使用 \underbrace 对齐

Question

我想将四个方程的右侧放在彼此之下，因此 e^(...) 应该在每一列中彼此之下。由于 ，我无法理解这一点\underbrace。

\documentclass{scrartcl}

\usepackage{amsmath}

\begin{document}
\begin{equation}\label{eqn:4}
\begin{alignedat}{4}
&\eta(z+1)^{4}&&=e^{\frac{\pi i}{3}}\eta(z)^{4}, &\quad\quad&\phantom{~}& \eta(2(z+1))^{8}&=e^{\frac{4\pi i}{3}}\eta(2z)^{8},\\
&\eta(3(z+1))^{4}&&=\underbrace{e^{\pi i}}_{=-1}\eta(3z)^{4}, &&\phantom{~}& \eta(6(z+1))^{8}&=\underbrace{e^{4\pi i}}_{=1}\eta(6z)^{8}.
\end{alignedat}
\end{equation}
\end{document}

Answer 1

您可以使用eqparbox，并进行一些调整：

\documentclass{scrartcl}

\usepackage{amsmath}
\usepackage{eqparbox}                   

\newcommand{\matheqparbox}[2]{%
  \eqparbox[t]{#1}{\hfil$\displaystyle#2$\hfil}%
}

\begin{document}
\begin{equation}\label{eqn:4}
\begin{alignedat}{4}
&\eta(z+1)^{4}&&=\matheqparbox{1}{e^{\frac{\pi i}{3}}}\eta(z)^{4},
  &\quad\quad&
  & \eta(2(z+1))^{8}&=\matheqparbox{2}{e^{\frac{4\pi i}{3}}}\eta(2z)^{8},\\
&\eta(3(z+1))^{4}&&=\matheqparbox{1}{\underbrace{e^{\pi i}}_{=-1}\,}\eta(3z)^{4},
  &&
  & \eta(6(z+1))^{8}&=\matheqparbox{2}{\underbrace{e^{4\pi i}}_{=1}\,}\eta(6z)^{8}.
\end{alignedat}
\end{equation}
\end{document}

该命令\eqparbox需要一个标签来计算具有相同标签的框的最大宽度。这里我使用了1和2，但它们可以是任何（唯一）字符串。

另一方面，我更喜欢等号前右对齐：

\documentclass{scrartcl}

\usepackage{amsmath}
\usepackage{eqparbox}

\newcommand{\matheqparbox}[2]{%
  \eqparbox[t]{#1}{\hfil$\displaystyle#2$\hfil}%
}

\begin{document}
\begin{equation}\label{eqn:4}
\begin{aligned}
\eta(z+1)^{4}&=\matheqparbox{1}{e^{\frac{\pi i}{3}}}\eta(z)^{4},
  & \eta(2(z+1))^{8}&=\matheqparbox{2}{e^{\frac{4\pi i}{3}}}\eta(2z)^{8},\\
\eta(3(z+1))^{4}&=\matheqparbox{1}{\underbrace{e^{\pi i}}_{=-1}\,}\eta(3z)^{4},
  & \eta(6(z+1))^{8}&=\matheqparbox{2}{\underbrace{e^{4\pi i}}_{=1}\,}\eta(6z)^{8}.
\end{aligned}
\end{equation}
\end{document}

Answer 2

为了获得最大的灵活性，您可能需要使用array环境。在下面的解决方案中，我假设您希望术语e^{...}在行间居中。如果您更喜欢左对齐或右对齐，请随意切换C到 或。LR

\documentclass{scrartcl}
\usepackage{amsmath}  % for general math typesetting support
\usepackage{array}    % for "\newcolumntype" macro
\newcolumntype{C}{>{\displaystyle}c}
\newcolumntype{L}{>{\displaystyle}l}
\newcolumntype{R}{>{\displaystyle}r}

\begin{document}
\begin{equation}\label{eqn:4}
\setlength\arraycolsep{0pt}
\renewcommand\arraystretch{1.5}
\begin{array}{ R@{{}={}}C@{\kern1pt}L @{\hspace{2em}} 
               R@{{}={}}C@{\kern1pt}L }
\eta(z+1)^{4}    & e^{\frac{\pi i}{3}}          & \eta(z)^{4},  &
\eta(2(z+1))^{8} & e^{\frac{4\pi i}{3}}         & \eta(2z)^{8}, \\
\eta(3(z+1))^{4} & \underbrace{e^{\pi i}}_{=-1} & \eta(3z)^{4}, &
\eta(6(z+1))^{8} & \underbrace{e^{4\pi i}}_{=1} & \eta(6z)^{8}.
\end{array}
\end{equation}
\end{document}

Answer 3

\documentclass{scrartcl}
\usepackage{amsmath}

\begin{document}
\begin{align}\label{eqn:4}
\begin{aligned}
\eta(z+1)^{4}   &= e^{\frac{\pi i}{3}} \eta(z)^{4},    & 
\eta(2(z+1))^{8}&= e^{\frac{4\pi i}{3}}\eta(2z)^{8}, \\
\eta(3(z+1))^{4}&= \mkern-5mu\underbrace{e^{\pi i}}_{=-1} \eta(3z)^{4},   &
\eta(6(z+1))^{8}&= \underbrace{e^{4\pi i}}_{=1} \eta(6z)^{8}.
\end{aligned}
\end{align}
\end{document}