Conformal map for regular polygon in circle.

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The discussion focuses on finding a conformal mapping that transforms the unit circle into a regular polygon with n vertices using complex functions. The author, Patrick, initially struggled with the complexity of the Schwarz-Christoffel transformation for arbitrary polygons, particularly for a polygon with 120 vertices. He discovered that the symmetry of a regular polygon simplifies the mapping, leading to a specific formula for the derivative of the mapping function. The derived formula, dw/dz = 1/(z^m - 1)^(2/m), integrates to produce a mapping that accurately reproduces the desired polygon shape. This approach demonstrates a successful method for achieving the mapping of the unit circle to a regular polygon.
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Hi All,

I'm looking for the conformal mapping (using complex functions) that maps the unit circle (or the upper half plane) into a REGULAR polygon with n vertices. I know the Schwarz-Christoffel transformation for an ARBITRARY polygon, but that doesn't help me because the expression is way too complex to be integrated (I'm trying to find the mapping for a polygon with 120 vertices). I was hoping that the fact that the polygon is REGULAR would simplify the problem. I used the mapping on the unit circle in the S-C transform because out of the symmetry of the problem, that allowed me (I would guess) to fix the unknown images of the vertices: they should also be on a regular polygon. But nevertheless, I cannot solve the integral beyond n = 4.
 
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Ok, I think I found it. For a regular m-polygon, the mapping between the unit circle (z) and the polygon (w) is given by:

dw/dz = 1/(z^m - 1)^(2/m)

which, according to Mathematica, integrates to:

z (1-z^m)^(2/m) (-1 + z^m)^(-2/m) Hypergeometric2F1[1/m, 2/m, 1+1/m,z^m]

and numerically this does indeed reproduce a polygon...

cheers,
Patrick.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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