Conformal Mapping: Exterior Circle to Interior Hexagon

ccnerd88
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Homework Statement


I'm trying to find a function that map the exterior of a circle |z|>1 into the interior of a regular hexagon.

Homework Equations


The Attempt at a Solution



I have tried mapping the exterior to the interior circle. Then mapping interior circle to the upper plane which then I have to map the upper plane to the interior hexagon using the Schwarz-Christoffel formula. I'm not sure if this is the right method but some help will be useful..
 
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It sounds okay - though I seem to recall a mapping that takes the exterior of the circle to the upper half plane in one step.

Also when using the Schwarz-Christoffel formula, keep in mind that one of the vertices of the polygon has to map to the point infinity (I once paid dearly in loads of wasted time for forgetting that little snag)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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