答案1
欢迎!本网站上有不少类似的图表。现在又多了一个。可以找到与intersections
库的交点、与decorations.pathreplacing
(和calligraphy
) 的括号以及与库的直角符号angles
。
\documentclass[tikz,border=3mm]{standalone}
\usepackage{dsfont}
\usetikzlibrary{angles,decorations.pathreplacing,calligraphy,intersections}
\begin{document}
\begin{tikzpicture}[mbrace/.style={decorate,decoration={calligraphic brace,
amplitude=7pt,raise=1ex},thick},
dot/.style={circle,fill=red,inner sep=0pt,minimum size=2.5pt}]
\draw (-3,0) node[left]{$\mathds{C}$} -- (4.5,0);
\draw[name path=circ] (0,0) node[dot,label=below:{$0$}] (O){}
circle[radius=2.5cm];
\draw[name path=line] (0,2.5) node[dot,label=above:{$\zeta$}] (zeta) {}
-- (4,0)node[dot,label=above:{$z$}] (z) {};
\draw[name intersections={of=circ and line}] (intersection-2) node[dot,label=above
right:{$z^*$}](z*){} (z*) -- (O|-z*) node[dot](h){}
(O) -- (zeta) (z*) edge[dashed] (O-|z*) (O-|z*)node[dot](z'){} ;
\draw[mbrace,decorate] (O) -- node[left=2ex] {$1$} (zeta);
\draw[mbrace,decorate] (O-|z*) -- node[left=2ex] {$x_3$} (z*);
\path[angle radius=4pt] pic[draw]{right angle = z'--O--zeta}
pic[draw]{right angle = z*--h--zeta}
pic[draw]{right angle = z--z'--z*};
\end{tikzpicture}
\end{document}