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{\bf Question}

\begin{itemize}
\item[a)]
Find the Mobius transformation

$$w=\frac{az+b}{cz+d}$$

which maps the disc $|z|<1$ onto the disc $|w-1|<1$, and which
maps the points $z=0, \,\, z=1$ onto the points $w=\frac{1}{2},
\,\, w=0$ respectively.

\item[b)]
Find the real and imaginary parts of the function $\cos z$, where
$z=x+iy$.

Find the images of the lines $x=$constant under the transformation
$w=\cos z$, and use this to show that the transformation maps the
infinite strip

$$0<x<\pi, \hspace{0.3in} y>0$$

onto the lower half-plane ${\rm im} w<0$.
\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]

${}$

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\put(7.3,1.7){\makebox(0,0){$w$-plane}}
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The transformation must map the boundaries of the discs onto one
another.  Inverse pairs map to inverse pairs of points, so

$z=0\to w=\frac{1}{2} \Rightarrow z=\infty\to w=-1$

Thus $\frac{b}{d}=\frac{1}{2}$ and $\frac{a}{c}=-1$.

Now $z=1\to w=0 \Rightarrow a+b=0$

So we have $b=-a, \,\, c=-a, \,\, d=-2a$.

Hence $\ds w=\frac{z-1}{-z-2}$


\item[b)]
$\cos(x+iy)=\cos x\cos iy-\sin x\sin iy=\cos x\cosh y-i\sin x\sinh
y$

So $u=\cos x\cosh y$ and $v=-\sin x\sinh y$

For each $x$=constant this gives parametric equations for
hyperbolae.  Now consider $0<x<\pi$.

For $x=\frac{\pi}{2} \,\,\, \cos x=0 \,\,\, \sin x=1$

So $u=0, \,\,\, v=-\sinh y<0$ for $y>0$

So $x=\frac{\pi}{2} \,\,\, y<0$ maps to the negative imaginary
axis $u=0, \,\,\, v<0$.

For $0<x<\frac{\pi}{2} \,\,\, \cos x>0$ and $\sin x>0$.

For $y>0 \,\,\, \cosh y>0$ and $\sinh y>0$.

So lines $x$=constant and $y>0$ in this range map to hperbolas in
the fourth quadrant.

Similarly for $\frac{\pi}{2}<x<\pi$ the images are parts of
hyperbolas in the third quadrant.


DIAGRAMS


\end{itemize}

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