<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Find the real and imaginary parts of the function $\sinh z$, where

$z=x+iy$.  Find the images of the lines $x$=constant and $y$=
constant under the transformation $w=\sinh z$, identifying what
kinds of curves they are.

\item[b)]
Show that the points $z=1, \,\,\, z=-\frac{1}{2}$ are inverse
points with respect to the circle $C_1$ with centre $z=-1$ and
radius 1.

Denote the circle with centre $z=+1$ and radius 1 by $C_2$.

Find a Mobius transformation

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

which maps $z=1$ to $w=-1$, the circle $C_1$ to the line
$Re(w)=\frac{1}{2}$, and the circle $C_2$ to the line
$Re(w)=-\frac{1}{2}$.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
$\sin(x+iy)=\sinh x\cos y+i\cosh x\sin y=u+iv$

Therefore $u=\sinh x\cos y \hspace{0.5in} v=\cosh x\sin y$

So $x$=constant gives parametric equations for ellipses.

$\hspace{0.2in}y$=constant gives parametric equations for
hyperbolas.


\item[b)]

DIAGRAM

Now $A_1B_1=\frac{1}{2}, \,\,\,\, A_1A_2=2, \,\,\,\, A_10=1$
therefore $A_2$ and $B_1$ are inverse with respect to $C_1$.

Similarly $A_1$ and $B_2$ are inverse with respect to $C_2$.

$C_1$ maps to $L_1$ so $A_2,B_1$ map to image points in $L_1$.

$A_2$ maps to $A_1$ so $B_1$ maps to $w=2$.

$C_2$ maps to $L_2$ so $A_1,B_2$ map to image points in $L_2$.

$A_1$ maps to $A_2$ so $B_2$ maps to $w=-2$.

${}$

So we have

$\begin{array}{rr} z&amp;w\\ 1&amp;-1\\ -\frac{1}{2}&amp;2\\
\frac{1}{2}&amp;-2\end{array}$

So since $czw+dw-az-b=0$, we have

\begin{eqnarray} -c-d-a-b&amp;=&amp;0\\ -c+2d+\frac{1}{2}a-b&amp;=&amp;0\\
-c-2d-\frac{1}{2}a-b&amp;=&amp;0\end{eqnarray}

add 2 and 3, so $c+b=0$, then 1 and 2 give $a+d=0, \,\,\,
2d+\frac{1}{2}a=0$ giving $a=d=0$.  So the transformation is

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

\end{itemize}

\end{document}
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