Let $r \in \mathbb{N}-\{0\}$.

**Commutative case:**
Let $f : (x,y) \mapsto (p,q)$ be a map from $\mathbb{C}[x,y]$ to $\mathbb{C}[x^{1/r},x^{-1/r},y]$ satisfying the following two conditions:

**(i)** $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$.

**(ii)** $l_{1,-1}(p)=\lambda x^{\frac{n}{r}+1}y$ and $l_{1,-1}(q)=\mu x^{-\frac{n}{r}}$, where $n \in \mathbb{Z}-\{0\}$ and $\lambda,\mu \in \mathbb{C}-\{0\}$.

Then we can write $p=\lambda x^{\frac{n}{r}+1}y + A$ and $q=\mu x^{-\frac{n}{r}}+B$, where $v_{1,-1}(A) < \frac{n}{r}$ and $v_{1,-1}(B) < -\frac{n}{r}$.

Question 1:Is necessarily $A \in k[x^{1/r},x^{-1/r}]$ and $B=0$?

**Non-commutative case:**

Let $A_1(\mathbb{C})$ be the first Weyl algebra, namely, the associative non-commutative $k$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$. Let $A_1(\mathbb{C})^{-\frac{1}{r}}$ be a ring extension of the first Weyl algebra, generated by $x^{\frac{1}{r}}$, $x^{-\frac{1}{r}}$ and $y$ subject to the relations $yx-xy=1$, $x (x^{-\frac{1}{r}})^r= (x^{-\frac{1}{r}})^r x=1$, and $[y,x^{-\frac{1}{r}}]:=-\frac{1}{r} x^{-\frac{1}{r}-1}$.

Let $f : (x,y) \mapsto (p,q)$ be a map from $A_1(\mathbb{C})$ to $A_1(\mathbb{C})^{-\frac{1}{r}}$ satisfying the following two conditions:

**(i)** $[q,p]:=qp-pq \in \mathbb{C}-\{0\}$.

**(ii)** $l_{1,-1}(p)=\lambda x^{\frac{n}{r}+1}y$ and $l_{1,-1}(q)=\mu x^{-\frac{n}{r}}$, where $n \in \mathbb{Z}-\{0\}$ and $\lambda,\mu \in \mathbb{C}-\{0\}$.

Then we can write $p=\lambda x^{\frac{n}{r}+1}y + A$ and $q=\mu x^{-\frac{n}{r}}+B$, where $v_{1,-1}(A) < \frac{n}{r}$ and $v_{1,-1}(B) < -\frac{n}{r}$.

Question 2:Is necessarily $A \in k[x^{1/r},x^{-1/r}]$ and $B=0$?

**Remark:** A special case of the above question can be found here (actually, the later edit of that question is the same as the above question).

Thank you veru much!