Past exam of the mathematics course of the University of Cambridge 2009 ib Paper 2 14D ii Solution Created 2026-09-24 Updated 2026-10-14
HereThis entire function vanishes exactly at , , and those points map to . The second derivative there is , so each is a simple critical point of the conformal map:Take the half-stripWriting , its imaginary part maps to . More precisely, maps bijectively onto , with inverse using arguments in . ThenPart (i) with , followed by positive real scaling, proves that this second map is a conformal bijection onto the upper half-plane. Their composition proves the required result and gives the half-strip conformal map by hyperbolic cosine.