\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Show that if $\alpha$ has a positive imaginary part then the
transformation

$$z\to\frac{z-\alpha}{z-\bar{\alpha}}$$

maps the upper half plane onto the unit disc $D=\{z|\,|z|<1\}.$

Hence find a transformation $T$ which maps the half plane
$\{z=x+iy|x\leq\frac{1}{2}\}$ onto $D$ and maps 0 to 0 and
$\infty$ to $-1$.

Find the image of the strip $\{z=x+iy|0\leq x\leq\frac{1}{2}\}$
under $T$.




\vspace{0.25in}

{\bf Answer}

Let $\ds w=\frac{z-\alpha}{z-\bar{\alpha}}$, if $z=x$ - real then
$\ds w=\frac{x-\alpha}{x-\bar{\alpha}}=
\frac{x-\alpha}{\bar{x}-\bar{\alpha}}$, so $|w|=1$.

Conversely if $|w|=1$ then $|z-\alpha|=|z-\bar{\alpha}|$  i.e. $z$
is equidistance from $\alpha$ and $\bar{\alpha}$ and is therefore
real.

So the transformation maps the real axis to the unit circle.

Now $w=0$ is the image of $z=\alpha$, so if im$\alpha>0$, the
interior of $U$ maps to the interior of $D$.


\setlength{\unitlength}{0.5in}
\begin{picture}(5,2.5)
\multiput(2,1)(0.2,0){10}{\line(1,0){0.1}}

\multiput(3,0)(0,0.2){10}{\line(0,1){0.1}}

\put(3.5,0){\line(0,1){2}}

\put(3.6,0.75){\makebox(0,0){$\frac{1}{2}$}}

\put(2,1.7){\line(1,1){0.3}}

\put(2,1.3){\line(1,1){0.7}}

\put(2,0.9){\line(1,1){1.1}}

\put(2,0.5){\line(1,1){1.5}}

\put(2,0.1){\line(1,1){1.5}}

\put(2.3,0){\line(1,1){1.2}}

\put(2.7,0){\line(1,1){0.8}}

\put(3.1,0){\line(1,1){0.4}}
\end{picture}

$z\to z-\frac{1}{2} \hspace{0.5in} z\to-iz$  The composite of
these two maps sends $H$ to$U$.

So $\ds w=\frac{-i\left(z-\frac{1}{2}\right)-\alpha}
{-i\left(z-\frac{1}{2}\right)-\bar{\alpha}}$ maps $H$ to $D$.

$\ds z=0 \rightarrow w=0 \Leftrightarrow \frac{1}{2}i-\alpha=0$
i.e. $\ds\alpha=\frac{1}{2}i$.

Then $\ds w=\frac{-iz}{-iz+\frac{1}{2}i}$

Now under this transformation $z=\infty\rightarrow w=1$

So a reflection will make $z=\infty\rightarrow w=-1$

i.e. $w\rightarrow -w$

So $\ds w=\frac{iz}{-iz+\frac{1}{2}i}$

does all that is required.

\end{document}