Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is about the connection between linear algebra and complex analysis. Coming from a two real dimensional domain a transformation matrix geometrically transforms a set of points (e.g. a circle) in that plane to another set of points in the two real dimensional codomain. The same holds true for a complex valued function which does the transformation from a one complex dimensional domain to the complex dimensional codomain.

I have two questions:
1. Do you know of any software (or add-ons) that let you "paint" some images and have them transformed by an arbitrary transformation matrix and/or complex function?
2. What is the connection between transformation matrices and complex functions? How can you get from one to the other, i.e. finding corresponding transformation matrices to a given complex function and vice versa to produce the same transformation on the plane?

Any answers concerning the above questions will be appreciated! Thank you!

share|improve this question
    
Geometer's Sketchpad 5 can probably apply almost any transformation (or complex function that can be used as a transformation) to an arbitrary image, including a hand-drawn one, but it's not necessarily a simple thing to achieve in Sketchpad. –  Isaac Jan 6 '11 at 15:39

1 Answer 1

up vote 1 down vote accepted
  1. For complex functions, try Bombelli or Complex visualizer (both are Java applets).

  2. In general, there is no connection. Transformation matrices correspond to linear mappings, and complex functions are in general nonlinear. Exception: the function $f(z)=(a+ib)z$ which simply multiplies $z$ by the constant $a+ib$ (where $a$ and $b$ are real) corresponds to the linear transformation given by the matrix $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.

(Close to a point $z_0$ where $f'(z_0) \neq 0$, the formula $f(z)-f(z_0) \approx f'(z_0) (z-a)$ shows that the function $f$ can be approximated using a linear function of the above type, but that's a different story.)

share|improve this answer

Your Answer

 
discard

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.