# Schwartz - Christoffel transformation: point at infinity confusion

1. Jul 20, 2014

### D_Tr

We know that the derivative of the general Schwartz - Christoffel map (function) is:
$$f'(z) = λ(z - x_1)^{a_1}.....(z-x_n)^{a_n}$$
Question: In various sources around the web, it is mentioned that $x_n$ can be taken to be the "point at infinity", and the last factor can be removed from the above derivative expression. Why is this helpful? What is the difference between taking a $x_n$ to infinity and just having n-1 points? And why can the last factor be removed in the first place if a term inside it becomes infinitely large??

Your help will be greatly appreciated!

Last edited: Jul 20, 2014
2. Aug 11, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Aug 13, 2014

### D_Tr

Thank you for your interest :) I think I have figured out the answer. I am a bit surprised that I haven't got any kind of feedback after so many views. Is it unclear? Does it make any sense at all? I had two questions:

1) How can moving $x_n$ to infinity make the last factor vanish?
The answer to this is found in the last few lines of the 3rd page of the paper found here: http://www.mth.kcl.ac.uk/~shaww/web_page/books/complex/Chapter21Excerpt.pdf.

2) Why do we do this?
My confusion was probably due to the fact that when dealing with this transformation you usually have a given closed polygon to which you want to map the upper half complex plane and you search for the correct formula to achieve this. I initially thought the usual problem was: "I have this Schwartz-Christoffel transformation formula, how does it distort the upper half complex plane?"
Each factor in the Schwartz-Christoffel formula corresponds to a vertex. If you want to get a n-sided polygon and you search for a formula with n factors, you need to force each of the n factors to map to one of the vertices of your desired polygon and additionally you need to force the point at infinity map to a point on your polygon, for example to a point between two vertices. If, now you take $x_n$ to be the point at infinity, you have one less pre-image to deal with because the factor containing $x_n$ is gone (see link above) and you now need to deal with n-1 finite pre-images and the point at infinity. So it's just a matter of convenience.