Plotting the branches of a complex function











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I'm trying to plot a kind of Riemann's surface of a function (I'm not sure if it's the right name for the thing), as shown below:



Desired result



Here's what I tried:



r = (0:1:15)';                           % create a matrix of complex inputs
theta = pi*(-1:0.05:1);
z = r*exp(1i*theta);
w = z.^(1/2) ; % calculate the complex outputs

figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
subplot(121)

surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
xlabel('Real(z)')
ylabel('Imag(z)')
zlabel('Real(u)')
cb = colorbar;
colormap jet; % gradient from blue to red
cb.Label.String = 'Imag(v)';

subplot(122)
surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
xlabel('Real(z)')
ylabel('Imag(z)')
zlabel('Imag(v)')
cb = colorbar;
colormap jet; % gradient from blue to red
cb.Label.String = 'Real(u)';


Which gives me the following:



Current result



My questions are:




  1. I thought I would plot what is on the first image, but I got something else. What did I plot if it isn't a Riemann's surface?


  2. How can I change my code to get the top plot?


  3. Would it be possible have a scale in radians on the first graph?











share|improve this question




























    up vote
    2
    down vote

    favorite












    I'm trying to plot a kind of Riemann's surface of a function (I'm not sure if it's the right name for the thing), as shown below:



    Desired result



    Here's what I tried:



    r = (0:1:15)';                           % create a matrix of complex inputs
    theta = pi*(-1:0.05:1);
    z = r*exp(1i*theta);
    w = z.^(1/2) ; % calculate the complex outputs

    figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
    subplot(121)

    surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
    xlabel('Real(z)')
    ylabel('Imag(z)')
    zlabel('Real(u)')
    cb = colorbar;
    colormap jet; % gradient from blue to red
    cb.Label.String = 'Imag(v)';

    subplot(122)
    surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
    xlabel('Real(z)')
    ylabel('Imag(z)')
    zlabel('Imag(v)')
    cb = colorbar;
    colormap jet; % gradient from blue to red
    cb.Label.String = 'Real(u)';


    Which gives me the following:



    Current result



    My questions are:




    1. I thought I would plot what is on the first image, but I got something else. What did I plot if it isn't a Riemann's surface?


    2. How can I change my code to get the top plot?


    3. Would it be possible have a scale in radians on the first graph?











    share|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I'm trying to plot a kind of Riemann's surface of a function (I'm not sure if it's the right name for the thing), as shown below:



      Desired result



      Here's what I tried:



      r = (0:1:15)';                           % create a matrix of complex inputs
      theta = pi*(-1:0.05:1);
      z = r*exp(1i*theta);
      w = z.^(1/2) ; % calculate the complex outputs

      figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
      subplot(121)

      surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
      xlabel('Real(z)')
      ylabel('Imag(z)')
      zlabel('Real(u)')
      cb = colorbar;
      colormap jet; % gradient from blue to red
      cb.Label.String = 'Imag(v)';

      subplot(122)
      surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
      xlabel('Real(z)')
      ylabel('Imag(z)')
      zlabel('Imag(v)')
      cb = colorbar;
      colormap jet; % gradient from blue to red
      cb.Label.String = 'Real(u)';


      Which gives me the following:



      Current result



      My questions are:




      1. I thought I would plot what is on the first image, but I got something else. What did I plot if it isn't a Riemann's surface?


      2. How can I change my code to get the top plot?


      3. Would it be possible have a scale in radians on the first graph?











      share|improve this question















      I'm trying to plot a kind of Riemann's surface of a function (I'm not sure if it's the right name for the thing), as shown below:



      Desired result



      Here's what I tried:



      r = (0:1:15)';                           % create a matrix of complex inputs
      theta = pi*(-1:0.05:1);
      z = r*exp(1i*theta);
      w = z.^(1/2) ; % calculate the complex outputs

      figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
      subplot(121)

      surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
      xlabel('Real(z)')
      ylabel('Imag(z)')
      zlabel('Real(u)')
      cb = colorbar;
      colormap jet; % gradient from blue to red
      cb.Label.String = 'Imag(v)';

      subplot(122)
      surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
      xlabel('Real(z)')
      ylabel('Imag(z)')
      zlabel('Imag(v)')
      cb = colorbar;
      colormap jet; % gradient from blue to red
      cb.Label.String = 'Real(u)';


      Which gives me the following:



      Current result



      My questions are:




      1. I thought I would plot what is on the first image, but I got something else. What did I plot if it isn't a Riemann's surface?


      2. How can I change my code to get the top plot?


      3. Would it be possible have a scale in radians on the first graph?








      matlab plot visualization complex-numbers surface






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      edited Nov 11 at 11:19









      Dev-iL

      16.3k64075




      16.3k64075










      asked Nov 9 at 23:16









      Marine Galantin

      1428




      1428
























          1 Answer
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          up vote
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          accepted










          Your first plot shows multiple branches of a multiple-valued "function". It's not really a function in the usual sense, since for a given z you have more than one function value. You can only reproduce this by going around more than once around the origin, i.e. more than 2*pi in your angular variable. What you plot is the principal branch of that function, i.e. the sheet that corresponds to complex phases ranging from -pi to pi.



          Furthermore, there's a more fundamental issue. Once you switch to complex numbers represented as doubles, you lose any information about additional phase around the origin (complex numbers represented as a real + imaginary part will only result in the principal value of their complex phase, which falls between -pi and pi). So you need to compute the square root "manually", from the trigonometric form of the complex number:



          r = (0:1:15)';                           % create a matrix of complex inputs
          theta = pi*(-2:0.05:2);
          z = r*exp(1i*theta);
          %w = z.^(1/2) ; % calculate the complex outputs
          w = sqrt(r)*exp(1i*theta/2);

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)

          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          result from polar solution



          As you can see, the function acts as it should. Note that it doesn't make sense to have "the scale in radians" in the figure. Everything you plot has "linear" dimensions: real parts and imaginary parts. Radians would only make sense for angles, i.e. theta-like quantities.



          Also, you may note that the above figure has round edges, since we're plotting using polar coordinates. It's possible to create a rectangular plot but it takes a lot more work. Here's a partial solution. The idea is to sew together the same mesh twice in order to plot the two branches of the function:



          r0 = 15;
          re = linspace(-r0, r0, 31).'; % create a matrix of complex inputs
          im = linspace(-r0, r0, 31);
          z = re + 1j*im;
          theta = angle(z); % atan2(imag(z), real(z));
          r = abs(z);

          % concatenate the same mesh twice (plotting trick) back to back, insert nan in between
          w1 = sqrt(r).*exp(1i*theta/2); % first branch
          w2 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch

          z = [z, nan(size(w1,1),1), z(:,end:-1:1)];
          w = [w1, nan(size(w1,1),1), w2(:,end:-1:1)];

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)
          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          Here's the result:



          result from Cartesian solution



          As you can see the complex part looks weird. This is because the phase of complex numbers jumps along the negative real half axis. This could be remedied but takes a lot more work, this is left as an exercise to the reader. The reason I injected a column of nans into the data is to prevent a similar jump artifact to be present in the first plot. The other option is to plot the two branches of the function separately, with hold on in between, but then extra work would have to be done to normalize the colormap on the figures.



          Finally, do consider not using jet but the default parula colormap instead. Jet is very bad for people with impaired colour vision, and parula is close to perceptually uniform. For a short introduction to the problem I suggest watching this great talk from the scipy guys.






          share|improve this answer

















          • 1




            nice talk. Thank you for the help !
            – Marine Galantin
            Nov 16 at 22:44











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          1 Answer
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          active

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          active

          oldest

          votes






          active

          oldest

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          up vote
          3
          down vote



          accepted










          Your first plot shows multiple branches of a multiple-valued "function". It's not really a function in the usual sense, since for a given z you have more than one function value. You can only reproduce this by going around more than once around the origin, i.e. more than 2*pi in your angular variable. What you plot is the principal branch of that function, i.e. the sheet that corresponds to complex phases ranging from -pi to pi.



          Furthermore, there's a more fundamental issue. Once you switch to complex numbers represented as doubles, you lose any information about additional phase around the origin (complex numbers represented as a real + imaginary part will only result in the principal value of their complex phase, which falls between -pi and pi). So you need to compute the square root "manually", from the trigonometric form of the complex number:



          r = (0:1:15)';                           % create a matrix of complex inputs
          theta = pi*(-2:0.05:2);
          z = r*exp(1i*theta);
          %w = z.^(1/2) ; % calculate the complex outputs
          w = sqrt(r)*exp(1i*theta/2);

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)

          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          result from polar solution



          As you can see, the function acts as it should. Note that it doesn't make sense to have "the scale in radians" in the figure. Everything you plot has "linear" dimensions: real parts and imaginary parts. Radians would only make sense for angles, i.e. theta-like quantities.



          Also, you may note that the above figure has round edges, since we're plotting using polar coordinates. It's possible to create a rectangular plot but it takes a lot more work. Here's a partial solution. The idea is to sew together the same mesh twice in order to plot the two branches of the function:



          r0 = 15;
          re = linspace(-r0, r0, 31).'; % create a matrix of complex inputs
          im = linspace(-r0, r0, 31);
          z = re + 1j*im;
          theta = angle(z); % atan2(imag(z), real(z));
          r = abs(z);

          % concatenate the same mesh twice (plotting trick) back to back, insert nan in between
          w1 = sqrt(r).*exp(1i*theta/2); % first branch
          w2 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch

          z = [z, nan(size(w1,1),1), z(:,end:-1:1)];
          w = [w1, nan(size(w1,1),1), w2(:,end:-1:1)];

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)
          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          Here's the result:



          result from Cartesian solution



          As you can see the complex part looks weird. This is because the phase of complex numbers jumps along the negative real half axis. This could be remedied but takes a lot more work, this is left as an exercise to the reader. The reason I injected a column of nans into the data is to prevent a similar jump artifact to be present in the first plot. The other option is to plot the two branches of the function separately, with hold on in between, but then extra work would have to be done to normalize the colormap on the figures.



          Finally, do consider not using jet but the default parula colormap instead. Jet is very bad for people with impaired colour vision, and parula is close to perceptually uniform. For a short introduction to the problem I suggest watching this great talk from the scipy guys.






          share|improve this answer

















          • 1




            nice talk. Thank you for the help !
            – Marine Galantin
            Nov 16 at 22:44















          up vote
          3
          down vote



          accepted










          Your first plot shows multiple branches of a multiple-valued "function". It's not really a function in the usual sense, since for a given z you have more than one function value. You can only reproduce this by going around more than once around the origin, i.e. more than 2*pi in your angular variable. What you plot is the principal branch of that function, i.e. the sheet that corresponds to complex phases ranging from -pi to pi.



          Furthermore, there's a more fundamental issue. Once you switch to complex numbers represented as doubles, you lose any information about additional phase around the origin (complex numbers represented as a real + imaginary part will only result in the principal value of their complex phase, which falls between -pi and pi). So you need to compute the square root "manually", from the trigonometric form of the complex number:



          r = (0:1:15)';                           % create a matrix of complex inputs
          theta = pi*(-2:0.05:2);
          z = r*exp(1i*theta);
          %w = z.^(1/2) ; % calculate the complex outputs
          w = sqrt(r)*exp(1i*theta/2);

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)

          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          result from polar solution



          As you can see, the function acts as it should. Note that it doesn't make sense to have "the scale in radians" in the figure. Everything you plot has "linear" dimensions: real parts and imaginary parts. Radians would only make sense for angles, i.e. theta-like quantities.



          Also, you may note that the above figure has round edges, since we're plotting using polar coordinates. It's possible to create a rectangular plot but it takes a lot more work. Here's a partial solution. The idea is to sew together the same mesh twice in order to plot the two branches of the function:



          r0 = 15;
          re = linspace(-r0, r0, 31).'; % create a matrix of complex inputs
          im = linspace(-r0, r0, 31);
          z = re + 1j*im;
          theta = angle(z); % atan2(imag(z), real(z));
          r = abs(z);

          % concatenate the same mesh twice (plotting trick) back to back, insert nan in between
          w1 = sqrt(r).*exp(1i*theta/2); % first branch
          w2 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch

          z = [z, nan(size(w1,1),1), z(:,end:-1:1)];
          w = [w1, nan(size(w1,1),1), w2(:,end:-1:1)];

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)
          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          Here's the result:



          result from Cartesian solution



          As you can see the complex part looks weird. This is because the phase of complex numbers jumps along the negative real half axis. This could be remedied but takes a lot more work, this is left as an exercise to the reader. The reason I injected a column of nans into the data is to prevent a similar jump artifact to be present in the first plot. The other option is to plot the two branches of the function separately, with hold on in between, but then extra work would have to be done to normalize the colormap on the figures.



          Finally, do consider not using jet but the default parula colormap instead. Jet is very bad for people with impaired colour vision, and parula is close to perceptually uniform. For a short introduction to the problem I suggest watching this great talk from the scipy guys.






          share|improve this answer

















          • 1




            nice talk. Thank you for the help !
            – Marine Galantin
            Nov 16 at 22:44













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Your first plot shows multiple branches of a multiple-valued "function". It's not really a function in the usual sense, since for a given z you have more than one function value. You can only reproduce this by going around more than once around the origin, i.e. more than 2*pi in your angular variable. What you plot is the principal branch of that function, i.e. the sheet that corresponds to complex phases ranging from -pi to pi.



          Furthermore, there's a more fundamental issue. Once you switch to complex numbers represented as doubles, you lose any information about additional phase around the origin (complex numbers represented as a real + imaginary part will only result in the principal value of their complex phase, which falls between -pi and pi). So you need to compute the square root "manually", from the trigonometric form of the complex number:



          r = (0:1:15)';                           % create a matrix of complex inputs
          theta = pi*(-2:0.05:2);
          z = r*exp(1i*theta);
          %w = z.^(1/2) ; % calculate the complex outputs
          w = sqrt(r)*exp(1i*theta/2);

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)

          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          result from polar solution



          As you can see, the function acts as it should. Note that it doesn't make sense to have "the scale in radians" in the figure. Everything you plot has "linear" dimensions: real parts and imaginary parts. Radians would only make sense for angles, i.e. theta-like quantities.



          Also, you may note that the above figure has round edges, since we're plotting using polar coordinates. It's possible to create a rectangular plot but it takes a lot more work. Here's a partial solution. The idea is to sew together the same mesh twice in order to plot the two branches of the function:



          r0 = 15;
          re = linspace(-r0, r0, 31).'; % create a matrix of complex inputs
          im = linspace(-r0, r0, 31);
          z = re + 1j*im;
          theta = angle(z); % atan2(imag(z), real(z));
          r = abs(z);

          % concatenate the same mesh twice (plotting trick) back to back, insert nan in between
          w1 = sqrt(r).*exp(1i*theta/2); % first branch
          w2 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch

          z = [z, nan(size(w1,1),1), z(:,end:-1:1)];
          w = [w1, nan(size(w1,1),1), w2(:,end:-1:1)];

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)
          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          Here's the result:



          result from Cartesian solution



          As you can see the complex part looks weird. This is because the phase of complex numbers jumps along the negative real half axis. This could be remedied but takes a lot more work, this is left as an exercise to the reader. The reason I injected a column of nans into the data is to prevent a similar jump artifact to be present in the first plot. The other option is to plot the two branches of the function separately, with hold on in between, but then extra work would have to be done to normalize the colormap on the figures.



          Finally, do consider not using jet but the default parula colormap instead. Jet is very bad for people with impaired colour vision, and parula is close to perceptually uniform. For a short introduction to the problem I suggest watching this great talk from the scipy guys.






          share|improve this answer












          Your first plot shows multiple branches of a multiple-valued "function". It's not really a function in the usual sense, since for a given z you have more than one function value. You can only reproduce this by going around more than once around the origin, i.e. more than 2*pi in your angular variable. What you plot is the principal branch of that function, i.e. the sheet that corresponds to complex phases ranging from -pi to pi.



          Furthermore, there's a more fundamental issue. Once you switch to complex numbers represented as doubles, you lose any information about additional phase around the origin (complex numbers represented as a real + imaginary part will only result in the principal value of their complex phase, which falls between -pi and pi). So you need to compute the square root "manually", from the trigonometric form of the complex number:



          r = (0:1:15)';                           % create a matrix of complex inputs
          theta = pi*(-2:0.05:2);
          z = r*exp(1i*theta);
          %w = z.^(1/2) ; % calculate the complex outputs
          w = sqrt(r)*exp(1i*theta/2);

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)

          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          result from polar solution



          As you can see, the function acts as it should. Note that it doesn't make sense to have "the scale in radians" in the figure. Everything you plot has "linear" dimensions: real parts and imaginary parts. Radians would only make sense for angles, i.e. theta-like quantities.



          Also, you may note that the above figure has round edges, since we're plotting using polar coordinates. It's possible to create a rectangular plot but it takes a lot more work. Here's a partial solution. The idea is to sew together the same mesh twice in order to plot the two branches of the function:



          r0 = 15;
          re = linspace(-r0, r0, 31).'; % create a matrix of complex inputs
          im = linspace(-r0, r0, 31);
          z = re + 1j*im;
          theta = angle(z); % atan2(imag(z), real(z));
          r = abs(z);

          % concatenate the same mesh twice (plotting trick) back to back, insert nan in between
          w1 = sqrt(r).*exp(1i*theta/2); % first branch
          w2 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch

          z = [z, nan(size(w1,1),1), z(:,end:-1:1)];
          w = [w1, nan(size(w1,1),1), w2(:,end:-1:1)];

          figure('Name','Graphique complexe','units','normalized','outerposition',[ 0.08 0.1 0.8 0.55]);
          subplot(121)
          surf(real(z),imag(z),real(w),imag(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Real(u)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Imag(v)';

          subplot(122)
          surf(real(z),imag(z),imag(w),real(w)) % visualize the complex function using surf
          xlabel('Real(z)')
          ylabel('Imag(z)')
          zlabel('Imag(v)')
          cb = colorbar;
          colormap jet; % gradient from blue to red
          cb.Label.String = 'Real(u)';


          Here's the result:



          result from Cartesian solution



          As you can see the complex part looks weird. This is because the phase of complex numbers jumps along the negative real half axis. This could be remedied but takes a lot more work, this is left as an exercise to the reader. The reason I injected a column of nans into the data is to prevent a similar jump artifact to be present in the first plot. The other option is to plot the two branches of the function separately, with hold on in between, but then extra work would have to be done to normalize the colormap on the figures.



          Finally, do consider not using jet but the default parula colormap instead. Jet is very bad for people with impaired colour vision, and parula is close to perceptually uniform. For a short introduction to the problem I suggest watching this great talk from the scipy guys.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 11 at 16:35









          Andras Deak

          20.3k63770




          20.3k63770








          • 1




            nice talk. Thank you for the help !
            – Marine Galantin
            Nov 16 at 22:44














          • 1




            nice talk. Thank you for the help !
            – Marine Galantin
            Nov 16 at 22:44








          1




          1




          nice talk. Thank you for the help !
          – Marine Galantin
          Nov 16 at 22:44




          nice talk. Thank you for the help !
          – Marine Galantin
          Nov 16 at 22:44


















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