and then

= − 1 √ (b − a′·b − a′), dw= − m,
du 2n(u − b) √ (a − a·u − a′) duπu

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the formulas by which the conformal representation is obtained.

For the Ω polygon has a right angle at u = a, a′, and a zero angle at u = b, where θ changes from 0 to ½π/n and Ω increases by ½iπ/n; so that

= A, where A = √ (b − a·b − a′).
du (u − b) √ (u − a·u − a′)2n

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And the w polygon has a zero angle at u = 0, ∞, where ψ changes from 0 to m and back again, so that w changes by im, and

dw= B, where B = − m.
du uπ

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