's lemma pins down every map of the disc to itself.

Why the exam asks it

Because the geometry is concrete enough to be examined without heavy machinery, and because two facts get misremembered in predictable directions: that a line is a kind of circle once $\infty$ is included, and that Schwarz's lemma needs a fixed point at the origin.

The idea

Conformal maps

$f$ is conformal at $a$ if it preserves angles between curves through $a$, including orientation. For holomorphic $f$ this happens exactly where $f'(a) \ne 0$ — the derivative acts as multiplication by a complex number, which rotates by $\arg f'(a)$ and scales by $|f'(a)|$, and both operations leave angles alone.

Where $f'(a) = 0$ conformality fails in a specific way: if $f(z) - f(a)$ vanishes to order $m$, angles at $a$ are multiplied by $m$. At $z \mapsto z^2$ the origin sends a right angle to a straight one.

$\overline{z}$ preserves the magnitude of angles and reverses their sense; it is anticonformal, and not holomorphic. Options use it to test whether "preserves angles" was read as including orientation.

Möbius transformations

$$T(z) = \frac{az+b}{cz+d}, \qquad ad - bc \ne 0.$$

The condition $ad-bc\ne0$ is exactly what stops $T$ collapsing to a constant, and $T'(z) = \dfrac{ad-bc}{(cz+d)^2}$, so $T$ is conformal wherever it is defined.

Viewed on the Riemann sphere $\mathbb{C}\cup\{\infty\}$, $T$ is a bijection of the sphere: $T(-d/c) = \infty$ and $T(\infty) = a/c$. Composition corresponds to matrix multiplication of $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, so the Möbius group is $PGL_2(\mathbb{C})$ — the matrix is determined up to a scalar, which is why $2\times2$ determinant facts transfer directly.

Circles to circles. Every Möbius map is a composition of translations, rotation-scalings and the single inversion $1/z$; each of those sends circles-or-lines to circles-or-lines, so all of them do. On the sphere this is one statement, because a line is precisely a circle through $\infty$. That is why $1/z$ can turn a line into a circle: the line passed through $\infty$, the image does not.

Three points determine the map. Given distinct $z_1,z_2,z_3$ and $w_1,w_2,w_3$, there is exactly one Möbius map sending each $z_k$ to $w_k$. The cross-ratio is the invariant behind this, and it is preserved by every Möbius transformation.

Two maps worth knowing by name:

• Cayley, $\dfrac{z-i}{z+i}$: upper half plane onto the unit disc. It sends $0 \mapsto -1$, $1 \mapsto -i$, $\infty \mapsto 1$ — the real line onto the circle, as required.
• Disc automorphisms: every conformal bijection of the unit disc to itself is $$\varphi(z) = e^{i\theta}\,\frac{z-a}{1-\overline{a}z}, \qquad |a|<1,$$ which sends $a$ to $0$ and preserves $|z|=1$. Every one of them; there are no others.

Schwarz's lemma

$f : \mathbb{D}\to\mathbb{D}$ holomorphic with $f(0) = 0$. Then $|f(z)|\le|z|$ for all $z$, and $|f'(0)|\le1$. If equality holds in either at a single interior point, $f(z) = cz$ with $|c|=1$.

The proof is the maximum principle applied to $f(z)/z$, which has a removable singularity at $0$.

The rigidity is the examinable part: one equality anywhere forces the map to be a rotation everywhere. Nothing in real analysis behaves like that.

Without the fixed point at the origin the correct statement is Schwarz–Pick, which says every holomorphic self-map of the disc is non-expanding in the hyperbolic metric, with equality exactly for the automorphisms above.

Riemann mapping theorem

Every simply connected domain that is not all of $\mathbb{C}$ is conformally equivalent to the unit disc. Both exclusions matter: $\mathbb{C}$ itself is not (by Liouville, a bijection $\mathbb{C}\to\mathbb{D}$ would be bounded entire), and a domain with a hole is not, since conformal equivalence preserves simple connectedness.

Where intuition breaks

"Möbius maps send circles to circles and lines to lines." They mix the two. $1/z$ sends the line $\operatorname{Re}z = 1$ to a circle through the origin. The correct statement joins the classes: circles-or-lines to circles-or-lines.

"A conformal map preserves angles, so $\overline{z}$ is conformal." It reverses orientation. Conformal includes preserving the sense of the angle, and $\overline{z}$ is not holomorphic anyway.

"$f' \ne 0$ makes $f$ conformal, so it is injective." Locally conformal, not globally injective. $e^{z}$ again.

"Schwarz's lemma applies to any self-map of the disc." It needs $f(0)=0$. Without it, use Schwarz–Pick or pre-compose with an automorphism moving the relevant point to the origin — which is the standard trick and often the intended route.

"$|f'(0)| \le 1$, so $f$ is a rotation." Only if equality holds. Strict inequality says nothing beyond the bound.

"Every simply connected domain is conformally a disc." Except $\mathbb{C}$ itself. That exclusion is not a technicality; it is Liouville.

"$\mathbb{C}$ and $\mathbb{D}$ are conformally equivalent because both are simply connected." No — that is exactly the excluded case, and the reason the theorem has to name it.

The exam's angle

1. To build a Möbius map, use three points. Send three convenient points where they must go and the map is forced; no other conditions are needed.
2. Track $\infty$. Where $\infty$ goes, and what maps to it, usually determines the geometry — whether a line becomes a circle, and where the image is centred.
3. For self-maps of the disc, normalise to $0$ first. Compose with $\frac{z-a}{1-\overline a z}$ so Schwarz applies.
4. If an option asserts a bound at one point with equality, expect rigidity. Schwarz turns one equality into a complete description.
5. Check simple connectedness before invoking Riemann, and remember $\mathbb{C}$ is excluded.

The night before

• Conformal at $a$ $\iff$ holomorphic with $f'(a) \ne 0$. A zero of order $m$ multiplies angles by $m$.
• $\overline z$ is anticonformal: same angles, reversed orientation, not holomorphic.
• Möbius $\frac{az+b}{cz+d}$ with $ad-bc\ne0$; $T' = \frac{ad-bc}{(cz+d)^2}$; group is $PGL_2(\mathbb{C})$, composition is matrix multiplication.
• Circles-or-lines to circles-or-lines. A line is a circle through $\infty$.
• Three points determine a Möbius map; cross-ratio is the invariant.
• Cayley $\frac{z-i}{z+i}$: upper half plane $\to$ disc.
• Disc automorphisms are exactly $e^{i\theta}\frac{z-a}{1-\overline a z}$.
• Schwarz: $f(0)=0$, $f:\mathbb{D}\to\mathbb{D}$ $\Rightarrow |f(z)|\le|z|$, $|f'(0)|\le1$; equality anywhere forces a rotation. Without the fixed point, Schwarz–Pick.
• Riemann mapping: simply connected and not $\mathbb{C}$ $\Rightarrow$ conformally a disc.