to derive. We move to the next. Let $$X_i:=\sum\nolimits_{j>i}x_j$$ and note that $X_iz_i=q^2z_iX_i$. It follows by induction on $k\geq 1$ that $$\label{eq:zstarzi}
[z_i^*,z_i^k] = (1-q^{2k})\,z_i^{k-1}X_i \;.$$ For $k=1$ this is just , and from the algebraic identity $$\begin{aligned}
[z_i^*,z_i^{k+1}] &=
[z_i^*,z_i^k]z_i+z_i^k[z_i^*,z_i]
=(1-q^{2k})z_i^{k-1}X_iz_i+(1-q^2)z_i^kX_i
\\
&=(1-q^{2k})q^2z_i^kX_i+(1-q^2)z_i^kX_i
=(1-q^{2k+2})z_i^kX_i\end{aligned}$$ the inductive step follows. From and $[x_i,z_i^{k+1}]=z_i[z_i^*,z_i^{k+1}]$ we get for $k=p_i$.

For $k\geq 1$, it follows from that $$z_i^{k+1}(z_i^*)^{k+1}=
z_iz_i^*z_i^k(z_i^*)^k-z_i[z_i^*,z_i^k](z_i^*)^k=
\big\{x_i+(1-q^{-2k})X_i\big\}z_i^k(z_i^*)^k \;.$$ That is $$\label{eq:Yk}
Y_i(k)=\big\{x_i+(1-q^{-2(k-1)})X_i\big\}Y_i(k-1) \;,$$ where $Y_i(k):=z_i^k(z_i^*)^k$ if $k\geq 1$ and $Y_i(0)=1$. By iterated use of we find: $$Y_i(k)=\prod_{j=0}^{k-1}\big\{x_i+(1-q^{-2j})X_i\big\} \;.$$ This gives when $k=p_i$. Note that the order in the product does not matter, since it follows from that $x_i$ and $X_i$ commute.

Similarly, using the conjugate of : $$-[z_i,(z_i^*)^k] = (1-q^{2k})X_i(z_i^*)^{k-1} \;,$$ and $$(z_i^*)^{k+1}z_i^{k+1}=z_i^*z_i(z_i^*)^kz_i^k-z_i^*[z_i,(z_i^*)^k]z_i^k \;,$$ for $Z_i(k):=(z_i^*)^kz_i^k$, we find $$\begin{aligned}
Z_i(k+1)&=\Big\{z_i^*z_i+(1-q^{2k})q^2X_i\Big\}Z_i(k) \\
&=\Big\{x_i+(1-q^2)X_i+(1-q^{2k})q^2X_i\Big\}Z_i(k)
=\Big\{x_i+(1-q^{2k+2})X_i\Big\}Z_i(k) \;.\end{aligned}$$ By iterated use of this equation we arrive at: $$Z_i(k)=\prod_{j=1}^k\big\{x_i+(1-q^{2j})X_i\big\} \;.$$ This implies . Last relation is simply .

From and one also computes for all $i=0,\ldots,n$, the commutator: $$[\zeta_i^*,\zeta_i]=(q-q^{-1})\sum_{k=0}^{p_0}{\genfrac{[}{]}{0pt}{}{p_0}{k}_{\!q}}[\hspace{1pt}p_0k]_q
\left(-q\sum\nolimits_{j\geq i}x_j\right)^k
\left(\sum\nolimits_{j\geq i+1}x_j\right)^{p_0-k} \;,$$ where ${\genfrac{[}{]}{0pt}{}{p_0}{k}_{\!q}}$ is the $q$-binomial (cf. ). In the equations we separated a first group of relations, reducing to the property of the algebra being commutative when $q=1$, and a second group 