= Solution
Write $x,y$ for the two coordinates. The PDF gives $\mu=e^{\pi i/3}$, a primitive sixth root of unity. The matrices satisfy
$$
\sigma^6=1,\qquad \tau^2=-I=\sigma^3,\qquad
\tau\sigma\tau^{-1}=\sigma^{-1}.
$$
These relations reduce every word to $\sigma^j$ or $\tau\sigma^j$, $0\le j<6$. The six first matrices are diagonal and distinct, and the six second matrices are antidiagonal and distinct. Thus \b[$G$ has exactly twelve elements]; it is a <binary dihedral group>.
First calculate the <polynomial invariant ring> of $\langle\sigma\rangle$. A monomial $x^a y^b$ is invariant exactly when $a-b$ is divisible by $6$. Removing the smaller exponent shows that it is a product of powers of
$$
A=x^6,\qquad B=y^6,\qquad C=xy.
$$
The sole relation is $AB=C^6$. Indeed, reduction by this relation leaves monomials $C^jA^a$ and $C^jB^b$ with $b>0$, whose images in $\mathbb C[x,y]$ have distinct exponent pairs. Hence
$$
\mathbb C[x,y]^{\langle\sigma\rangle}
\cong\mathbb C[A,B,C]/(AB-C^6).
$$
On these generators, $\tau$ interchanges $A,B$ and sends $C$ to $-C$. Put $S=A+B$ and $T=A-B$. Since $2$ is invertible, the preceding <ring> is
$$
\mathbb C[S,C,T]/(T^2-S^2+4C^6).
$$
Every element has a unique form $P(S,C)+TQ(S,C)$. The induced involution fixes $S$ and negates both $C$ and $T$. Its invariants are therefore exactly the expressions
$$
P_0(S,C^2)+CTQ_0(S,C^2).
$$
Set
$$
U=x^6+y^6,\qquad V=x^2y^2,\qquad W=xy(x^6-y^6).
$$
We obtain the relation
$$
W^2=V(U^2-4V^3).
$$
The unique normal form above also proves that there are no further relations: $P_0(U,V)+WQ_0(U,V)=0$ forces both polynomials to vanish. Thus the <polynomial invariant ring> is
$$
\boxed{\mathbb C[x,y]^G\cong
\mathbb C[U,V,W]/(W^2-VU^2+4V^4).}
$$
For completeness, the <algebraic quotient by a finite group> is the <affine variety> with this <coordinate ring>. Each element $h$ of $\mathbb C[x,y]$ satisfies the monic orbit polynomial $\prod_{g\in G}(Z-g h)$ with invariant coefficients, so the quotient map is a <finite morphism>. Invariants separate distinct finite orbits: interpolate a <polynomial> taking value $0$ on one orbit and $1$ on the other, and average it over $G$. Thus its fibres are precisely the orbits. The defining polynomial and resulting <algebraic quotient by a finite group> are
$$
\boxed{p(U,V,W)=W^2-VU^2+4V^4,\qquad
\mathbb A^2/G\cong V(p)\subset\mathbb A^3.}
$$
Using the converted TeX's fourth root would instead give a different group and a different <binary dihedral invariant hypersurface>; the sixth root from the original PDF is essential.
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