这是我目前拥有的代码：
\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{graphicx}
\usepackage{setspace}
\onehalfspacing
\DeclareMathOperator{\Log}{Log}
\begin{document}
\title{Homework Chapter 5}
\author{}
\maketitle
\section*{Section 5.2: 4, 5, and 7}
\paragraph{4.}
Let $\alpha$ be a complex number. Show that if $(1+z)^{\alpha}$ is taken
as $e^{\alpha \Log(1+z)}$, then for $|z|< 1$
\begin{equation*}
(1+z)^{\alpha} = 1 + \frac{\alpha}{1}z +
\frac{\alpha(\alpha -1)}{1\cdot2}z^{2} +
\frac{\alpha(\alpha-1)(\alpha-2)}{1\cdot2\cdot3}z^{3} + \dotsb
\end{equation*}
In general,
\begin{equation}
\frac{d^{j}}{dz^{j}}(1+z_0)^{\alpha}=
\frac{\alpha! \, (1+z_0)^{\alpha - j}}{(\alpha - j)!}
\end{equation}
We can prove result (1) by induction. For our base case, we let $j=1$. Hence,
\begin{equation}
\frac{d}{dz}(1+z_0)^{\alpha} = \alpha (1+z_0)^{\alpha - 1}
\end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
\begin{equation}
\frac{d^{k}}{dz^{k}}(1+z_0)^{\alpha}=
\frac{\alpha! \, (1+z_0)^{\alpha - k}}{(\alpha - k)!}
\end{equation}
And so,
\begin{equation}
\frac{d}{dz}\Big[\frac{d^{k}}{dz^{k}}(1+z_0)^{\alpha}\Big] = \frac{d}{dz}\Big[\frac{\alpha! \, (1+z_0)^{\alpha - k}}{(\alpha - k)!}\Big] = \frac{\alpha !\, (1+z_0)^{\alpha -k - 1}}{(\alpha -k -1)!} = \frac{\alpha !\, (1+z_0)^{\alpha-(k+1)}}{[\alpha - (k+1)]!}
\end{equation}
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
Expanding (1), we have
\begin{equation}
\frac{\alpha! \, (1+z_0)^{\alpha - j}}{(\alpha - j)!} = \frac{\alpha (\alpha-1) \dots (\alpha-j)! \, (1+z_0)^{\alpha - j}}{(a-j)!} = \alpha (\alpha - 1) \dots (\alpha - j + 1) (1+z_0)^{\alpha - j}.
\end{equation}
The Taylor series for $e^{\alpha \Log(1+z)}$ about $z_0=0$ is thus
\begin{equation}
1 + \sum_{j=1}^{\infty} \frac{\alpha \, (\alpha - 1) \dots (\alpha - j + 1)}{j!}z^{j}
\end{equation}
\end{document}
如何在
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED
和
Expanding (1), we have
答案1
我认为没有必要在“Expanding (1), we have”上方创建额外的垂直空白。段落分隔符（缩进字符串“Expanding (1), we have”）应该足以让读者注意到动作中的“中断”。
\dots但是，请用\dotsb或替换 的两个实例\cdots。
\documentclass{article}
\usepackage[hmargin=1.5in, top=0.5in]{geometry}
%%\usepackage{amsmath}
\usepackage{mathtools} % loads "amsmath" automatically
\DeclareMathOperator{\Log}{Log}
\usepackage{amsthm,graphicx}
\usepackage{setspace}
\onehalfspacing
\begin{document}
\title{Homework Chapter 5}
\author{}
\maketitle
\section*{Section 5.2: 4, 5, and 7}
\paragraph{4.}
Let $\alpha$ be a complex number. Show that if $(1+z)^{\alpha}$ is taken
as $e^{\alpha \Log(1+z)}$, then for $|z|< 1$,
\begin{equation*}
(1+z)^{\alpha} = 1 + \frac{\alpha}{1}z +
\frac{\alpha(\alpha -1)}{1\cdot2}z^{2} +
\frac{\alpha(\alpha-1)(\alpha-2)}{1\cdot2\cdot3}z^{3} + \dotsb
\end{equation*}
In general,
\begin{equation} \label{eq:1}
\frac{d^{j}}{dz^{j}}(1+z_0)^{\alpha}=
\frac{\alpha! \, (1+z_0)^{\alpha - j}}{(\alpha - j)!}
\end{equation}
% A blank line creates a paragraph break:
We can prove result \eqref{eq:1} by induction. For our base case,
we let $j=1$. Hence,
\begin{equation}
\frac{d}{dz}(1+z_0)^{\alpha} = \alpha (1+z_0)^{\alpha - 1}
\end{equation}
Now for our induction hypothesis, we let $j=k$ and assume that
\begin{equation}\label{eq:3}
\frac{d^{k}}{dz^{k}}(1+z_0)^{\alpha}=
\frac{\alpha! \, (1+z_0)^{\alpha - k}}{(\alpha - k)!}\,.
\end{equation}
And so,
\begin{equation}
\frac{d}{dz}\Bigl[\frac{d^{k}}{dz^{k}}(1+z_0)^{\alpha}\Bigr]
= \frac{d}{dz}\Bigl[\frac{\alpha! \, (1+z_0)^{\alpha - k}}{(\alpha - k)!}\Bigr]
= \frac{\alpha !\, (1+z_0)^{\alpha -k - 1}}{(\alpha -k -1)!}
= \frac{\alpha !\, (1+z_0)^{\alpha-(k+1)}}{[\alpha - (k+1)]!}
\end{equation}
Since \eqref{eq:3} holds for $j = k + 1$, it follows that \eqref{eq:1}
holds for all positive integers~$j$. QED
% A blank line creates a paragraph break:
Expanding \eqref{eq:1}, we have
\begin{equation}
\frac{\alpha! \, (1+z_0)^{\alpha - j}}{(\alpha - j)!}
= \frac{\alpha (\alpha-1) \dotsb (\alpha-j)! \, (1+z_0)^{\alpha - j}}{(a-j)!}
= \alpha (\alpha - 1) \dotsb (\alpha - j + 1) (1+z_0)^{\alpha - j}.
\end{equation}
The Taylor series for $e^{\alpha \Log(1+z)}$ about $z_0=0$ is thus
\begin{equation}
1 + \sum_{j=1}^{\infty} \frac{\alpha \, (\alpha - 1) \dotsb (\alpha - j + 1)}{j!}z^{j}
\end{equation}
\end{document}