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{\bf Question}

 Let $m$ be a parabolic M\"obius transformation with fixed point
$x\ne \infty$.  Show that there exists a unique complex number $p$
so that
\[ m(z) =\frac{(1+px)z - px^2}{pz + 1 -px}. \]
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{\bf Answer}

Since we are given a formula, we can check it directly: suppose
there are two such $p$s, call them $p_1,p_2$; and write:

$$m(z)=\ds\frac{(1+p_1x)z-p_1x^2}{p_1z+1-p_1x}=\ds\frac{(1+p_2x)z-p_2x^2}{p_2z+1-p_2x}$$

Then,

$$mm^{-1}(z)=\ds\frac{(1+p_1x-p_2x)z+p_2x^2-p_1x^2}{(p_1-p_2)z+(1-p_1x+p_2x)}$$

Since $mm^{-1}(\infty)=\infty$, the coefficient of $z$ in the
denominator is 0, and so $p_1=p_2$.

As for existence: for all $p \in \bf{C}$, all $x \in \bf{C}$.

$$m(z)=\ds\frac{(1+px)z-px^2}{pz+1-px}$$ is parabolic fixing $x$
(except $p=0$ when $m(z)=z$).

(Could also explicitly derive the formula for $m(z)$ directly from
the information given.)
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