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Problem: Show that any four distinct points can be carried by a linear transformation to positions $1, -1, k, -k$, where the value of $k$ depends on the points. How many solutions are there, and how are they related?

Attempt at a solution:

So I know that given any three points $z_2, z_3, z_4$ I can find a linear transformation that carries these points to some other points $w_2, w_3, w_4$ this can be done since the ratio is preserved; that is $$(w, w_2, w_3, w_4)=(z, z_2, z_3,z_4).---(1)$$

So, in our case we have $w_2=-1, w_3=k$ and $w_4=-k$, and we want this transformation to be such that it will also take $z_1$ to $1$.

After performing the necessary algebraic steps in $(1)$ I found

$$w(z)=\frac{(1+k)(z-z_3)(z_2-z_4)+(1-k)(z-z_4)(z_2-z_3)}{(1-k)(z-z_4)(z_2-z_3)-(1+k)(z-z_3)(z_2-z_4)}k$$

This transformation takes the $z_2$ to $-1$, $z_3$ to $k$ and $z_4$ to $-k.$

So now, do I just force the transformation so that $w(z_1)=1$?

I am not sure how the whole $k$ being dependent on the points we choose is coming into play?

Any hints??

Thanks!

Amzoti
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sualia
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  • It zooks like the question is about Mobius transformations, not linear ones. – Dan Shved Mar 08 '13 at 07:51
  • Also, are we in $\mathbb{C}$ or in $\mathbb{R}$? The statement seems to be true for rational functions on $\mathbb{R}$ with degree 1, but it is false for Mobius transformations of $\mathbb{C}$. – Dan Shved Mar 08 '13 at 07:56
  • Also, you might want to take a look at this question: http://math.stackexchange.com/questions/238566/mobius-map-from-circles-to-lines/238588 – Dan Shved Mar 08 '13 at 07:57
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    It is in $\mathbb{C}$...well, by linear transformation they mean Mobius transformations (i.e. linear fractional transformations). – sualia Mar 08 '13 at 08:05
  • Is $k$ bound to be real, or can it be complex too? – Dan Shved Mar 08 '13 at 08:08
  • It does not say on the problem! – sualia Mar 08 '13 at 08:08
  • Well, if $k$ must be real, then this is only true for concyclic points, as in the question linked above. If $k$ can be arbitrary, then it is possible, and the solution is similar. – Dan Shved Mar 08 '13 at 08:20
  • Can you elaborate on Step 2? – sualia Mar 08 '13 at 08:33
  • Do you just mean by that: Let $A=(z_1,z_1,z_3,z_4)$, then I found $k=\frac{\sqrt{A}-1}{\sqrt{A}+1}$?? So as a result, there are also infinitely many solutions, right?? – sualia Mar 08 '13 at 08:37
  • Yes, $k$ should look something like this. But why infinitely many solutions? For the given 4 points with a fixed order there are only 1 or 2 possible values of $k$, so only finitely many transformations will achieve the goal. – Dan Shved Mar 08 '13 at 10:04

1 Answers1

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Starting from scratch, let's consider the cross-ratio $$(z_1,z_2;z_3,z_4) = \frac{z_1-z_3}{z_2-z_3}\frac{z_2-z_4}{z_1-z_4}$$ The map $$f(z)=\frac{z_1-z_3}{z_2-z_3}\frac{z_2-z }{z_1-z }$$ sends $z_1$ to $\infty$, $z_2$ to $0$, $z_3$ to $1$, and $z_4$ to $(z_1,z_2;z_3,z_4)$. Therefore, two quadruples of points with the same cross-ratio are related by a Möbius map. The converse is also true, because cross-ratio is preserved by $z\mapsto az$, $z\mapsto z+b$, and $z\mapsto 1/z$, which generate all Möbius maps.

It remains to count the quadruples $1,-1;k,-k$ with given cross-ratio $r\in\mathbb C\setminus \{0,1\}$. We have $$( 1,-1;k,-k) = \frac{1-k}{-1-k}\frac{-1+k}{1+k} = \left(\frac{1-k}{1+k}\right)^2$$ Here $k\mapsto \frac{1-k}{1+k}$ is a bijection from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{-1,0,1\}$. Therefore, $k\mapsto ( 1,-1;k,-k)$ is a 2-to-1 map from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{0,1\}$. Conclusion: there are precisely two suitable values of $k$ for every quadruple. (They are reciprocals of each other).

user98130
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