If \(f: \DD_R(a)\to\DD_M(0)\) with \(f(a) = 0\), then \[ \abs{f(z)}\leq {M\over R}\abs{z-a} .\]
Solution
Set \(g(z) \da {f(Rz + a) \over M}\), then \(g: \DD\to \DD\) with \(g(0) = f(a)/M = 0\), so unwinding Schwarz yields \[ \abs{g(z)} \leq \abs{z} \implies \abs{f(Rz+a)\over M} &\leq \abs{z} \\ \implies \abs{f(Rz+a)} &\leq M\abs{z} \\ \implies \abs{f(w)} &\leq M\abs{w-a\over R} \qquad w = Rz+a\implies z={w-a\over R} .\]