This is what has been used to make the more recent activities. This works for every equation, with a few caveats. The full "theorem" or whatever you'd like to call it, essentially says that f(x,n)=(x+i*f(x,n))*g(y,n). In most equations I use y^(1/n-1) since it relates directly to every subsequent simple x^n, and makes really nice-looking planes that still support the expression. However, g(y,n) is best the more closely related it is to f(x,n). In other words, the "g(y,n)" in the activity is essentially just chosen, for the most part, because it keeps the graphs clean. If I had, say, n^(x+y*i), then we are already given g(y,n) as (cos(yln(n))+i*sin(yln(n)). Real equations relate to polynomials, imaginary equations are radial, and complex equations are products of both. Also, it is important to note that f(x,n) does not equal g(y), y, or g(y,n). It is specifically either equal to z (an initial complex number) multiplied by g(y,n), or a complex function of x (in the form of (x+i*f(x,n))*g(y,n).
Lastly, you can actually derive and integrate these. Just ensure to only integrate/derive the imaginary part by "y", and the real part by "x". In the final equation given, that just turns a circle into an ellipse with the same imaginary intercepts. Nothing here is groundbreaking, but rather to just point out an extremely simple relationship that can be used to solve equations where a real function is equal to a complex, or imaginary one. Honestly, that sounds useless, but you never know.
