An elliptical equation derived from circles
Change "n" to increase the "y" radius of the ellipse.
Additional Details for those Interested.
The main trick you do is notice that the expression for both e^x and e^xi can be rewritten as series of integrals. With e^x, they're all simple polynomials, but e^ix results in all the even integrals being real, and the odd ones being imaginary. So you take some positive integral of you're function, and add the integral of that multiplied by i. What I do next is completely neglect the real part because I can use f(1+xi) to find (or at the very least, estimate) all curves of f(x+yi). So I divide the imaginary by the real. Doing this for any even integral, and its following integral, will result in the same factor of variables regardless. Next, because in this case I have no variables to relate to the real axis anymore (because I only have "r"), I say the imaginary expression is the sin of some theta. Next just do inverse trig, derive, yadda yadda. Look, if you're interested enough to read this, I'd expect those sorts of things to be easy for you. Figuring the rest out isn't hard. In case you want to get the same solution, I used the second and third integrals of the AREA of a circle, those are the ones that will net you the same integers at least. The n is just added to make up for the fact I have no y in my equation.
Now, I have two axis, and I basically just threw out the x axis, and if you notice on the graph, the x diameter never changes. Coolest part is because this was solved from an expression for the imaginary portion, the equation only changes across the y axis... Even though it isn't complex, or even imaginary! Nor did I ever have to treat it as such!
If you want some interpretations, this is nothing more than a 2d circle rotating in 3d. Looks just like a complex number rotating over the w axis, basically implicating in some small way that complex numbers have the ability to relate higher dimensional objects to lower dimensional, correlate-able, objects. Sounds correct enough for me.