Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 4 / c

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.

Either equality condition says that the holomorphic function $h$ from part (i) has modulus one at an interior point: at zero in the derivative case, or at the given nonzero point in the other case. Since $|h|\le 1$ everywhere, the maximum modulus principle makes $h$ a constant $w$ with $|w|=1$. ThereforeThese maps are rotations of the unit disk and do achieve both equalities, so this also identifies all equality cases of the Schwarz lemma.