[see https://www.geogebra.org/m/zwsenrug for details on why this is the pythagorean theorem]
This is adapted for complex numbers by replacing the arccos term in the sin(arccos()) function with it's complex counterpart (-i*ln(z-sqrt(z^2-1)). This gives the same general surface as the original function, but rotated slightly. I consider these equal only because the radial form of the same equation would (or can be) rotated as well, with the same absolute value. However, using just sqrt(a-b*i) and writing the complex form of the pythagorean identity does give you the exact same value complex number
