Problems with putting $r=e^{-\theta \cot(\theta)}$ into rectangular form Recently, I found a polar curve which contains all(?) complex numbers $z$, which satisfy $\Im(z^z)=0$; or in other words, all complex numbers which raised to itself give a real number.
The curve is described by the equation $r=e^{-\theta \cot(\theta)}$. The following image shows the graph of the equation for $0\leq \theta\leq 2\pi$:

The function has infinite amount of lines going away into the infinity, as $\theta$ is increased below $0$ or above $2\pi$. As far as I have been able to find, only the section $0\leq \theta\leq 2\pi$ actually satisfies $\Im(re^{\,i\theta\, re^{\,i\theta}})=\Im(z^z)=0$. I may be wrong, because I haven't looked at this closely, but that's not the important part to the question.
It is easy to observe, that the curves formed on the graph form an interesting curve, one which would be nice to analyze in the rectangular form. However, the farthest I have been able to convert between systems is this:
$$x^2+y^2=e^{-{2x\over y}\arctan\big({y\over x}\big)}$$
There are however many problems with this equation:
1. The rectangular equation is lossy
When I graphed the resultant rectangular equation, not surprisingly I found that the graph is only true for the first quadrant, the rest were mere artifact reflections of it:

Now, I'll admit that that's a beautiful eye, but this is not really what I wanted to get. ;)
2. Equation is (at least in my effort so far) insolvable for $x$ or $y$
Not as big of a problem, but it would be nice to solve for one of the variables.
Most importantly I want to fix problem number 1. I feel like I should know how to fix this issue, but I'm having a huge brainfart. Can anyone point me in the right direction?
EDIT: I have considered by now, that the nature of $\arctan$ or the square root may cause this problem, but I have not been able to somehow change the expression to one that views the second quadrant correctly at the least.
 A: I explored further the idea, that the $\arctan$ may be causing said problems. I figured that since the codomain of inverse tangent is $Y=\{y:-{\pi\over 2}<y\leq {\pi\over 2}\}$, and since $\tan(x+\pi)=\tan x$, then all values from the second quadrant, should actually spit out values of respective angles from the fourth quadrant (quadrant that is $\pi$ away from the second quadrant).
So I went on, and subtracted $\pi$ from the value of inverse tangent in the rectangular formula, to get the resulting angle from the second quadrant to be converted into the fourth quadrant. This is the result:

So, as can be seen, the new (blue) formula, made it possible to produce a reflection of the second quadrant in the first quadrant.
I, then, negated the argument of the arctangent, and negated the whole exponent, to reflect the curve along the y axis, and I was able to get an overlap over the actual curve for the polar equation:

Notice, that the blue curve also provided correct resemblance of the actual third quadrant. That is because the third quadrant is also $\pi$ bigger than the first quadrant, which falls in the codomain of the inverse tangent. Hence, subtracting $\pi$ from it, also resulted in the correct curve.
Subtracting $2\pi$ instead of $\pi$ also corrected the fourth quadrant:

Putting this all together, I have managed to scramble a piecewise rectangular definition of the curve ($0\leq\theta\leq 2\pi$):
$$C:
\begin{cases}
 x^2+y^2=e^{{2x\over y}\arctan\big(-{y\over x}\big)}&\text{if}\, 0\,\leq\theta\leq {\pi\over 2}\\
 x^2+y^2=e^{{2x\over y}\big(\arctan\big(-{y\over x}\big)-\pi\big)}&\text{if}\, {\pi\over 2}\,\leq\theta\leq{3\pi\over 2}\\
 x^2+y^2=e^{{2x\over y}\big(\arctan\big(-{y\over x}\big)-2\pi\big)}&\text{if}\, {3\pi\over 2}\,\leq\theta\leq 2\pi\\
\end{cases}$$
In general, subtracting $\pi$ successively "unlocks" the next two quadrants of the curve.
So, I have figured out the answer to my own question :)
P.S. What will happen to the bounty I set for this question now?
