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

Suppose $T$ is a linear fractional transformation such that $T(0)=1, T(1)=i,$ and $T(\infty)=0$. Find $T(i)$ and describe $T(R)$, where R is the real line.

I ended up getting $T(z)=\frac{1}{(-i-1)z+1}$, and I am pretty sure this is correct. I am lost on what $T(Z)$ does to the real line.

share|cite|improve this question
I don't see that your function satisfies $T(1) = i$. Am I missing something? – Tunococ Aug 14 '12 at 18:47
The way you formulated, $T(1) = \frac{1}{2+i}$. Are you sure you didn't mean $T(i)=i$? – gt6989b Aug 14 '12 at 18:47
Just fixed it - it reads correctly now. – Frank White Aug 14 '12 at 18:56
I believe $T(R)$ is a circle centered at $\frac 12 + \frac 12i$ with radius $\frac 1{\sqrt 2}$. – Tunococ Aug 14 '12 at 19:00
Every Moebius transformation takes a line or circle into a line or circle. The point at $\infty$ is a part of every line. As $T(\infty) = 0$ is not $\infty,$ it follows that any line, including the real line, is mapped to a circle passing through the origin. Very carefully, find $T(2), T(3), T(100), T(-1), T(-2), T(-100).$ You already know $T(0), T(1).$ Draw a careful picture. – Will Jagy Aug 14 '12 at 19:25

As Will Jagy notes in the comments, $T$ takes a line to a line or a circle. Since $T$ takes $0,1,\infty$ to $1,i,0$, it takes the real line to a circle through $1$, $i$, and $0$. Think about that and you'll see we're talking about the circle that goes through the 4 corners of that little square, so it's centered at the center of that square, and its radius is half the diagonal of that square (in agreement with Tunococ's comment).

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.