 Solution

The Schwarz lemma states that a holomorphic function $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$ satisfiesIf equality holds in the first inequality at any nonzero point, or in the derivative inequality, then $f(z)=\lambda z$ for a constant $|\lambda |=1$. Conversely those rotations give equality.

To prove it, the function $g(z)=f(z)/z$ extends holomorphically at zero with value $f^{\prime }(0)$, by the power series of $f$. On $|z|=r<1$, $|g(z)|\le 1/r$. The maximum modulus principle gives this same bound inside that circle. For each fixed $z$, let $r\uparrow 1$ to conclude $|g(z)|\le 1$. This proves both inequalities. Equality in either makes $g$ attain its maximum modulus at an interior point, so the maximum modulus principle makes $g$ a constant of modulus one. This proves the equality cases as well.

For the classification of automorphisms of the unit disk, choose $a\in \mathbb{D}$ and defineDirect expansion givesThus $T_a$ is a biholomorphism of the disk and sends $a$ to zero. If $F$ is any conformal map of the disk onto itself, take $a=F^{-1}(0)$ and put $h=F\circ T_a^{-1}$. Both $h$ and $h^{-1}$ fix zero. The Schwarz lemma gives $|h(z)|\le |z|$ and, applied to $h^{-1}$ at $h(z)$, $|z|\le |h(z)|$. Equality holds, so $h(z)=e^{i\theta }z$. Consequently all the holomorphic conformal automorphisms are preciselyEvery displayed map is indeed a conformal automorphism by the verified inverse and a rotation. Here conformal maps have the usual holomorphic, orientation-preserving meaning; if orientation reversal is allowed, their complex conjugates give the antiholomorphic alternatives.