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Let $X$ be the unit circle centered at 1 and let $F: X\rightarrow {\Bbb C}$ satisfy $F(0)=0$ and, for $a,b,c,d\in X$, if $(a-b)/(c-d)$ is real, then so is $(F(a)-F(b))/(F(c)-F(d))$. Is it true that $F(z)=uz+v\overline z$ for some complex $u,v$ and all $z\in X$ (i.e., $F(X)$ is an ellipse)?

A strategy I have tried: 1) extend $F$ to a function $G$ on the whole right half-plane by defining $G(z)=F(cz)/c$ where $c$ is unique positive real so that $cz$ is in $X$; 2) use the main hypothesis to show that $G$ takes lines to lines; 3) conclude $G$ is linear as a function from ${\Bbb R}^2$ to itself and so $G(z)=uz+v\overline z$ (and the image $F(X)$ is an ellipse). Step (2) is holding me up.

Thanks for any help.

Sam
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    I believe $\dfrac{a-b}{c-d}$ being real is equivalent to $AB || CD$. – Scaramouche Aug 04 '13 at 03:46
  • Is $F$ supposed to have any properties at all? Like, does it need to be continuous? – Daniel McLaury Aug 04 '13 at 04:42
  • Yes, I'm assuming F is continuous – Sam Aug 04 '13 at 09:33
  • I'm not sure I believe in your step 1) yet. I'd treat everything in $\mathbb R^2$ from the start, and extend $F$ to the plane by representing a line in the plane as a pair of points on the circle, in the spirit of Hesse's Übertragungsprinzip. Then in 2) you'd have to show that if three lines defined this way are concurrent, then so are their images. One way (again according to Hesse) would be requiring that these six points form a quadrilateral set. I don't see how to do this from the parallelity yet, so I, too, am stuck at 2), but perhaps someone else can build on this to make it work. – MvG Aug 05 '13 at 16:32

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