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What does it mean if this message appears:

{Im[(1-E^Times[<<3>>] f)/(1-Power[<<2>>] f)]-0,Im[(1-E^Times[<<3>>] f)/(1-Power[<<2>>] f)]-0} must be a list of equalities or real-valued functions. >>

while Iam trying to plot this complex function

(I α)/π (Log[(1 - E^(-((I π(1 - α))/α)) f) / (1 - E^((I π(1 - α))/α) f)])

How can I plot this function for the range {α, 0.1, 1} and {f, 0.2, 1}?

Edit

Corrected errors in the expression to be plotted.

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  • $\begingroup$ Something is really messed up here, can you cleanup the expression? $\endgroup$
    – Spawn1701D
    Apr 15, 2013 at 2:35
  • $\begingroup$ Can you provide the mathematical problem itself? $\endgroup$
    – Spawn1701D
    Apr 15, 2013 at 2:43
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    $\begingroup$ The factor π[1 - α] in your expression is a function call in Mathematica. Did you mean π(1 - α)? $\endgroup$
    – m_goldberg
    Apr 15, 2013 at 2:44
  • $\begingroup$ Could you post the entire Plot expression you tried? $\endgroup$
    – m_goldberg
    Apr 15, 2013 at 2:45
  • $\begingroup$ 1-(I[Alpha])/[Pi](Log[( 1 - E^(-((I[Pi][1-[Alpha]])/[Alpha]))f)/( 1 - E^((I[Pi][1-[Alpha]])/[Alpha])f)]) $\endgroup$
    – sana
    Apr 15, 2013 at 2:47

1 Answer 1

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Try this:

for the real part of the expression

ContourPlot[Re[(I α)/π(Log[(1 - E^(-((I π (1 - α))/α))f)/(1 - E^((I π (1 - α))/α)f)])], 
            {α, 0.1,1}, {f, .2, 1}]

enter image description here

for the imaginary part of the expression

ContourPlot[Im[(I α)/π(Log[(1 - E^(-((I π (1 - α))/α))f)/(1 - E^((I π (1 - α))/α)f)])], 
            {α, 0.1,1}, {f, .2, 1}]

enter image description here

If you simplify the expression using ComplexExpand you will find out that this is is in fact a real function

$$ -\frac{\alpha \text{Arg}\left[\frac{1+e^{-\frac{i \pi }{\alpha }} f}{1+e^{\frac{i \pi }{\alpha }} f}\right]}{\pi } $$

Using this instead its Plot3D is:

enter image description here

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  • $\begingroup$ thank u very much but may I have the 2D Plotting of the function plz? $\endgroup$
    – sana
    Apr 15, 2013 at 8:15
  • $\begingroup$ @sana the function depends on two variables so it defines a surface in 3D. In 2D you can have its contour plot which I have given in the first plot. $\endgroup$
    – Spawn1701D
    Apr 15, 2013 at 8:25
  • $\begingroup$ understood now.. thank you again $\endgroup$
    – sana
    Apr 15, 2013 at 9:31

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