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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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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.2012-12-30