# Mobiust transformations

• Apr 22nd 2008, 03:03 PM
kathrynmath
Mobiust transformations
Find a Mobius Transformation that takes the circle modulus z=1 to the straight line x+y=1.
I know I need to pick 3 points, but I don't know what to do then. Like if I pick out the point (0,0) or (0,i) do I plug these values into x+y=1. I'm just really confused on this step.
• Apr 22nd 2008, 07:44 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
Find a Mobius Transformation that takes the circle modulus z=1 to the straight line x+y=1.
I know I need to pick 3 points, but I don't know what to do then. Like if I pick out the point (0,0) or (0,i) do I plug these values into x+y=1. I'm just really confused on this step.

Here is an easier way to do it. The function $f_1(z) = (1-z)/(1+z)$ takes the circle $|z| = 1$ to the line $x=0$. The function $f_2(z) = e^{i\pi/4}z$ takes the line $x=0$ to $y=-x$. The function $f_3(z) = z+1$ takes the line $y=-x$ to $y=1-x$, which is the desired line. Thus, $f_3\circ f_2\circ f_1(z)$ is a Mobius transformation which does this mapping.

Here is an interesting thread on this topic.