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{\bf Question}

The boundary of a domain $D_1$ in the $z=x+iy$ domain is the
ellipse
$\left(\ds\frac{X}{5}\right)^2+\left(\ds\frac{y}{3}\right)^2=\ds\frac{1}{4}$
together with the line $y=0,\ |x|<2$. Show that an application of
the Joukowski transform $z=w+\ds\frac{1}{w}$, where $w=u+iv$ maps
$D_1$ onto two concentric circles in the $w$-plane, centred on the
origin, of radius 1 and 2. (Hint: Note that the notation of the
Joukowski transform is here the reverse of the lectures. Start
with the $w$-circles and work back to the $z$-plane).


\medskip

{\bf Answer}

The Joukowski tranform $z=w+\ds\frac{1}{w}$ maps $w$-circles to
$z$-ellipses. (NB notation is opposite of lecture.)

So if $z=w+\ds\frac{1}{w}$ then the circle $|w|=1$ corresponds to
the segment $y=0,\ |x|<2$ in the $z$-plane, as shown in the
lectures:

$\left.\begin{array} {rcl} x & = &
\left(R+\ds\frac{1}{R}\right)\cos \theta\\y & = &
\left(R-\ds\frac{1}{R}\right)\sin \theta \end{array} \right\}
\hspace{0.25in} \stackrel{R=1}{\longrightarrow} \hspace{.25in}
\left\{\begin{array}{rcl} x & = & 2\cos\theta\\ y & = & 0
\end{array} \right.$

The circle $|w|=R$ maps to the ellipse

$$\left(\ds\frac{x}{R+\frac{1}{R}}\right)^2+
\left(\ds\frac{y}{R-\frac{1}{R}}\right)^2=1$$

To get the given ellipse, we need
$R+\ds\frac{1}{R}=\ds\frac{5}{2},\ R-\ds\frac{1}{R}=\ds\frac{3}{2}
\Rightarrow R=2$, \un{as required}.

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