Let \(D=\{z\in\mathbb C:|z|<1\}\) and \(H=\{w\in\mathbb C:\operatorname{Im}w>0\}\). Let \(f:D\to H\) be holomorphic.
Suppose that \(f ( 0 ) = 3 i\) . Find the maximal possible value of \(\left| f ^ { \prime } ( 0 ) \right|\)
Solution
The maximum is \(\boxed{6}\).
1. Schwarz’s lemma gives \(|f'(0)|\leq6\).
Define \[ g(z)=\frac{f(z)-3i}{f(z)+3i}. \] The denominator is nonzero since \(f(z)\in H\). For every \(w\in H\), the identity \[ |w+3i|^2-|w-3i|^2=12\operatorname{Im}w>0 \] shows that the same fractional transformation has modulus less than one. Thus \(g:D\to D\) is holomorphic and \(g(0)=0\). Differentiation at zero yields \[ g'(0)=\frac{6i}{(f(0)+3i)^2}f'(0)=\frac{f'(0)}{6i}. \] Schwarz’s lemma gives \(|g'(0)|\leq1\), hence the asserted bound [@SS03].
2. Explicit maps attain the bound.
For \(|\lambda|=1\), set \[ f_\lambda(z)=3i\frac{1+\lambda z}{1-\lambda z}. \] Its denominator has no zero in \(D\), and direct computation gives \[ \operatorname{Im}f_\lambda(z) =3\frac{1-|z|^2}{|1-\lambda z|^2}>0. \] Hence \(f_\lambda:D\to H\) is holomorphic, with \(f_\lambda(0)=3i\) and \(f_\lambda'(0)=6i\lambda\). Its derivative has modulus six, so the upper bound is a maximum. Moreover, equality in step 1 forces \(g(z)=\lambda z\) by the equality case of Schwarz’s lemma [@SS03]; inversion gives precisely the displayed maps.