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I'm looking for some software that my help me to graph some complex functions on unit circle. I.e. let say if I have $\ f(z)=1/(1-z)$ I want to see to give an input an image with unit circle and want to get the transformed image of unit circle with $\ f(z)$ function.

Can anybody suggest some grapher for this, or something similar ?

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In the case of Möbius transformations, you don't need any software. Consider the map $f : \overline{\mathbb{C}}_z \to \overline{\mathbb{C}}_w$ given by $w = (1-z)^{-1}$. It follows that $z = (w-1)w^{-1}$. If $|z| = 1$ then $|(w-1)w^{-1}| = 1$ and so $|w| = |w-1|$. The image of the unit circle is the perpendicular bisector of $w=0$ and $w=1$, i.e. the line parallel to the imaginary axis that passes through $w = \frac{1}{2}$.

In general, if $f : \overline{\mathbb{C}}_z \to \overline{\mathbb{C}}_w$ is given by

$$w = \frac{az + b}{cz + d} \, . $$

where $(a:b:c:d) \in \mathbb{CP}^3$ then

$$z = \frac{dw-b}{cw-a} \, . $$

The image of the unit circle is given by setting $|z| = 1$ and so $|cw-a| = |dw-b|$ is the equation of the image in the $w$-sphere.

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  • $\begingroup$ the example function in my question is just a sample. I need visually see the transformation of unit circle with any complex function. So that why I need software $\endgroup$
    – deimus
    Sep 11 '12 at 9:55
  • $\begingroup$ math.stackexchange.com/questions/204229/… $\endgroup$
    – User3910
    May 9 '15 at 23:06
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For that, you can use Reflex4You.com. Normally it is a ray-tracing engine, but you can make it work to find the image of your circle. The identity function looks like this:

identity

A function describing the unit circle with colors on a black background is the following:

if((abs(z)>0.9)*(abs(z)<1.1),z,0)

if((abs(z)>0.9)*(abs(z)<1.1),z,0)

The function you are invoking has the following representation:

1/(1-z)

To see what the circle becomes through the function, you can use the following code:

set k = 0; let n = 100; let f = 1/(1-z); set result = 0; let threshold = 0.1; repeat n in set theta = exp(i*2*pi*k/n); set result = if(abs(f(theta)-z) < threshold, theta, result); set k = k + 1; result

Function described above

Edit it yourself: Click on the image above.

If you know the inverse of the function (if it exist) then you can print it faster. In our case, $f^{(-1)}=1-\frac{1}{z}$, so that you can write the formula:

let w = 1-1/z; let c= if((abs(z)>0.9)*(abs(z)<1.1),z,0); o(c,w)

Smooth version of the image of the circle

DISCLAIMER: I am one of the authors of this website which can also serves the purpose of customized object store.

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Follow this guide to Sage: to using Sage Online if you don't want to install Sage on your computer.

Graphing $\frac{1}{1-z}$ that way yeilds:

enter image description here

Graphing $\frac{1}{1-z^2}$ that way yields:

enter image description here

It would be nice to see it in 3D instead of merely color coded. The y-axis is coming out of the picture toward us and instead of seeing the surface that is desired we see a color-graph. I have asked that question.

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  • $\begingroup$ If I am not mistaken, your are plotting 1/(1+z) instead of 1/(1-z). $\endgroup$ May 5 '15 at 9:47
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    $\begingroup$ Amazing that this is the accepted answer when it does not answer the question at all... $\endgroup$
    – user65203
    May 5 '15 at 9:50
  • $\begingroup$ @Dale I kindly mean that the formula you plotted 1/(1+z) is not the same as the one you introduce (1/(1-z)). The freeware is not faulty there, just the input formula is. $\endgroup$ May 6 '15 at 7:45
  • $\begingroup$ @MikaëlMayer wolframalpha.com/input/?i=plot%28+1%2F%281%2Bz%29%29 $\endgroup$
    – User3910
    May 9 '15 at 22:35
  • $\begingroup$ @Dale Your link is still wrong compared to what you introduce. You say 1/(1-z) but both your representation and the link are 1/(1+z) (the zero of your function is -1). Please have a closer look. $\endgroup$ May 11 '15 at 4:44

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