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\begin{document}

{\bf Question}

Consider the element $n(z) = \frac{i\overline{z} +
2}{-3\overline{z} + 6}$ of the general M\"obius group ${\rm
M\ddot{o}b}$.

Determine the set of fixed points of $n(z)$.

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{\bf Answer}

Set $n(z)=z$ and solve for $z$:

$\ds\frac{i\bar z + 2}{-3\bar z+6}=z \Rightarrow -3z\bar
z+6z=-\bar z+2$

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Set $z=x+iy$ and simplify:

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$-3(x^2+y^2)+6x+6iy-i(x-iy)-2=0$

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$-3(x^2+y^2)+6x-y-2+6iy-ix=0$

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Take imaginary parts: $6y-x=0 \Rightarrow x=6y$ and substitute
into the equation for the real parts:

$-3(36y^2+y^2)+6 \cdot 6y-y-2=0$

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$-111y^2+35y-2=0$

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\begin{eqnarray*} y & = & \ds\frac{-1}{222}(-35 \pm
\sqrt{(35)^2-4(-111)(-2)})\\ & = & \ds\frac{-1}{222}(-35 \pm
\sqrt{337})=y_1 {\rm \ and \ }y_2 \end{eqnarray*}

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\un{Two points}: $6y_1+iy_1$ and $6y_1+iy_2$ are the fixed points
of $n$.

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