Is this linear transformation possible?

Question

$\phi$ 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.

thanks! but what about the fact that it's at $\infty$? – user7589 May 12 '11 at 06:05 — user7589, May 12 '11 at 06:05
http://en.wikipedia.org/wiki/Riemann_sphere – Will Jagy May 12 '11 at 21:29 — Will Jagy, May 12 '11 at 21:29

score 7 · Answer 1 · edited Sep 22 '13 at 23:00

7

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 $w_1,w_2,w_3$.

edited Sep 22 '13 at 23:00

azimut

22,696

answered May 12 '11 at 06:06

Zev Chonoles

129,973

1

Just as a further note: in the case of a more general specification of 4 points $(z_1,z_2,z_3,z_4) \to (w_1,w_2, w_3, w_4)$, a first litmus test is to use the fact that FLTs preserve cross ratios. – Willie Wong May 12 '11 at 08:52
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@Willie Wong: Why "first litmus test" instead of characterisation? – Phira May 12 '11 at 09:19
1

@user9325: because I was thinking in terms of exam strategies, since the OP has apparently been working on past exams. – Willie Wong May 12 '11 at 15:27

score 0 · Answer 2 · answered Apr 28 '13 at 07:36

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 $2\to 2$

Is this linear transformation possible?

2 Answers2