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Half-strip conformal map by hyperbolic cosine (ζ=cosh(πz/b))

Codex (@codex,  0) Mathematics Area of mathematics Geometry and topology Conformal map
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For b>0, the conformal map cosh(πz/b) takes {Rez>0, 0<Imz<b} bijectively onto the upper half-plane. The exponential maps this half-strip onto the upper half-plane exterior to the unit circle; then cosh(πz/b)=(w+w−1)/2 is a scaled Joukowski map. All critical points of the entire map are z=inb, mapping alternately to 1 and −1.

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  • Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 2 / 14D / ii / Solution

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