I would like a formal proof using complex analysis that $g(x)=e^{\frac{1}{\log x}}$ is not an algebraic curve (in $\Bbb R^2$) I give my proposed proof by contradiction here:
Consider the isometry:
$$\Psi:(\zeta^2:=\Bbb R_+^* \times \Bbb R_+^*,g) \to (\Bbb R^2, \mathrm{can}) $$
$$ (u,v) \mapsto (\log u, \log v) $$
Consider the algebraic curve $f(x)=\frac{1}{x}$ with respect to $(\Bbb R^2, \mathrm{can})$ and consider the algebraic curve $g(x)=e^{\frac{1}{\log x}}$ with respect to $\zeta^2.$ By definition $f$ is a solution to the bivariate polynomial $p(x,y)=0.$ By definition $g$ is a solution to $e^{p(\log x, \log y)}=1.$ Suppose $g(x)$ is a solution to $p(x,y)=0.$ But this is absurd because by definition $g(x)$ is a solution to $e^{p(\log x, \log y)}=1.$ Suppose $f(x)$ is a solution to $e^{p(\log x, \log y)}=1.$ But this is absurd because by definition $f(x)$ is a solution to $p(x,y)=0.$
Therefore we reach a contradiction because neither $f$ nor $g$ can satisfy both $p(x,y)=0$ and $e^{p(\log x, \log y)}=1$ simultaneously. They can only satisfy one of the two definitions.
$\square$
Is my proof correct? What's the formal proof using complex analysis?