以下代码：
In order to find $A_{0}^1$:
\begin{equation}
A_{0}^{1} = \lim_{z \rightarrow -1}\dfrac{d^0}{dz^0}\Bigg[\dfrac{(z+1)z}{(z+1)(z-2)}\Bigg]=\lim_{z \rightarrow -1}\dfrac{z}{(z-2)}=\dfrac{1}{3}
\end{equation}
And in order to find $A_{0}^2$:
\begin{equation}
A_{0}^2 = \lim_{z \rightarrow 2} \dfrac{d^0}{dz^0}\Bigg[\dfrac{(z-2)z}{(z+1)(z-2)}\Bigg]=\lim_{z \rightarrow 2}\dfrac{z}{(z+1)}=\dfrac{2}{3}
\end{equation}
From (38) we get:
\begin{equation}
\dfrac{1}{3}\Bigg[\dfrac{1}{z+1}\Bigg]+\dfrac{2}{3}\Bigg[\dfrac{1}{z-2}\Bigg]
\end{equation}
where we have that $\Bigg[\dfrac{1}{z+1}\Bigg]$ can be rewritten as $\dfrac{1}{z}\dfrac{1}{1-\Big(-\dfrac{1}{z}\Big)}$
给出以下内容：
以下是我的序言：
\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{amsfonts}
\onehalfspacing
\DeclareMathOperator{\Log}{Log}
\DeclareMathOperator{\Arg}{Arg}
\begin{document}
\title{Homework Chapter 5}
\author{}
\maketitle
如何才能使最后一行重写后显示的术语更加成比例？
答案1
我会用1-(-z^{-1})。
我还做了一些修复：
- 删除了几个多余的括号；
\Bigg太大而错误，应该是\biggl和\biggr；\Big同样也是错误的和不必要的；- 交叉引用应该使用
\eqref而不是明确的数字。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
In order to find $A_{0}^1$
\begin{equation}\label{A01}
A_{0}^{1}
= \lim_{z \rightarrow -1}\frac{d^0}{dz^0}\biggl[\frac{(z+1)z}{(z+1)(z-2)}\biggr]
= \lim_{z \rightarrow -1}\frac{z}{z-2}=\frac{1}{3}
\end{equation}
And in order to find $A_{0}^2$
\begin{equation}
A_{0}^2
= \lim_{z \rightarrow 2} \frac{d^0}{dz^0}\biggl[\frac{(z-2)z}{(z+1)(z-2)}\biggr]
= \lim_{z \rightarrow 2} \frac{z}{z+1}=\frac{2}{3}
\end{equation}
From~\eqref{A01} we get
\begin{equation}
\frac{1}{3}\biggl[\frac{1}{z+1}\biggr]+\frac{2}{3}\biggl[\frac{1}{z-2}\biggr]
\end{equation}
where we have that $\dfrac{1}{z+1}$ can be rewritten as
$\dfrac{1}{z}\dfrac{1}{1-(-z^{-1})}$.
\end{document}
