我需要创建以下内容：
我的代码：
\begin{enumerate}[(i).]
\item $\begin{aligned}[t]
\sigma(0,\alpha,1,z;q)&+\frac{z}{\alpha(1+z)}\mu(0,\alpha,1,z;q)=\frac{1}{(1-\alpha q)}\frac{(1/\alpha z^6q;q^2)_{\infty} (q^2z^2\alpha;q)_{\infty}}{(-1/z^3,-1/\alpha q z^3,q^3z^3\alpha;q)_{\infty}}\\
&\times\sum_{r=-\infty}^{\infty}\frac{(1-\alpha z^3q^{2r+2})(-zq;q)_r(\alpha z^6q^3;q^2)_r(-1)^rz^{2r}\alpha^{2r}q^{(3r^2+5r+2)/2}}{(1-\alpha z^3 q^2)(\alpha^2q^3;q^2)_r(-z^2\alpha q^2;q)_r},
\end{aligned}$
\item $\begin{aligned}[t]\phi(0,\alpha,1,z;q)&+\frac{1+\alpha}{(\alpha)}\phi_{-}(0,\alpha,1,z;q)=\frac{(-1/ z^3,-q^2/z^3,qz^2\alpha;q^2)_{\infty}}{(\alpha q/z^3,q^2/\alpha z^3,q^2z^3\alpha;q^2)_{\infty}}\\
&\times\sum_{r=-\infty}^{\infty}\frac{(1-\alpha z^3q^{4r})(zq;q^2)_r(-z^3q,-z^3;q^2)_rz^{2r}\alpha^{2r}(-1)^rq^{3r^2-r}}{(1-\alpha z^3 )(-\alpha q;q)_{2r}(z^2\alpha q;q^2)_r},
\end{aligned}$
\item $\begin{aligned}[t]\psi(0,\alpha,1,z;q)&+\frac{1+\alpha}{\alpha^2}\psi_{-}(0,\alpha,1,z;q)=\frac{1}{(1+\alpha q)}\frac{(-1/z^3,-q/z^3,q^3z^2\alpha;q^2)_{\infty}}{(q\alpha/z^3,1/\alpha z^3,q^4z^3\alpha;q^2)_{\infty}}\\
&\times\sum_{r=-\infty}^{\infty}\frac{(1-\alpha z^3q^{4r+2})(zq,- z^3q^2,-z^3q;q^2)_rz^{2r}\alpha^{3r}(-1)^rq^{3r^2+3r+1}}{(1-\alpha z^3 q^2)(-\alpha q^2,-\alpha q^3,zq^3\alpha;q^2)_r},
\end{aligned}$
\item $\begin{aligned}[t]\rho(0,\alpha,1,z;q)&+\frac{z}{\alpha(1+z)}\lambda(0,\alpha,1,z;q)=\frac{1}{(1-\alpha q)}\frac{(q/\alpha z^6;q^2)_{\infty} (-qz^2\alpha;q)_{\infty}}{(-q/ z^3,1/\alpha z^3,q^2z^3\alpha;q)_{\infty}}\\
&\sum_{r=-\infty}^{\infty}\frac{(1-\alpha z^3q^{2r+1})(-zq;q)_r(z^6\alpha q;q^2)_rz^{2r}\alpha^{2r}q^{(3r^2+3r)/2}}{(1-\alpha z^3 q)(\alpha q^3;q^2)_r(-z^2q\alpha;q)_r}.
\end{aligned}$
\end{enumerate}
答案1
如果编号列表中的每个项目都只包含一个显示的方程式（可能覆盖两行或更多行），我认为用项目编号标记方程式是一种可疑的印刷做法和方程编号。选择一个标记系统或另一个标记系统，但并非两者兼而有之。应该不惜一切代价地避免无意义的冗余，不是吗？
因此，下面我将展示两种不同的解决方法。第一种方法采用enumerate使用小写罗马数字的环境，但不创建方程编号。第二种方法采用普通方程编号，但不创建环境enumerate。
无论哪种方式，我认为有必要为每个方程式创建一个额外的换行符，以避免使某些数学材料突出到右边距。
\documentclass{article}
\usepackage{mathtools,amsthm,enumitem}
\newtheorem{thm}{Theorem}[section]
\allowdisplaybreaks
\begin{document}
\setcounter{section}{3} % just for this example
\begin{thm} \hspace*{\fill} % force a line break
\begin{enumerate}[label=\textnormal{(\roman*)}]
\item $\begin{aligned}[t] \sigma&(0,\alpha,1,z;q)
+\frac{z}{\alpha(1+z)}\mu(0,\alpha,1,z;q)\\
&=\frac{1}{(1-\alpha q)}\frac{(1/\alpha z^6q;q^2)^{}_{\infty}
(q^2z^2\alpha;q)^{}_{\infty}}{(-1/z^3,-1/\alpha q z^3,q^3z^3\alpha;q)^{}_{\infty}}\\
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}
\frac{(1-\alpha z^3q^{2r+2})(-zq;q)^{}_r(\alpha z^6q^3;q^2)^{}_r
(-1)^rz^{2r}\alpha^{2r}q^{(3r^2+5r+2)/2}}{%
(1-\alpha z^3 q^2)(\alpha^2q^3;q^2)^{}_r(-z^2\alpha q^2;q)^{}_r},
\end{aligned}$
\item $\begin{aligned}[t] \phi&(0,\alpha,1,z;q)
+\frac{1+\alpha}{(\alpha)}\phi^{}_{-}(0,\alpha,1,z;q)\\
&=\frac{(-1/ z^3,-q^2/z^3,qz^2\alpha;q^2)^{}_{\infty}}{
(\alpha q/z^3,q^2/\alpha z^3,q^2z^3\alpha;q^2)^{}_{\infty}}\\
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}
\frac{(1-\alpha z^3q^{4r})(zq;q^2)^{}_r(-z^3q,-z^3;q^2)^{}_r
z^{2r}\alpha^{2r}(-1)^rq^{3r^2-r}}{(1-\alpha z^3 )(-\alpha q;q)_{2r}
(z^2\alpha q;q^2)^{}_r},
\end{aligned}$
\item $\begin{aligned}[t] \psi&(0,\alpha,1,z;q)
+\frac{1+\alpha}{\alpha^2}\psi^{}_{-}(0,\alpha,1,z;q)\\
&=\frac{1}{(1+\alpha q)}\frac{(-1/z^3,-q/z^3,q^3z^2\alpha;q^2)^{}_{\infty}}{
(q\alpha/z^3,1/\alpha z^3,q^4z^3\alpha;q^2)^{}_{\infty}}\\
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}\frac{(1-\alpha z^3q^{4r+2})
(zq,- z^3q^2,-z^3q;q^2)^{}_rz^{2r}\alpha^{3r}(-1)^rq^{3r^2+3r+1}}{%
(1-\alpha z^3 q^2)(-\alpha q^2,-\alpha q^3,zq^3\alpha;q^2)^{}_r},
\end{aligned}$
\item $\begin{aligned}[t] \rho&(0,\alpha,1,z;q)
+\frac{z}{\alpha(1+z)}\lambda(0,\alpha,1,z;q)\\
&=\frac{1}{(1-\alpha q)}\frac{(q/\alpha z^6;q^2)^{}_{\infty}
(-qz^2\alpha;q)^{}_{\infty}}{(-q/ z^3,1/\alpha z^3,q^2z^3\alpha;q)^{}_{\infty}}\\
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}\frac{(1-\alpha z^3q^{2r+1})
(-zq;q)^{}_r(z^6\alpha q;q^2)^{}_rz^{2r}\alpha^{2r}q^{(3r^2+3r)/2}}{%
(1-\alpha z^3 q)(\alpha q^3;q^2)^{}_r(-z^2q\alpha;q)^{}_r}.
\end{aligned}$
\end{enumerate}
\end{thm}
\clearpage
\begin{thm}
\begin{align}
\sigma&(0,\alpha,1,z;q)
+\frac{z}{\alpha(1+z)}\mu(0,\alpha,1,z;q)\notag \\*
&=\frac{1}{(1-\alpha q)}\frac{(1/\alpha z^6q;q^2)^{}_{\infty}
(q^2z^2\alpha;q)^{}_{\infty}}{(-1/z^3,-1/\alpha q z^3,q^3z^3\alpha;q)^{}_{\infty}}\notag \\*
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}
\frac{(1-\alpha z^3q^{2r+2})(-zq;q)^{}_r(\alpha z^6q^3;q^2)^{}_r
(-1)^rz^{2r}\alpha^{2r}q^{(3r^2+5r+2)/2}}{%
(1-\alpha z^3 q^2)(\alpha^2q^3;q^2)^{}_r(-z^2\alpha q^2;q)^{}_r}, \\[2ex]
\phi&(0,\alpha,1,z;q)
+\frac{1+\alpha}{(\alpha)}\phi^{}_{-}(0,\alpha,1,z;q)\notag \\*
&=\frac{(-1/ z^3,-q^2/z^3,qz^2\alpha;q^2)^{}_{\infty}}{
(\alpha q/z^3,q^2/\alpha z^3,q^2z^3\alpha;q^2)^{}_{\infty}}\notag \\*
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}
\frac{(1-\alpha z^3q^{4r})(zq;q^2)^{}_r(-z^3q,-z^3;q^2)^{}_r
z^{2r}\alpha^{2r}(-1)^rq^{3r^2-r}}{(1-\alpha z^3 )(-\alpha q;q)_{2r}
(z^2\alpha q;q^2)^{}_r},\\[2ex]
\psi&(0,\alpha,1,z;q)
+\frac{1+\alpha}{\alpha^2}\psi^{}_{-}(0,\alpha,1,z;q)\notag \\*
&=\frac{1}{(1+\alpha q)}\frac{(-1/z^3,-q/z^3,q^3z^2\alpha;q^2)^{}_{\infty}}{
(q\alpha/z^3,1/\alpha z^3,q^4z^3\alpha;q^2)^{}_{\infty}}\notag \\*
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}\frac{(1-\alpha z^3q^{4r+2})
(zq,- z^3q^2,-z^3q;q^2)^{}_rz^{2r}\alpha^{3r}(-1)^rq^{3r^2+3r+1}}{%
(1-\alpha z^3 q^2)(-\alpha q^2,-\alpha q^3,zq^3\alpha;q^2)^{}_r}, \\[2ex]
\rho&(0,\alpha,1,z;q)
+\frac{z}{\alpha(1+z)}\lambda(0,\alpha,1,z;q)\notag \\*
&=\frac{1}{(1-\alpha q)}\frac{(q/\alpha z^6;q^2)^{}_{\infty}
(-qz^2\alpha;q)^{}_{\infty}}{(-q/ z^3,1/\alpha z^3,q^2z^3\alpha;q)^{}_{\infty}}\notag \\*
&\quad\times\sum_{\mathclap{r=-\infty}}^{\infty}\frac{(1-\alpha z^3q^{2r+1})
(-zq;q)^{}_r(z^6\alpha q;q^2)^{}_rz^{2r}\alpha^{2r}q^{(3r^2+3r)/2}}{%
(1-\alpha z^3 q)(\alpha q^3;q^2)^{}_r(-z^2q\alpha;q)^{}_r}.
\end{align}
\end{thm}
\end{document}