the distortion theorem

A conformal map magnifies the neighbourhood of each point by the local factor |f'(z)| — small disks near z become small disks magnified by that much. So a natural question for a normalized univalent map is: how unequal can this magnification be across the disk? Can one spot be stretched a thousand times more than another? The distortion theorem caps the disparity: at each radius the magnification |f'(z)| is squeezed between an explicit floor and ceiling.

Precisely, for f in the class S and |z| = r < 1, the distortion theorem states (1 - r)/(1 + r)^3 <= |f'(z)| <= (1 + r)/(1 - r)^3. The name 'distortion' refers exactly to this local magnification factor |f'|, the amount by which f distorts lengths near z. The proof is a clean piece of machinery: apply the second-coefficient bound |a_2| <= 2 not to f itself but to the Mobius-transplanted map g(w) = f((z + w)/(1 + z-bar w)) renormalized into S, read off a bound on f''(z)/f'(z), and integrate. Both bounds are sharp, with equality again given by rotations of the Koebe function.

Integrating the distortion bounds along a radius yields the growth theorem, so distortion is the more fundamental of the pair. Practically, the distortion theorem is what guarantees that a normalized conformal map does not develop arbitrarily thin or arbitrarily fat regions — vital for the stability of numerical conformal mapping and for estimates in potential theory. A caveat to keep honest: like the growth theorem, these constants are tied to the S-normalization f(0) = 0, f'(0) = 1; they bound |f'|, the size of the derivative, not its argument (the rotation part of the local map is unconstrained).