Recipe

Compute pseudo-hyperbolic distance and read what it implies

Use when: you have a point $a\in\mathbb D$ to recenter to $0$ with $\varphi_a$ , two points needing a pseudo-hyperbolic distance $\rho(z_1,z_2)$ , or an equality $\rho(f(z_1),f(z_2))=\rho(z_1,z_2)$ to interpret

Goal: a confirmed $\varphi_a(a)=0$ , a distance between $0$ and $1$ measuring how far apart two points sit inside $\mathbb D$ , and, when equality holds, the conclusion that $f$ is a disk automorphism everywhere

Taught in The Schwarz-Pick theorem

Quick recipe

Build $\varphi_a$ and confirm it sends $a$ to $0$
Compute the numerator $|z_1-z_2|$
Compute the denominator $|1-\bar z_1z_2|$
Divide to get $\rho(z_1,z_2)$
Compare against a second $\rho$ , or against equality

1
Write $\varphi_a(z)=\dfrac{z-a}{1-\bar az}$ for the given $a$ . Put in $z=a$ and simplify. The result must be exactly $0$ .

Example

Take $a=0.5+0.2i$ . Then $\varphi_a(a)=\dfrac{a-a}{1-\bar aa}=\dfrac{0}{1-|a|^2}$ . Here $|a|^2=0.5^2+0.2^2=0.29$ . So the bottom is $1-0.29=0.71$ , a positive real number. The whole fraction is $0$ .

Why: $\varphi_a$ only works as a recentering tool once you confirm it sends $a$ where it should. Do not assume it from the form.
2
Subtract the two points. Then take the modulus with $\sqrt{(\text{real part})^2+(\text{imaginary part})^2}$ .

Example

Take $z_1=0.5$ and $z_2=0.5i$ . Then $z_1-z_2=0.5-0.5i$ . So $|z_1-z_2|=\sqrt{0.5^2+0.5^2}=\sqrt{0.5}\approx0.707$ .

Why: this is the top of $\rho(z_1,z_2)=\left|\dfrac{z_1-z_2}{1-\bar z_1z_2}\right|$ . It is the plain Euclidean distance, before the bottom reweights it.
3
Conjugate $z_1$ by flipping the sign of its imaginary part only. Multiply by $z_2$ . Subtract the product from $1$ . Take the modulus.

Example

Here $\bar z_1=0.5$ , since $z_1$ is real. Then $\bar z_1z_2=0.5\cdot0.5i=0.25i$ . So $1-\bar z_1z_2=1-0.25i$ . Its modulus is $\sqrt{1^2+0.25^2}=\sqrt{1.0625}\approx1.031$ .

Why: this is the exact bottom of the distance formula. If you forget to conjugate $z_1$ , you compute a different quantity.
4
Divide the Step 2 number by the Step 3 number. The result should lie between $0$ and $1$ , since both points lie in $\mathbb D$ .

Example

$\rho(z_1,z_2)=\dfrac{0.707}{1.031}\approx0.686$ . That is a real distance strictly between $0$ and $1$ .

Why: a value outside $[0,1]$ signals an arithmetic error. The theorem keeps this quantity in that range for any two points of the disk.
5
If an analytic self-map $f$ is given, compare $\rho(z_1,z_2)$ with $\rho(f(z_1),f(z_2))$ .
- $\rho(f(z_1),f(z_2))=\rho(z_1,z_2)$ for even one pair $z_1\neq z_2$ — equality holds for every pair. So $f$ is a disk automorphism on all of $\mathbb D$ , not just at the tested pair.
- $\rho(f(z_1),f(z_2))<\rho(z_1,z_2)$ strictly — $f$ shrinks the distance at that pair. Nothing forces $f$ to be an automorphism.
Why: one value of $\rho$ is just a distance. The theorem's real content is in this comparison. It turns one number into a classification of $f$ .

Common mistakes

Forgetting to conjugate $z_1$ , not $z_2$ , before multiplying in Step 3.
Reporting a value of $\rho$ outside $[0,1]$ without hunting for the slip.
Deciding $f$ is "close to" an automorphism from one equal pair. The theorem is stronger: equality at one pair forces it everywhere.
Mixing up the number $\rho(z_1,z_2)$ with its role as a bound on how much any analytic self-map can shrink it.