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Post by Admin on Oct 19, 2016 6:09:27 GMT
At the end of paragraph two of subsection 1 of section VII Needham asks
"If we think of $M(z)$ as a mapping $z\mapsto w=M(z)$ of this figure to itself, then each member of $\mathcal{C}_1$ is mapped to another member of $\mathcal{C}_1$. Why?"
"this figure" refers to figure 29 on page 163.
One way to see this is to say: A Möbius transformation maps circles to circles and because $\xi_+$ and $\xi_-$ are fixed points of $M(z)$, any circle through these two points must be mapped to another circle through these two points. Done.
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 25, 2016 3:20:12 GMT
Vasco,
I somehow missed this, but it's a nice proof.
Gary
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