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I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?

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You can think of a function from C to C as a vector field on C, although I don't know of any software that does this. – Qiaochu Yuan Jul 28 '10 at 17:59
up vote 12 down vote accepted

One way that functions from C to C can be represented is to show the image of a grid. That is, plot the images of the lines x = constant and y = constant under your function, where z = x + yi.

Another is what Wikipedia calls domain coloring (see this article by Hans Lundmark for a more detailed exposition. The idea here is to color the range of the complex function -- since color space is three-dimensional there are multiple ways to do this -- and then color each point in the domain by the color of the corresponding point in the range.

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+1 for domain coloring, which I use all the time. – Darsh Ranjan Oct 2 '10 at 7:00
New address: – Hans Lundmark Aug 14 '15 at 10:22

For Moebius transformations, check out this nice YouTube video.

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Is that from C to C? – Casebash Jul 28 '10 at 21:05
Yep. See wikipedia:öbius_transformation – Kevin H. Lin Jul 28 '10 at 21:19
That is one extremely cool video! – MathematicalOrchid May 7 '12 at 18:51

Open two browser windows side-by-side and use wolfram alpha. Recast your function f(z) as f(x+iy) and plot the real part in one window and the imaginary part in another. My example links provide plots for z3.

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Oh yes, there's a way to do this. Here is my exploration into the topic about a month ago using Mathematica. The easiest thing to do is to plot the vector field and let the direction of the arrows represent the phase and let the color represent the magnitude. This is a great way to get all four dimensions on a plane and I think it's very enlightening.

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The graphs in the middle of the MathWorld page on Conformal Mapping show examples of the first method in Michael Lugo's answer as well as something somewhat similar to the second method in that answer.

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