<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
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\begin{document}
%PREVIOUSLY QUESTION 5%
{\bf Question}

Show that the transformation

$$w=\ds\frac{i-z}{i+z}$$

maps the upper half plane $Im(z)&gt;0$ into the interior of the unit
circle, centred on the origin of the $w$-plane. Hence, solve
Laplace's equation $\nabla ^2 F=0$ inside the circle
$w=exp(i\theta)$, when the values of $F$ on the circumference are

$$F=\left\{\begin{array} {rccccl} 1, &amp; 0 &amp; &lt; &amp; \theta &amp; &lt; &amp;
\pi\\0, &amp; \pi &amp; &lt; &amp; \theta &amp; &lt; &amp; 2\pi \end{array} \right.$$


\medskip

{\bf Answer}

$w=\ds\frac{i-z}{i+z}$

Consider the \un{inverse} mapping $w \rightarrow z$.

If $|w|=1$,\ i.e., the unit circle,

then

$$|w|=\left|\ds\frac{i-z}{i+z}\right|=1$$

i.e., $|i-z|=|i+z|$

This is the locus of points $z$ which are equidistant from $i$ and
$-i$, i.e., the line $Im(z)=0$.

(w)

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\put(1,1){\makebox(0,0)[bl]{$Im$}}

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\end{picture} \hspace{.25in}
$\stackrel{z(w)}{\longrightarrow}$ \hspace{.25in}
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\vspace{1in}

So the map $z \to w$ must take $Im(z)=0$ onto $|w|=1$ as required.

What about interior points? Pick one $z$-value $z=+i$ say $w(i)=0$
i.e., $|w(i)|&lt;1$

Therefore upper $Im(z0 \geq 0 \longrightarrow |w| \leq 1$

\newpage

In particular, consider $z=x+iy$ with $y=0,\ x&gt;0$

$w(0)=1,\ w(+\infty)=-1$

$w(1)=\ds\frac{i-1}{i+1}=\ds\frac{-(1-i)(1-i)}{(1+i)(1-i)}=-\ds\frac{1}{2}(-2i)=i$

Now for $y=0,\ x&lt;0$

$w(0)=1,\ w(-\infty)=-1$

$w(-1)=\ds\frac{i+1}{i-1}=-i$

Thus

PICTURE \vspace{2in}

Thus if $\phi_{(x,y)}=1$ on $OAB$,\ $F_{(u,v)}=1$ on $0'A'B'$

and if $\phi(x,y)=0$ on $OCD$,\ $F(u,v)=0$ on $0'C'D'$

Thus the boundary conditions on $F$ are equivalent to solving the
boundary conditions of $\phi$ of Q4A in (z).

Thus if we can solve $\nabla^2\phi=0$ in $z$ we map back to get
back $\nabla^2 F=0$ in $w$.

But we have solved $\nabla^2\phi=0$ + boundary conditions in Q4A.

Hence

$$\phi(x,y)=1-\ds\frac{1}{\pi}\tan^{-1}\left(\ds\frac{y}{x}\right)$$

If we can find $x(u,v),\ y(u,v)$ we have solved the problem
\un{OR} we can say

$\phi(x,y)=Im\left(1-\ds\frac{1}{\pi}\log z\right)$

So \begin{eqnarray*} F(u,v) &amp; = &amp; Im\left(1-\ds\frac{1}{\pi}\log
z(w)\right)\\ &amp; = &amp;
Im\left(1-\ds\frac{1}{\pi}\log\left[i\left(\ds\frac{1-w}{1+w}\right)\right]\right)
\end{eqnarray*}

\un{or}

if $z=i\ds\frac{(1-w)}{(1+w)}$

a little algebra gives

$$x=\ds\frac{2v}{(1+u)^2+v^2},\
y=\ds\frac{1-(u^2+v^2)}{(1+u)^2+v^2}$$

so $F(u,v)=\phi(x(u,v),\ y(u,v))$

$\Rightarrow
\un{F(u,v)=1-\ds\frac{1}{\pi}\arctan\left[\ds\frac{2v}{1-(u^2+v^2)}\right]}$

\end{document}
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