Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 8 1 iii Solution Created 2026-10-03 Updated 2026-10-14
Let , the inverse Cayley transform between the half-plane and disk. Direct calculation givessoOn the unit disc, choose the holomorphic square root of equal to one at zero. Both sides define nonvanishing holomorphic derivatives, and their quotient has constant square. The branch chosen in part ii gives . ConsequentlyAn affine change of image coordinate preserves conformality, so the lemniscatic integral maps the disc bijectively onto a square.
The normalization specifies the square exactly. Put . The substitution gives . The integral satisfies and , so its four corner values are . ThusIn particular the image is a square rotated through relative to the one in part ii, and its side length is . These symmetries also give from the affine identity.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 9 1 b Solution Created 2026-10-03 Updated 2026-10-14
The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius : the Poisson kernel on the circle lies between and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with ; let .
A harmonic conjugate on the simply connected unit disc makes holomorphic. The Carathéodory derivative bound for the right half-plane gives . The Cauchy-Riemann equations give , so . For a fixed , use the disk automorphismThe chain rule and conformality multiply the gradient length by . Applying the estimate at zero to proves the sharp gradient bound for positive harmonic functions:Sharpness follows from , which has equality at zero, and then composing with . These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.
