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\begin{document}

{\bf Question}

\begin{description}
\item[(a)]
Define carefully what is meant by a \emph{conformal map}, $w=f(z)$

\item[(b)]
Let $z=x+iy,\ w=u+iv$ and consider the Joukowski transformation

$$w=z+\ds\frac{1}{z}.$$

Show that this transformation maps the region $Im(z)>0,\ |z|>1$ to
the region $Im \>0$ (i.e., the shaded portions on the diagram
below).

PICTURE  \vspace{2in}

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\put(5,0){\vector(1,2){1}}

\put(1,0){\line(5,2){5}}

\put(7,0){\makebox(0,0)[l]{$Re\ w$}}

\put(3,3){\makebox(0,0)[bl]{$Im\ w$}}

\put(6,2){\makebox(0,0)[bl]{$w=u+iv$}}

\put(1,0){\makebox(0,0)[b]{$-2$}}

\put(1,0){\makebox(0,0)[t]{$C'$}}

\put(3,0){\makebox(0,0)[t]{$B'$}}

\put(5,0){\makebox(0,0)[t]{$A'$}}

\put(5,0){\makebox(0,0)[b]{$2$}}
\end{picture}

\vspace{2in}

\item[(c)]
By considering the imaginary part of the complex function
$$\alpha\log(w+2)+\beta\log(w-2)+\gamma,$$ where $\alpha,\ \beta,\
\gamma$ are real constants to be found, write down a harmonic
function $\phi$ which satisfies the boundary conditions

$$\phi(u,0^+)=\left\{\begin{array}{l}0, |u|>2\\ 1,|u|<2
\end{array} \right.$$

(Hint: take $-\pi<\arg(w+2)\leq \pi$ and $-\pi<\arg(w-2)\leq
\pi$.)

\item[(d)]
Hence solve the equation $\bigtriangledown^2F(x,y)=0$ in the
region $y \geq 0,\ |x^2+y^2| \geq 1$, subject to the boundary
conditions, leaving your answer in terms of $z=x+iy$

$$F(x,y)=\left\{\begin{array}{ll} 0, & y=0,\ |x|>1\\ 1, & y=0,\
|x|<1 \end{array} \right.$$
\end{description}


\medskip

{\bf Answer}

\begin{description}
\item[(a)]
A conformal map $f$ on ???? is one which preserves angles (and
also the sense of the angle). A differentiable function gives
conformal transformations, provided $f'(z) \ne 0$.

\item[(b)]
Consider Joukowski:

$$w=f(z)=z+\ds\frac{1}{z}$$

Take $|z|=1,\ Im(z)>0$ with $z=e^{i\theta},\ 0<\theta<\pi$

$w=e^{i\theta}+e^{-i\theta}=2\cos\theta;\ 0<\theta<\pi$

so $ABC \longrightarrow A'B'C'$

since $-2<2\cos \theta<2$

Take $Im(z)=0,\ Re(z)=x<-1$

Therefore $w=x+\ds\frac{1}{x}$ with runs between
$w=-1-\ds\frac{1}{1}=-2$ and
$w=-\infty+\ds\frac{1}{-\infty}=-\infty$

so $-\infty c z \to -\infty c'$

Likewise for $Im(z)=0,\ Re(z)=x>1$

$$A\infty \to A'\infty$$

Pick point in $Im(z)>0,\ |z|>1$ and see where it goes, e.g.,

$z=2i
\Rightarrow=2i+\ds\frac{1}{2i}=\left(2-\ds\frac{1}{2}\right)i=\ds\frac{3}{2}i$

which has $Im(w)>0$.

Thus transformation is as stated in question.

\item[(c)]
Imaginary part of

$$\alpha \log(w+2)+\beta\log(w-2)+\gamma,\ \alpha,\ \beta\ \gamma\
\rm{real}$$

is $\alpha\theta_1+\beta\theta_2+\gamma$

where $\theta_1$ and $\theta_2$ are defined by

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\put(7,0){\makebox(0,0)[l]{$u$}}

\put(3,3){\makebox(0,0)[bl]{$v$}}

\put(6,2){\makebox(0,0)[bl]{$w$}}

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\put(1.5,0){\makebox(0,0)[bl]{$\theta_1$}}

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$$\Phi(w)=\alpha\log(w+2)+\beta\log(w-2)+\gamma$$

is analytic, except at $w=\pm 2$.

Thus $Im(\Phi(w))$ must be harmonic, except at those points and
hence satisfies Laplace's equation in $(u,v)$.

To satisfy

$$\phi(u,0^+)=\left\{\begin{array}{ll}0, & |u|>2\\ 1, & |v|<2
\end{array} \right\}$$

we have on

\begin{description}
\item[(A)] $Re(u)>0,\ |u|>2;\ \theta_1=0,\ \theta_2=0$

$$\rm{Therefore}\ 0=\alpha \cdot 0+\beta \cdot 0+\gamma
\Rightarrow \un{\gamma=0}$$

\item[(B)] $Re(u)>0,\ |u|<2;\ \theta_1=0,\ \theta_2=\pi$

$$\rm{Therefore}\ 1=\alpha \cdot 0+\beta \cdot \pi+0 \Rightarrow
\un{\beta=\ds\frac{1}{\pi}}$$

\item[(C)] $Re(u)>0,\ |u|<2;\ \theta_1=0,\ \theta_2=0$

$$\rm{So\ same\ as\ above} \un{\beta=\ds\frac{1}{\pi}}$$

\item[(D)] $Re(u)>0,\ |u|>2;\ \theta_1=\pi,\ \theta_2=\pi$

$$\rm{Therefore}\ 0=\alpha \cdot \pi+\beta \cdot \pi \Rightarrow
\un{\alpha=-\ds\frac{1}{\pi}}$$
\end{description}

Therefore
$$\un{\Phi(w)=\ds\frac{1}{\pi}\log(w-2)-\ds\frac{1}{\pi}\log(w+2)}$$

$$\un{\phi=\ds\frac{\theta_2}{\pi}-\ds\frac{\theta_1}{\pi}
=\ds\frac{1}{\pi}\arctan\left(\ds\frac{v}{u-2}\right)
-\ds\frac{1}{\pi}\arctan\left(\ds\frac{v}{u+2}\right)}$$

\item[(d)]
Avoiding $z=0$ we have from theorem in notes that image of
harmonic $\phi$ in $w=f(z)$ is also harmonic.

So given that Joukowski transform

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

we have the boundary conditions of $F(x,y)$ mapping onto the
boundary conditions of $\phi(x,y)$.

Hence we have that

$Im(\Phi(w(z)))$ satisfies $\nabla^2F(x,y)=0$ in given region of
$z$ with boundary conditions.

Therefore

$Im[\Phi(w(z))]=Im\left[\ds\frac{1}{\pi}\log\left(z+\ds\frac{1}{z}-2\right)
-\ds\frac{1}{i}\log\left(z+\ds\frac{1}{z}+2\right)\right]$

$=Im\left[\ds\frac{1}{\pi}\log\left(x+iy+\ds\frac{1}{x+iy}-2\right)
-\ds\frac{1}{\pi}\left(x+iy+\ds\frac{1}{x+iy}+2\right)\right]$
\end{description}
\end{document}