Plotting a complex function

Question

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.

Something is really messed up here, can you cleanup the expression? — Spawn1701D, Apr 15 '13 at 02:35
The factor π[1 - α] in your expression is a function call in Mathematica. Did you mean π(1 - α)? — m_goldberg, Apr 15 '13 at 02:44
1-(I[Alpha])/[Pi](Log[( 1 - E^(-((I[Pi][1-[Alpha]])/[Alpha]))f)/( 1 - E^((I[Pi][1-[Alpha]])/[Alpha])f)]) — sana, Apr 15 '13 at 02:47

Spawn1701D · Answer 1 · 2013-04-15T03:24:53.860

2

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}]

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}]

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:

edited Apr 15 '13 at 03:24

answered Apr 15 '13 at 02:56

Spawn1701D

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thank u very much but may I have the 2D Plotting of the function plz? – sana Apr 15 '13 at 08:15
@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. – Spawn1701D Apr 15 '13 at 08:25
understood now.. thank you again – sana Apr 15 '13 at 09:31

Plotting a complex function

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1 Answers1