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Plot graphs using Mathematica

  1. Aug 21, 2009 #1
    hi ,

    i am unable to draw the graphs of complex valued functions using mathematica,

    please help me .

    Ex:koebe function. z/(1-z)^2, z is a complex number
  2. jcsd
  3. Aug 21, 2009 #2


    Staff: Mentor

    How do you want to display the function? One common approach is to use ContourPlot and plot the Abs as the contours and use Arg for the colors.
  4. Aug 21, 2009 #3
    Hi, thnx for reply, please say how to plot in two ways so that i can use which is is required for me, and other one i may use whenever i needed.

    Thank you.
  5. Aug 21, 2009 #4


    Staff: Mentor

    Through[{Re, Im}[(x + I y)/(1 - (x + I y))^2]], {x, 0, 2}, {y, -1,
    1}, PlotRange -> {-1, 1}]

    Plot3D[Abs[(x + I y)/(1 - (x + I y))^2], {x, 0, 2}, {y, -1, 1},
    ColorFunction ->
    Function[{x, y},
    Hue[Arg[(x + I y)/(1 - (x + I y))^2]/(2 \[Pi]) + .5]],
    ColorFunctionScaling -> False]
  6. Aug 24, 2009 #5
    Thank you so much.

    Can you suggest me any book which is useful to plot this type of functions using mathematica
  7. Aug 24, 2009 #6
    Hi one more doubt regarding the above problem,

    The function z/(1-z)^2 maps the unit disk |z|<1 onto the entire plane except a line segment

    from (-infinity to -1/4) , how can we show that using the above function plot using mathematica.

    Thanking you
  8. Aug 24, 2009 #7


    Staff: Mentor

    I have found the online help (F1) to be quite thorough.

    Use the parametric plot version shown above, but map the complex plane using r Exp[-I theta] instead of x + I y
  9. Aug 24, 2009 #8
    I already work out at that time i have some doubt whether it is right or not, thank you now i conformed but here is a problem i m unable to interpret from the figure it mapping the unit disk

    that is |z|<1 (in polar form we are using r Exp(I*theta)

    r varies from 0 to 1

    and theta varies from 0 to 2 pi )

    to the entire XY plane except a line segment

    Attached Files:

  10. Aug 24, 2009 #9


    Staff: Mentor

    Yes, that looks correct.
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