Solution

Let

J (z) = \frac{1}{z}

be reflection in the unit circle, and represent a Möbius transformation

g

M = (α γ β δ) .

A direct calculation shows that

J g J

is represented by

τ (M) = (\overline{δ} \overline{β} \overline{γ} \overline{α}) .

Thus

g J = J g

exactly when

M

and

τ (M)

differ by a nonzero scalar. Applying the conjugate-linear involution

τ

twice shows that this scalar has modulus one. Rescaling

M

by a suitable complex scalar then makes

M = τ (M)

. Consequently all commuting maps, and only those maps, have the form

g (z) = \frac{a z + b}{b z + a}, ∣ a ∣^{2} - ∣ b ∣^{2} \neq = 0.

For such a map,

∣ a z + b ∣^{2} - ∣ \overline{b} z + \overline{a} ∣^{2} = (∣ a ∣^{2} - ∣ b ∣^{2}) (∣ z ∣^{2} - 1) .

It follows that

∣ g (z) ∣ < 1

whenever

∣ z ∣ < 1

precisely when

∣ a ∣ > ∣ b ∣ .

The inverse has the same property, so these and only these maps preserve the unit disc.

Solved by gpt-5.6-sol high.

 Ancestors (11)