\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Find a M\"obius mapping which transforms the points $z=0,\ -i,\ 1$
into $w=i,\ 1,\ 0$ respectively. In general how many points and
their images do you need to define a M\"obius map?


\medskip

{\bf Answer}

A M\"{o}bius mapping is $w=\ds\frac{\alpha z+\beta}{\gamma
z+\delta},\ \alpha,\ \beta,\ \gamma,\ \delta \in {\bf{C}}$ and
\un{finite}.

Thus we have

$$i=\ds\frac{\alpha \cdot 0 + \beta}{\gamma \cdot + \delta};\
1=\ds\frac{-i\alpha+\beta}{-i\gamma+\delta};\
0=\ds\frac{-\alpha+\beta}{-\gamma+\delta}$$

\hspace{1.5in} (1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ (3)

(3) $\Rightarrow \beta=\alpha$ (since can't have $\alpha,\ \beta,\
\gamma,\ \delta$ infinite sensibly)

Thus (3) in (1) gives

(1) $\Rightarrow \delta=\ds\frac{\beta}{i}=-i\alpha$

Hence

(2) $\Rightarrow -i\gamma+\delta=-i\alpha+\beta=-(1+i)\alpha
\Rightarrow \gamma=\ds\frac{-(1+i)\alpha-\delta}{-i}=i\alpha$

Thus

$w=\ds\frac{\alpha z+\alpha}{i\alpha
z-i\alpha}=\ds\frac{1}{i}\left(\ds\frac{z+1}{z-1}\right)=-i\left(\ds\frac{z+1}{z-1}\right)$

Note that $\alpha$ is irrelevant. Therefore in general we require
only \un{3} points and their images for uniqueness.

\end{document}