I have been working on transforming a series of different area circles to octagon using the Schwarz-Christoffel Transformation, hoping to preserve their areas while transitioning from circular to octagonal. I just want some confirmation about whether or not my equation for the derivative of the height to the top of the octagons is correct. Also, I made the assumption that Schwarz Christoffel transformations preserve area. Is this an assumption that I can make?
To explain my problem, I have a function S(x) which defines the radii of my original circles. I start with a unit circle to octagonal transformation: $$ z = wF\bigg(\frac{1}{8},\frac{1}{4},\frac{9}{8};w^8\bigg) $$ Then, I use a function $w(x) = R(x)$ to pick out certain concentric circle radii. Therefore to pick out the point at the "height" of the transformed shape, we use, $ w = R(x)\exp(i*\frac{\pi}{2}) = R(x)*i$, now giving us the equation $$ H(x) = R(x)F\bigg(\frac{1}{8},\frac{1}{4},\frac{9}{8};R(x)^8\bigg) $$ where H(x) is the height of the transformed octagons. However, in order to scale them so that the areas are the same, as my original circles, we have to multiply by a factor of $S(x)/R(x)$, making our final height equation $$ H(x) = S(x)F\bigg(\frac{1}{8},\frac{1}{4},\frac{9}{8};R(x)^8\bigg) $$ Now, using the fact that $$ \frac{dF(a,b,c;z)}{dz} = \frac{ab}{c}F(a+1,b+1,c+1;z) $$ we get $$ \frac{dH(x)}{dx} = \bigg[\frac{2}{9}F\bigg(\frac{9}{8},\frac{5}{4},\frac{17}{8};R(x)^8\bigg)\times R(x)^7\times R'(x)\times S(x)\bigg]\\+\bigg[F(\frac{1}{8},\frac{1}{4},\frac{9}{8};R(x)^8)\times S'(x)\bigg] $$