# Show Schwarz-Christoffel retains the same form

Set $w=\phi(z)=i\frac{1+z}{1-z}$ (which maps the unit disk in the complex plane to the upper half of the complex plane). Show that the Schwarz Christoffel formula, $$f(z)=A_1\int_0^z \frac 1{(w-x_1)^{\beta_1}(w-x_2)^{\beta_2}\cdots(w-x_n)^{\beta_n}}\ dw+A_2,\quad (z \in \mathbb H),$$ retains the same form.

Question 1: What does this question mean? What do they mean by CS "retains the same form"? What on earth are they asking for? What does SC formula have to do with $\phi(z)$?

Help! (Once again, sorry for the vaguely worded question - I know everyone hates when you just post "help").

EDIT: OK, based on math chatroom discussions and some comments in the text, I think this question is asking to show that $\phi(f(z))$ has the same form as $f(z)$. However, I can't figure out the next step after substituting $f(z)$ into the equation for $\phi(z)$. What is the next step after $$i\frac{1+(A_1\int_0^z \frac 1{(w-x_1)^{\beta_1}(w-x_2)^{\beta_2}\cdots(w-x_n)^{\beta_n}}\ dw+A_2)}{1-(A_1\int_0^z \frac 1{(w-x_1)^{\beta_1}(w-x_2)^{\beta_2}\cdots(w-x_n)^{\beta_n}}\ dw+A_2)}?$$ Thank you.

• Can SC formula be setup to map the unit disk to the upper half of the plane? If so, how? – Jeff May 3 '12 at 23:30
• The Schwarz-Christoffel formula shows how to map the upper half plane onto any convex polygon. By setting all exponents beta equal to 2/n, we get a mapping of the upper half plane to a regular n-gon. Letting n go to infinity, I would expect this formula to agree with your phi, since both would map the upper half plane to circles. – John Adamski May 4 '12 at 16:51
• @JohnAdamski That's it! That must be what they wanted. John, can you promote your comment to an answer? – Jeff May 5 '12 at 16:30
• If you are still interested in seeing a correct answer, it's here. – Post No Bulls Jan 5 '14 at 7:49

The Schwarz-Christoffel formula shows how to map the upper half plane onto any convex polygon. By setting all exponents $\beta = 2/n$, we get a mapping of the upper half plane to a regular $n$-gon. Letting $n$ go to infinity, I would expect this formula to agree with your $\phi^{-1}$, since both would map the upper half plane to circles.

• There is nothing about $n\to\infty$ in the question. The correct answer can be found here. – Post No Bulls Jan 5 '14 at 7:48