分段函数格式

Question

Let $f : \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}$, $f(z) = z + \frac{1}{z}$. Given that

$f(z) = \begin{cases} 
\sqrt{2 + \sqrt{n}} &\text{if } z \in \mathbb{R} \setminus \lbrace 0\rbrace \\
\sqrt{1 + \frac{\sqrt{n}}{2}} + i \sqrt{-1 + \frac{\sqrt{n}}{2}} &\text{if } z \in \mathbb{C} \setminus \mathbb{R}
\end{cases}
 
for some positive integer $n$ and that $f(z^8) = 2023$ for all $z \in \mathbb{C} \setminus \lbrace 0\rbrace$, find $n$.

这是我的分段函数的 latex 代码，但输出只有一行，而不是两行。我怎样才能让它看起来像一个正常的分段函数？谢谢！

Answer 1

正如评论中提到的，您的方程式代码并未以 结尾$。添加它并扩展您的代码片段以完成小文档（MWE：最小工作示例），我得到了：

\documentclass{article}
\usepackage{amsmath, amssymb}

\begin{document}
Let $f : \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}$, $f(z) = z + \frac{1}{z}$. Given that

$
f(z) = \begin{cases}
\sqrt{2 + \sqrt{n}}     & \text{if } z \in \mathbb{R} \setminus \lbrace 0 \rbrace \\
\sqrt{1 + \frac{\sqrt{n}}{2}} + i \sqrt{-1 + \frac{\sqrt{n}}{2}}
                        & \text{if } z \in \mathbb{C} \setminus \mathbb{R}
        \end{cases}
$

for some positive integer $n$ and that $f(z^8) = 2023$ for all $z \in \mathbb{C} \setminus \lbrace 0\rbrace$, find $n$.
\end{document}

不过，我会把你的条件方程写成

\documentclass{article}
\usepackage{mathtools, amssymb}

\begin{document}

Let $f : \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}$, $f(z) = z + \frac{1}{z}$. Given that
\[
f(z) = \begin{dcases}
\sqrt{2 + \sqrt{n}}     & \text{if } z \in \mathbb{R} \setminus \lbrace 0 \rbrace \\
\sqrt{1 + \frac{\sqrt{n}}{2}} + i \sqrt{-1 + \frac{\sqrt{n}}{2}}
                        & \text{if } z \in \mathbb{C} \setminus \mathbb{R}
\end{dcases}
\]
for some positive integer $n$ and that $f(z^8) = 2023$ for all $z \in \mathbb{C} \setminus \lbrace 0\rbrace$, find $n$.
\end{document}