Drawing complex integration

Question

Could someone help me to draw this picture Picture 1.

but in point (0,0) to have small circle negative oriented, like here

Picture 2.

I am beginner with tikz and this looks to complicated for me. Any help is appreciated.

UPDATE. I used your solution (thank you very much for all) and result is next

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.markings}
\begin{document}

\begin{tikzpicture}
% Configurable parameters
\def\gap{0.2}
\def\bigradius{3}
\def\littleradius{0.5}

% Axes
\draw [help lines,->] (-1.25*\bigradius, 0) -- (1.25*\bigradius,0);
\draw [help lines,->] (0, -1.25*\bigradius) -- (0, 1.25*\bigradius);
% Red path
\draw[line width=1pt,   decoration={ markings,
  mark=at position 0.2455 with {\arrow[line width=1.2pt]{>}},
  mark=at position 0.765 with {\arrow[line width=1.2pt]{>}},
  mark=at position 0.87 with {\arrow[line width=1.2pt]{>}},
  mark=at position 0.97 with {\arrow[line width=1.2pt]{>}}},
  postaction={decorate}]
  let
     \n1 = {asin(\gap/2/\bigradius)},
     \n2 = {asin(\gap/2/\littleradius)}
  in (\n1:\bigradius) arc (\n1:360-\n1:\bigradius)
  -- (-\n2:\littleradius) arc (-\n2:-360+\n2:\littleradius)
  -- cycle;

% The labels
\node at (3.6,-0.2){$x$};
\node at (-0.24,3.53) {$iy$};
\node at (-0.6,0.43) {$\gamma_{\varepsilon}$};
\node at (-1.8,2.8) {$\gamma_{R}$};
\node at (1.9,0.29) {$l_1$};
\node at (1.555,-0.32) {$l_2$};

\end{tikzpicture}
\end{document}

My work

Thank you one more time for helping me.

Welcome to TeX.sx! You might want to have a look at: complicated paths for complex integration. It contains almost what you want. — Alexander, Mar 18 '13 at 23:46
@Alexander Thanks. I saw this before 30 min, but this isn't too helpful for me (or I am wrong?) — Cortizol, Mar 18 '13 at 23:49
Maybe you can explain a bit more what you have problems with and what you already tried. A minimal working example is especially appreciated here. — Alexander, Mar 18 '13 at 23:53
@Alexander Yes I know for that and I understand it. Like I said, I'm am totally beginner, but I will show my "misery" work tomorrow, because now I must go to bed (it's pretty late in my country). Good night :) — Cortizol, Mar 18 '13 at 23:57
@Alexander I update my post. What you think, to ask new question (I wrote it) here or to make new topic for that? — Cortizol, Mar 19 '13 at 16:24
It might be better to create a new question instead of modifying an already answered one. In your example just add fill=gray, so \draw[fill=gray,line width=1pt, decoration={ markings, to get the inside filled. — Alexander, Mar 19 '13 at 17:17
@Alexander Thank you for that. It works. I will delete my new question. — Cortizol, Mar 19 '13 at 19:03

JLDiaz · Accepted Answer · 2013-03-19T00:32:50.187

I guess the difficult part is the red path. Here is an idea: use polar coordinates and compute the starting angle for each arc using a little of trigonometry.

Update. Another difficult part would be to put the small arrows along the red path. This requires the library decorations.markings but the specification of the position of each arrow is tricky. You can specify it as a fraction of the total length of the path (so that 0 will be the beginning and 1 the end) but it is difficult to guess the right values for the intermediate points, or you can specify them as distances in mm from the beginning of the path. These can be computed for particular values of the radii, but it is also difficult to have a general solution.

I choose the "fraction of the length" solution, and after some trial and error I got the right values:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.markings}
\begin{document}

\begin{tikzpicture}
% Configurable parameters
\def\gap{0.2}
\def\bigradius{3}
\def\littleradius{0.5}

% Axes
\draw (-1.1*\bigradius, 0) -- (1.1*\bigradius,0)
      (0, -1.1*\bigradius) -- (0, 1.1*\bigradius);
% Red path
\draw[red, thick,   decoration={ markings,
      mark=at position 0.17 with {\arrow{latex}}, 
      mark=at position 0.53 with {\arrow{latex}},
      mark=at position 0.755 with {\arrow{latex}},  
      mark=at position 0.955 with {\arrow{latex}}}, 
      postaction={decorate}]  
  let
     \n1 = {asin(\gap/2/\bigradius)},
     \n2 = {asin(\gap/2/\littleradius)}
  in (\n1:\bigradius) arc (\n1:360-\n1:\bigradius)
  -- (-\n2:\littleradius) arc (-\n2:-360+\n2:\littleradius)
  -- cycle;
\end{tikzpicture}
\end{document}

Result

Edit: my initial answer used atan2(radius,gap/2) to find the angles, but I realized that the right formula should be asin(gap/2/radius). For such small angles the difference is almost unnoticeable, but it what is right, is right :-)

score 11 · Answer 2 · edited Apr 13 '17 at 12:35

With PSTricks.

enter image description here

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pstricks-add,pst-eucl}

\psset
{
        PointName=none,
        PointSymbol=none,
        linecap=1,
        linecolor=red,
}

\def\Radius{1.8}
\def\radius{0.5}
\def\Angle{2.5}

\begin{document}
\begin{pspicture}[showgrid=false](-3,-3)(3,3)
    \psaxes[labels=none,ticks=none,linecolor=gray]{->}(0,0)(-3,-3)(2.7,2.7)[$\textbf{Re}\, z$,-90][$\textbf{Im}\, z$,180]
    \pstGeonode 
        (0,0){O}
        (\Radius;\Angle){A}
        (\Radius;90){B}
        (\Radius;-90){C}
        (\Radius;-\Angle){D}
        (\radius;0|D){E}
        (\radius;0|A){F}
    \pstArcOAB[arrows=->,arcsepB=-3pt]{O}{A}{B}
    \pstArcOAB[arrows=->,arcsepB=-3pt]{O}{B}{C}
    \pstArcOAB{O}{C}{D}
    \pstLineAB[ArrowInside=->]{D}{E}
    \pstArcnOAB{O}{E}{F}
    \pstLineAB[ArrowInside=->]{F}{A}
    %labeling
    \psset{labelsep=2pt}
    \uput[45](\Radius;45){$\Gamma$}
    \uput[135](\radius;135){$\gamma$}
    \uput[-45](\Radius;0){$R$}
    \uput[-135](\Radius;180){$-R$}  
\end{pspicture}
\end{document}

Note that actually this kind of question has been answered here.

Animated version

enter image description here

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pstricks-add,pst-eucl}

\psset
{
        PointName=none,
        PointSymbol=none,
        linecap=1,
        linecolor=red,
}


\def\Radius{1.8}

\begin{document}
\multido{\nr=0.25+0.05,\na=1.0+1.0}{15}{%
\def\Angle{\na}
\def\radius{\nr}
\begin{pspicture}[showgrid=false](-3,-3)(3,3)
    \psaxes[labels=none,ticks=none,linecolor=gray]{->}(0,0)(-3,-3)(2.7,2.7)[$\textbf{Re}\, z$,-90][$\textbf{Im}\, z$,180]
    \pstGeonode 
        (0,0){O}
        (\Radius;\Angle){A}
        (\Radius;90){B}
        (\Radius;-90){C}
        (\Radius;-\Angle){D}
        (\radius;0|D){E}
        (\radius;0|A){F}
    \pstArcOAB[arrows=->,arcsepB=-3pt]{O}{A}{B}
    \pstArcOAB[arrows=->,arcsepB=-3pt]{O}{B}{C}
    \pstArcOAB{O}{C}{D}
    \pstLineAB[ArrowInside=->]{D}{E}
    \pstArcnOAB{O}{E}{F}
    \pstLineAB[ArrowInside=->]{F}{A}
    %labeling
    \psset{labelsep=3pt}
    \uput[45](\Radius;45){$\Gamma$}
    \uput[135](\radius;135){$\gamma$}
    \uput[-45](\Radius;0){$R$}
    \uput[-135](\Radius;180){$-R$}  
\end{pspicture}}
\end{document}

Warning:

A special feature of ArrowInside is that it will not produce an arrow when <- (and its variants such as |<-) is assigned to it! In other words, -> (and its variants such as ->|) is the only working option.

[$\textbf{Re}, z$,-90][$\textbf{Im}, z$,180] What does -90 and 180 relative to? I change this value, the text place changes, but I don't know why. — an offer can't refuse, Jun 16 '15 at 13:09

Drawing complex integration

2 Answers2

Animated version

Warning:

Linked

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