Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Try Sage, – user31373 Sep 12 '12 at 4:30
I believe what you are describing is called Mapping. – Dale May 9 '15 at 23:00
up vote 0 down vote accepted

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.

share|cite|improve this answer
If I am not mistaken, your are plotting 1/(1+z) instead of 1/(1-z). – Mikaël Mayer May 5 '15 at 9:47
Amazing that this is the accepted answer when it does not answer the question at all... – Yves Daoust May 5 '15 at 9:50
@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. – Mikaël Mayer May 6 '15 at 7:45
@MikaëlMayer – Dale May 9 '15 at 22:35
@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. – Mikaël Mayer May 11 '15 at 4:44

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.

share|cite|improve this answer
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 – deimus Sep 11 '12 at 9:55

For that, you can use 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:


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



The function you are invoking has the following representation:


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.