\documentclass[12pt]{article}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Find the images, in the $w$-plane, of lines parallel to the real
and imaginary axes in the $z$-plane, under the transformation
$w=e^z$.  Explain how this illustrates the concept of
conformality.


\item[b)]
Show that any Mobius transforamtion mapping the upper half plane
im$(z)\geq0$ into the upper half plane im$(w)\geq0$ must be of the
form

$$w=\frac{\alpha z+\beta}{\gamma z+\delta},$$

where $\alpha, \,\, \beta, \,\, \gamma, \,\, \delta$ are all real
and $\alpha\delta-\beta\gamma>0$.  Deduce the general form of
Mobius transformation mapping im$(z)\geq0$ onto the right hand
half plane re$(w)\geq0$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $w=e^z$ and write $z=x+iy$.  Then $w=e^{x+iy}=e^xe^{iy}$.

For $x$ constant, as $y$ varies over {\bf R}, $w$ traces round the
circle centre $O$ radius $e^x$ in the $w$ plane, infinitely many
times.

For $y$ constant, as $x$ varies over {\bf R}, $w$ traces the ray
from $O$ (but not including $O$) making an angle $y$ (mod$2\pi$)
with the positive real axis.

Lines parallel to the real and imaginary axes in the $z$-plane are
orthogonal, as are circles centre $O$ and rays from $O$ in the
$w$-plane.

This illustrates the angle-preserving property which is that of
conformality.


\item[b)]
The required transformations must map the real axis to the real
axis (boundary $\rightarrow$ boundary).

Thus a pair of finite non-real conjugate points map to a pair of
finite non-real conjugate points.  Hence $z=\infty$ maps onto the
real axis and $w=\infty$ is the image point on the real axis.

\begin{itemize}
\item[i)]
If $\infty\to\infty$ then $C=0$, so $d\not=0$ and $w=\alpha
z+\beta$.

$z=0\to$ real $w$ so $\beta$ is real.

then $z=1\to$ real $w$ so $\alpha$ is real.

\item[ii)]
If $C\not=0$, $\ds w=\frac{Az+B}{z+D}$

$z=-D\to w=\infty$ so $-D$ must be real.

$\ds D=0 \Rightarrow w=A+\frac{B}{z}$

$z=\infty\to$ real $w$ so $A$ is real.

then $z=1\to$ real $w$ so $B$ is real.

$D\not=0 \Rightarrow z=0\to\frac{B}{D}$ - real so $B$ is real.

then $z=1\to w$ real so $A$ is real.

Thus in all cases the transformation has the form

$\ds w=\frac{\alpha z+\beta}{\gamma z+\delta}, \,\,\, \alpha, \,\,
\beta, \,\, \gamma, \,\, \delta$ real

when $z=i$, im$w>0$,

$\ds\frac{\alpha z+\beta}{\gamma z+\delta}=
\frac{(\alpha\delta-\beta\gamma)i+
(\alpha\gamma+\beta\delta)}{\gamma^2+\delta^2}$ so
$\alpha\delta-\beta\gamma>0$.

\end{itemize}


Now $\ds w=e^{-i\frac{\pi}{2}}\frac{\alpha z+\beta}{\gamma
z+\delta}$ maps im$z\geq0$ onto re$z\geq0$.

Conversely if $\ds w=\frac{az+b}{cz+d}$ maps im$z\geq0$ to
re$w\geq0$ then $\ds w=e^{i\frac{\pi}{2}}\frac{az+b}{cz+d}$ maps
im$z\geq0$ to im$w\geq0$.

So $e^{i\frac{\pi}{2}}a=\alpha$- real

$a=\alpha e^{-i\frac{\pi}{2}}$

So $\ds w=e^{-i\frac{\pi}{2}}\frac{\alpha z+\beta}{\gamma
z+\delta}$

$\alpha, \,\, \beta, \,\, \gamma, \,\, \delta$ are all real and
$\alpha\delta-\beta\gamma>0$.

\end{itemize}

\end{document}