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Let $h\in \mathbb{R}[x,y]$ be a nonzero polynomial and define a plane curve in polar coordinates as $r(\theta) = h(\cos\theta,\sin\theta)$. For all the examples I've looked at, it seems like we can describe this curve as a zero set of a polynomial $f\in \mathbb{R}[x,y]$.

If we use the standard parametrization for the unit circle, we see that the curve above is essentially defined by

$$t\mapsto \left(h\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)\frac{1-t^2}{1+t^2},h\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)\frac{2t}{1+t^2}\right).$$

If the components were both polynomials, then it would be easy to show that we have a zero set of a polynomial. However, the components are rational expressions in terms of $t$. Is there a way to show that these points are the roots of some polynomial?

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2 Answers 2

Expand the parameter to two dimensions, $tx$ and $ty$. Use the complex variable $$z^2=(tx+ity)^2=tx^2-ty^2+2i(tx)(ty)$$ $$|z^2|=tx^2+ty^2$$ Normalize $$\frac{z^2}{|z^2|}= \frac{tx^2-ty^2}{tx^2+ty^2} +\frac{2i(tx)(ty)}{tx^2+ty^2}$$ Use the Pythagorean theorem and Euler's theorem. $$cos() \rightarrow \frac{tx^2-ty^2}{tx^2+ty^2}$$ $$sin() \rightarrow \frac{2(tx)(ty)}{tx^2+ty^2}$$ Let $tx->1$ and $ty->t$. $$cos() \rightarrow \frac{1-t^2}{1+t^2}$$ $$sin() \rightarrow \frac{2t}{1+t^2}$$ The first roots are $t=0$ for $sin()$ and $t=\pm 1$ for $cos()$, the same roots as the standard parameterization for the unit circle.

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Hi BJ Henderson. Welcome to MSE. You might want to consider using latex coding to make your responses more readable. In the current format the arguments are not very easy to follow. –  Thomas E. Dec 30 '12 at 5:05

If $h$ is of degree $d$, then the terms in the expansion $\big(r^d h(\cos\theta,\sin\theta)\big)^2$ are equivalent to either a monomial in $x$ and $y$, when the degree (of the sines and cosines) of the term is even or as $r$ times a monomial in $x$ and $y$, when the degree is odd. Let's denote the even degree terms $g_{even}(x,y)$ and the odd degree terms $r\,g_{odd}(x,y)$. The previous remark implies both $g_{even},g_{odd}$ are polynomials.

From $r=h((\cos\theta,\sin\theta)$, we deduce $r^{2d+2}=g_{even}(x,y)+r\,g_{odd}(x,y)$. Isolating the $g_{odd}$ term and squaring yields $$r^2 \left(g_{odd}(x,y)\right)^2=\left(r^{2d+2}+g_{even}(x,y)\right)^2$$ or $$(x^2+y^2) \left(g_{odd}(x,y)\right)^2=\left((x^2+y^2)^{d+1}+g_{even}(x,y)\right)^2$$ which is a polynomial equation.

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