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{\bf Question}

Consider the stereographic projection map $\xi$ from ${\bf S}^1$
to ${\bf R}\cup\{\infty\}$, as defined in class. Determine the
images of the vertices of the regular pentagon with vertices
$\exp\left( \frac{2\pi i k}{n} \right)$ for $0\le k\le 4$.

\medskip
More generally, for each $n\ge 5$, determine the images under
$\xi$ of the vertices of a regular $n$-gon whose vertices lie on
${\bf S}^1$ (and where one of the vertices is at $1$).

\medskip

{\bf Answer}

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$\xi(z):$\ intersection of line through $i=N$ and $z$ with ${\bf
R}$.

\underline{line through $i$ and $z$}: slope $m=\ds\frac{{\rm
Im}(z)-1}{{\rm Re}(z)}$

equation:

$\begin{array} {ccl} y-1 & = & m(x-0)\\ y-1 & = &
\ds\frac{\rm{Im}(z)-1}{{\rm Re}(z)}x \end{array}$

Set $y=p$, to get \underline{$x=\ds\frac{{\rm Re}(z)}{1-{\rm
Im}(z)}$}.

\medskip
So, $\xi(z)=\ds\frac{{\rm Re}(z)}{1-{\rm Im}(z)}$
($\xi(n)=\infty$).

The vertices of the regular n-gon are

$$\rm{exp} \left(\ds\frac{2\pi i}{n}k\right)\ \ \ 0 \leq k < n$$

So, $\xi \left(\rm{exp} \left(\ds\frac{2\pi
i}{n}k\right)\right)=\ds\frac{\rm{Re} \left(\ds\frac{2\pi
i}{n}k\right)}{1-\rm{Im} \left(\ds\frac{2\pi
i}{n}k\right)}=\ds\frac{\cos\left(\ds\frac{2\pi
i}{n}k\right)}{1-\sin\left(\ds\frac{2\pi i}{n}k\right)}.$

(Note that $N$ is a vertex for the regular n-gon for all $n\equiv
0$ (mod 4))
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