draw[rotate=45]表示你把椭圆的线旋转了，但是切线没动，要想相对静止，应该考虑整体！

\documentclass[border=2pt]{standalone}
\usepackage{tikz}
\usepackage{xcolor}
\begin{document}
    \begin{tikzpicture}
    \foreach \N in {0,3,...,360}{
        \draw  (2,0) arc (0:\N:2 and 1)
            node[
                pos = 0.8,    % 设置切点在曲线上的位置
                sloped,    % 设置node按曲线斜率旋转
                anchor = south west,    % node的锚点设为左下角，即以此为切点
                fill opacity = 0.4,    % 将node填充为灰色半透明图形，与切线绘制无关
                fill = gray    % 将node填充为灰色半透明图形，与切线绘制无关
            ] (N) {};
        % 绘制切线（红色）
        \draw[red] (N.south west) -- (N.south east);
        % 绘制法线（蓝色）
        \draw[blue] (N.south west) -- (N.north west);
        }
    \end{tikzpicture}
\end{document} 

"[width=0.5textwidth]"改写成“[width=0.5textwidth]”或者“[width=0.5linewidth]”

因为这页你去掉页眉页脚了，去掉命令thispagestyle{empty}！line490,line495!

这个可以用pgfplots来实现，下面提供一个实现的方式（自己添加自己的图片）：

\documentclass{standalone}
\usepackage{tikz}
\usepackage{xcolor}
\usepackage{pgfplots}
\usetikzlibrary{spy}
\pgfplotsset{width=11cm,height=8cm}
\begin{document}

\begin{tikzpicture}[spy using outlines=
{circle, magnification=6, connect spies}]
\begin{axis}[no markers,grid=major,
hide axis,
every axis plot post/.append style={thick}]

\addplot graphics
[
includegraphics cmd=\pgfimage,
xmin=-5, xmax=90,
ymin=-0.5, ymax=1.1
] {back.pdf};

\coordinate (spypoint)
at (280,130);
\coordinate (magnifyglass) at (800,80);
\end{axis}
\spy [orange, size=2cm,thick] on (spypoint)
in node[fill=white] at (magnifyglass);
\end{tikzpicture}

\end{document}

坐标绘图的和矩形标记的：

\documentclass{standalone}
\usepackage{tikz}
\usepackage{xcolor}
\usepackage{pgfplots}
\usetikzlibrary{spy}
\pgfplotsset{width=11cm,height=8cm}
\begin{document}

\begin{tikzpicture}[spy using outlines=
{rectangle, magnification=6, connect spies}]
\begin{axis}[no markers,grid=major,
%hide axis,
every axis plot post/.append style={thick}]
\addplot
coordinates
{(0, 0) (0, 0.9) (1, 0.9) (2, 1) (3, 0.9) (80, 0)};
\addplot+ [line join=round] coordinates
{(0, 0) (0, 0.9) (2, 0.9) (3, 1) (4, 0.9) (80, 0)};
\addplot+ [line join=bevel] coordinates
{(0, 0) (0, 0.9) (3, 0.9) (4, 1) (5, 0.9) (80, 0)};
\addplot+ [miter limit=5] coordinates
{(0, 0) (0, 0.9) (4, 0.9) (5, 1) (6, 0.9) (80, 0)};

\coordinate (spypoint)
at (40,100);
\coordinate (magnifyglass) at (700,80);
\end{axis}
\spy [orange, size=2cm,thick] on (spypoint)
in node[fill=white] at (magnifyglass);
\end{tikzpicture}
\end{document}

这个可以用tikz直接绘制吧，下面是纯手工画的：

\documentclass[tikz,border=5pt]{standalone}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{tikz}
\usetikzlibrary{positioning,shadows,backgrounds}
\begin{document}

\newcommand\N[3]{\foreach \X in {0,...,#1}
 {\draw[draw=black,fill=black] (\X+#2,#3)circle (0.25cm);}}
\begin{tikzpicture}

\N{0}{3}{3.6}
\N{1}{2.5}{3}
\N{2}{2}{2.4}
\N{3}{1.5}{1.8}
\N{4}{1}{1.2}
\N{5}{0.5}{0.6}
\N{6}{0}{0}
\draw[line width=1pt,rounded corners=0.2cm](2.5,3.5)--(2.5,3.4)--(-0.3,0.06)--(-0.3,-3)
--(6.31,-3)--(6.31,0.06)--(3.6,3.3)--(3.6,3.6);
\foreach \M in {0,1,...,6}{\draw[line width=1pt](\M,0)--(\M, -2.8);}
\foreach \h in {1,2,...,6}{\draw (\h-0.5,-2.3) circle (0.25cm)node [inner sep=0.5pt]{\h};}
\draw[line width=1pt](0,-2.6)--(0,-2.8)--(6, -2.8)--(6, -2.6);
 
\end{tikzpicture}

\end{document}```
可以作为一个参考！

看了pgfplots宏包，感觉没有类似的，如果非要实现的话，也不是不行，去掉网线就会影响立体感！下面是一种组合形式的办法！若有大神更准确的绘制，请补充！

\documentclass[boder=5pt]{standalone}
\usepackage{tikz}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepackage{tikz-3dplot}

\begin{document}


\begin{tikzpicture}
\begin{axis}[axis lines=center,view={70}{60},
        ticks=none,
        enlargelimits=false,
        xmin=-3, xmax=6, ymin=-3, ymax=6, zmin=-80, zmax=100,
        xlabel=$x$, ylabel=$y$, zlabel=$z$,
        every axis x label/.style={
          at={(axis cs:6,0,0)},
          anchor=west,
        },
        every axis y label/.style={
          at={(axis cs:0,6,0)},
          anchor=west,
        },
        every axis z label/.style={
          at={(axis cs:0,0,100)},
          anchor=south,
        }]

\addplot3 [
surf,%white,
faceted color=%
%white,
orange,
samples=50,
domain=
0:4,
y domain=%0:4
0:4
] {x^2 - y^2};

\addplot3 [
surf,white,
faceted color=red,
samples=40,
domain=4:4.01,y domain=0:4
] {x^2 - y^2};

\addplot3 [
surf,white,
faceted color=red,
samples=20,
domain=0:4,y domain=4:4.01
] {x^2 - y^2};

\addplot3 [
surf,white,
faceted color=red,
samples=40,
domain=0:0.01,y domain=0:4
] {x^2 - y^2};
\addplot3[domain=0:4,samples y=0,red]{x^2};
\addplot3[domain=0:4,samples y=0,red]{x^2};
\end{axis}
\end{tikzpicture}

\end{document}

我试了几个效果：

$\mathop P^2\limits_{\substack{||\\\frac{E^2}{C^2}}}$

这个有点像是IEEE,直接用：

\section*{Nomenclature}

\subsection*{Indexes}

\medskip
\noindent
\begin{tabular}{@{}p{2.45cm}<{\raggedright}p{6.0cm}}
    $i$ & BalaBalaBalaBala. \\
    $j$ & BalaBalaBalaBala. \\
    $t$ & BalaBalaBalaBala. \\
\end{tabular}

就行了。

矩阵的尺寸不一样的时候看上去就是感觉很别扭；运行是没有问题的，但仍然不美观！

\documentclass{ctexart}
\usepackage{amsmath}
\begin{document}
\begin{align*}
    \begin{bmatrix}
        1 &x_{1} \\
        1 &x_{2} \\
        1 &x_{3} \\
        1 &x_{4} \\
        1 &x_{5}
    \end{bmatrix}
    \begin{bmatrix}
        a \\
        b
    \end{bmatrix}
    &=
    \begin{bmatrix}
        1/y_{1}^{2} \\
        1/y_{2}^{2} \\
        1/y_{3}^{2} \\
        1/y_{4}^{2} \\
        1/y_{5}^{2}
    \end{bmatrix}
    {\Rightarrow}
    \begin{bmatrix}
        1 &1 &1 &1 &1 \\
        x_{1} &x_{2} &x_{3} &x_{4} &x_{5}
    \end{bmatrix}
    \begin{bmatrix}
        1 &x_{1} \\
        1 &x_{2} \\
        1 &x_{3} \\
        1 &x_{4} \\
        1 &x_{5}
    \end{bmatrix}
    \begin{bmatrix}
        a \\
        b
    \end{bmatrix}\\
    &=
    \begin{bmatrix}
        1 &1 &1 &1 &1 \\
        x_{1} &x_{2} &x_{3} &x_{4} &x_{5}
    \end{bmatrix}
    \begin{bmatrix}
        1/y_{1}^{2} \\
        1/y_{2}^{2} \\
        1/y_{3}^{2} \\
        1/y_{4}^{2} \\
        1/y_{5}^{2}
    \end{bmatrix}\\
    &{\Rightarrow}
    \begin{bmatrix}
        5  &\sum_{i=1}^{5}x_{i} \\
        \sum_{i=1}^{5}x_{i} &\sum_{i=1}^{5}x_{i}^{2}
    \end{bmatrix}
    \begin{bmatrix}
        a \\
        b
    \end{bmatrix}\\
    &=
    \begin{bmatrix}
        \sum_{i=1}^{5}\frac{1}{y_{i}^{2}} \\
        \sum_{i=1}^{5}\frac{x_{i}}{y_{i}^{2}}
    \end{bmatrix}
    {\Rightarrow}
    \begin{bmatrix}
        5a + b\sum_{i=1}^{5}x_{i} \\
        a\sum_{i=1}^{5}x_{i} + b\sum_{i=1}^{5}x_{i}^{2}
    \end{bmatrix}\\
    &=
    \begin{bmatrix}
        \sum_{i=1}^{5}\frac{1}{y_{i}^{2}} \\
        \sum_{i=1}^{5}\frac{x_{i}}{y_{i}^{2}}
    \end{bmatrix}
\end{align*}
\end{document}

可以参看：https://latexstudio.net/index/details/index/mid/1372.html（nicematrix）宏包！

这样的确不规范！我没有想到全局设置！

这个可以修改一下王老师给的模板：

\documentclass{ctexart}

\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\tcbuselibrary{breakable,theorems,skins}
\definecolor{Back}{RGB}{70,87,121}
\newcounter{reidaibangou} 
\newtcolorbox{reidai}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--(interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}

\newcounter{reidaibangouA} 
\newtcolorbox{reidaiA}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--([xshift=0pt,yshift=-8pt]interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}

\newcounter{reidaibangouB}
\newtcolorbox{reidaiB}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=0pt,yshift=-8pt]interior.north west);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=-327pt,yshift=0pt]interior.south east);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}


\begin{document}
\begin{reidai}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以 

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidai}

\begin{reidaiA}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiA}


\begin{reidaiB}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiB}

\end{document} 

最后我们得到如下效果，当然也可以根据自己的要求随意修改。

稍作了修正：

\documentclass{ctexart}

\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\tcbuselibrary{breakable,theorems,skins}
\definecolor{Back}{RGB}{70,87,121}
\newcounter{reidaibangou} 
\newtcolorbox{reidai}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--(interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}

\newcounter{reidaibangouA} 
\newtcolorbox{reidaiA}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--([xshift=0pt,yshift=-16pt]interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}

\newcounter{reidaibangouB}
\newtcolorbox{reidaiB}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=0pt,yshift=-16pt]interior.north west);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=-327pt,yshift=0pt]interior.south east);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}


\begin{document}
\begin{reidai}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,
\tcblower 因为文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以 文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidai}

\begin{reidaiA}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiA}


\begin{reidaiB}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiB}


\end{document} 

没有具体函数图像的表达式，这里提供了一个大致图像：

\begin{tikzpicture}[xscale=3]
\draw [thick, ->] (-2,0) --(2,0) node[below]{$x$};
\draw [thick, ->] (0,-2) --(0,9) node[right]{$y$};
\draw [very thick,color=blue,domain=-1.6:1.3,samples=50,smooth,variable=\x] plot (\x,{2.5*sin(4*(\x+0.6) r)*exp((\x+0.6)/2.5)+3});
\draw [thick,color=purple,dashed](-1.6,{2.5*sin(4*(-1.6+0.6) r)*exp((-1.6+0.6)/2.5)+3})--(-1.6,0)node[below]{m};
\draw [thick,color=purple,dashed](1.3,{2.5*sin(4*(1.3+0.6) r)*exp((1.3+0.6)/2.5)+3})--(1.3,0)node[below]{n};
\draw [thick,color=magenta,dashed](-0.95,{2.5*sin(4*(-0.95+0.6) r)*exp((-0.95+0.6)/2.5)+3})--(-0.95,0)node[below]{a};
\draw [thick,color=magenta,dashed](-0.2,{2.5*sin(4*(-0.2+0.6) r)*exp((-0.2+0.6)/2.5)+3})--(-0.2,0)node[below]{b};
\draw [thick,color=orange,dashed](0.6,{2.5*sin(4*(0.6+0.6) r)*exp((0.6+0.6)/2.5)+3})--(0.6,0)node[above]{d};
\node[right=0.2cm,below=0.05cm] at (0,0){0};
\node[left=0.2cm,below=0.05cm] at (0.4,0){c};
\node[right=0.2cm,below=0.05cm] at (0.8,0){e};
\end{tikzpicture}

设置axis坐标环境，添加横坐标间距：“xtick distance=1,”即可

\documentclass{standalone}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{pgfplots}
%\pgfplotsset{compat=1.17}

\begin{document}

\pgfplotstableread{
x y error
CK-MB    -0.14654    0.02286
cTn-I    0.01203        0.06116
Myo        0.75388        0.16841
}\datatablemyo


\pgfplotstableread{
x y error
CK-MB    0.10632    0.07225
cTn-I    1.34    0.07211
Myo       -0.18961    0.06336
}\datatablectni


\pgfplotstableread{
x y error
CK-MB    0.89     0.01
cTn-I    0.03     0.04
Myo       -0.05     0.03
}\datatableckmb

\pgfplotsset{width=0.8\textwidth}

\begin{tikzpicture}
\begin{axis}[
    ybar=3pt,
    bar width=16pt,
    symbolic x coords={CK-MB,cTn-I,Myo},
    x tick label style={rotate=-30},
    xlabel=生物标志物,
    xtick distance=1,
    xlabel style={yshift=-4mm},
    ylabel=光谱漂移量（nm）,
    extra y ticks={0},
    extra y tick style={grid, grid style={black,thick}},
    ]
    \addplot+ [
    error bars/.cd,
    %y dir=both,
    %y explicit relative,
    ] table [x=x,y error=error] {\datatableckmb};
    \addplot+ [
    error bars/.cd,
    %y dir=both,
    %y explicit relative,
    ] table [x=x,y error=error] {\datatablectni};
    \addplot+ [
    error bars/.cd,
    %y dir=both,
    %y explicit relative,
    ] table [x=x,y error=error] {\datatablemyo};
    \legend{CK-MB,cTn-I,Myo}
  \end{axis}
\end{tikzpicture}


\end{document}

最后结果：