TITLE: Why is there no polynomial parametrization for the circle?
QUESTION [16 upvotes]: How does one show that the unit circle admits no polynomial parametrization?
What is needed for this, are there general criteria?
Thanks

REPLY [8 votes]: More generally, suppose $f$ and $g$ are analytic functions on a simply-connected domain $U$ with $f^2 + g^2 = 1$, i.e. 
$(f+ig)(f-ig) = 1$, so $f + i g$ is an analytic function in $U$ that is never 0.  This implies $f + i g = e^h$ for some function $h$ analytic in $U$.  Now $f - i g = 1/(f + i g) = e^{-h}$.  We then have $f = (e^h + e^{-h})/2 = \cos(h)$ and $g = (e^h - e^{-h})/(2 i) = \sin(h)$.  Conversely, of course, $f = \cos(h)$ and $g = \sin(h)$ satisfy the equation $f^2 + g^2 = 1$ for any function $h$.  
Now take the domain to be $\mathbb C$; it's easy to show that for any nonconstant function $h$ analytic in a neighbourhood of $\infty$, $\sin(h)$ and $\cos(h)$ have either removable or essential singularities at $\infty$, and thus can't be polynomials.