# Why is there no polynomial parametrization for the circle?

How does one show that the unit circle admits no polynomial parametrization? What is needed for this, are there general criteria? Thanks

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HINT $\rm\: \ f^{\:2} + g^2 = 1 \ \Rightarrow\ \gcd(f,g) = 1\:,\$ so $\rm\ f\ f\:' = - g\ g\:'\ \Rightarrow\ f\ |\ g'\ \Rightarrow f,\:g\:$ constant via $\rm\:\deg\: f\: =\: \deg\: g$

Alternatively $\rm\ \ (f-i\ g)\ (f+i\ g)\ =\ 1\ \Rightarrow\ f-i\ g\ =\ c,\ f+i\ g \ =\ d\$ where $\rm\:c\:,d\:$ are constants.

Therefore $\rm\ \ f\ =\ (c+d)/2,\ \ g \ =\ (c-d)\:i/2\$ are constants.

The generalization mentioned by Robert Israel goes back at least to Iyer, 1939, according to Ribenboim, $\:$ 13 Lectures on Fermat's Last Theorem, p. 265, $\:$ excerpted below.$\$ See also Burckel, $\$ An Introduction to Classical Complex Analysis, p. 433, where he proves additionally that $\rm\ f^n + g^n = 1\$ for entire $\rm\:f,\:g\:,\:$ and $\rm\: n > 2\ \Rightarrow\ f,\:g\:$ constant, using the Picard little theorem and the Open Map Theorem.

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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.

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It passes though the line at infinity twice (the homog. equation $x^2+y^2=z^2$ has 2 solutions with $z=0$ (up to rescaling): (1,i,0) and (1,-i,0)). Hence any rational parametrization $x(t),y(t)$ of the circle has to have 2 poles (that is, $x(t)$ has 2 poles and so does $y(t)$), so $x(t),y(t)$ can't be polynomials (as polynomials have only one pole, at $t=\infty$).

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