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$\phi$ is a linear fractional transformation such that $\phi(0) = 0, \phi(1) = 1$, and $\phi(2) = \infty$.

$\phi$ is a linear fractional transformation such that $\phi(0) = 0, \phi(1) = 1$, $\phi(2) = 2$, and $\phi(3) = \infty$.

I found this in a past final exam, so I don't have an answer for it.

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    http://en.wikipedia.org/wiki/Riemann_sphere2011-05-12

2 Answers 2

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The first one is yes, the second one no. This is because a fractional linear transformation (aka a Möbius transformation) is uniquely determined by where it sends any three distinct points. Thus, we can choose to send 0, 1, 2 to 0, 1, $\infty$, respectively, and this specifies a unique transformation; but because there is a unique transformation sending 0, 1, 2, to 0, 1, 2, (namely the identity transformation), we can't also have 3 sent to $\infty$.

The Wikipedia page also describes the method for coming up with the coefficients $a$, $b$, $c$, and $d$ in the unique transformation $\frac{az+b}{cz+d}$ that sends $z_1,z_2,z_3$ to $w_1,w_2,w_3$.

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    @user9325: because I was thinking in terms of exam strategies, since the OP has apparently been working on past exams.2011-05-12
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The Unique linearfractional transformation that sends $z_1,z_2,z_3\to\infty,0,1$ is given by $T(z)=\frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$

In the first case you can construct such map but for the second case your $T(z)$ is given by $T(z)=\frac{2z}{(3-z)}$ which does not send $2\to 2$