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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?2012-08-14
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    The way you formulated, $T(1) = \frac{1}{2+i}$. Are you sure you didn't mean $T(i)=i$?2012-08-14
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    Just fixed it - it reads correctly now.2012-08-14
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    I believe $T(R)$ is a circle centered at $\frac 12 + \frac 12i$ with radius $\frac 1{\sqrt 2}$.2012-08-14
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    @Tunococ Do you have any reasoning behind this?2012-08-14
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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.2012-08-14
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    While you are at it, do $$ T \left( \frac{1}{2} \right) $$ and include that in the picture.2012-08-14

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