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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.

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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
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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

1 Answer 1

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).

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