Mystery curve found here looks like this :
Was given by the complex formula : $$e^{it} – \frac{e^{6it}}{2} + i \frac{e^{-14it}}{3} $$
Is the parametric form simpler or the polar form would be simpler (leaving the definition of simpler open in terms of : to be any easier form to remember or what ever open interpretation that would make a nicer formula)
This is indeed a lovely curve, but there is nothing mysterious about it.
By all means leave it in the given parametric form $$\gamma:\quad t\mapsto z(t):=e^{it}-{e^{6it}\over2}+ i{e^{-14it}\over3}\qquad(0\leq t\leq 2\pi)\ ,$$ or written in the form $t\mapsto\bigl(x(t),y(t)\bigr)$ by separating real and imaginary parts.
What is surprising at first glance is the fivefold symmetry of $\gamma$, given that the number $5$ does appear nowhere in its definition. About this one can say the following: Any trigonometric polynomial $$f(t):=\sum_{k=-N}^N c_k\>e^{ikt},\qquad c_k\in{\mathbb C},$$ represents a smooth closed curve in the complex plane. When the right side contains only terms with $k=5n+1$, $\>n\in{\mathbb Z}$, then the function $f$ has the property $$f\left(t+{2\pi\over5}\right)\equiv e^{2\pi i/5}\>f(t)\ ,\tag{1}$$ because $$e^{i(5n+1)(t+2\pi/5)}\equiv e^{2\pi i/5}\>e^{i(5n+1)t}\ .$$ The functional equation $(1)$ at once gives rise to the fivefold symmetry of $\gamma$ we observe.