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{\bf Question}

Show that the transformation $\ds w=\frac{1+z}{1-z}$ maps the unit
disc $|z|<1$ conformally onto the half plane $u>0$, where
$w=u+iv$.  What is the image of the upper half of the unit disc?

Show that the transformation $w=\exp z$, where $z=x+iy$, maps the
half-strip $\Omega=\{(x,y):0<y<\pi, \, x<0\}$ conformally onto the
upper half of the unit disc.  how do the boundaries of $\Omega$
map?

Find a transformation which maps $\Omega$ conformally onto the
first quadrant.  Hence, or otherwise, obtain a solution $T(x,y)$
of Laplace's equation in $\Omega$ which satisfies the boundary
conditions

$T(x,0)=T(x,\pi)=0; \,\,\, T(0,y)=1.$


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{\bf Answer}

$\ds w=\frac{1+z}{1-z}=\frac{1+e^{i\theta}}{1-e^{i\theta}}=
\frac{e^{-i\frac{\theta}{2}}+e^{i\frac{\theta}{2}}}
{e^{-i\frac{\theta}{2}}-e^{i\frac{\theta}{2}}}=
\frac{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}}=$ $
i\cot\frac{\theta}{2}$

S0 $|z|=1\rightarrow$ imaginary axis.

Now $z=0\rightarrow w=1$, so $T$ maps $D$ to $U$.

For the upper half of $D$, $0<\theta<\pi$ on the boundary so
$\cot\frac{\theta}{2}>0$ thus the semicircle maps to the positive
imaginary axis.

Also $-1<z<1$ real $\Rightarrow w>0$ real.

So


DIAGRAM


$w=\exp z=e^xe^{iy}$

$y=0 \Rightarrow w=e^x$ real and $<1$ for $x<0$

$y=\pi \Rightarrow w=-e^x$ real and $>-1$ for $x<0$

$x=0, \hspace{0.2in} 0<y<\pi \Rightarrow w=e^{iy}$


DIAGRAM

$\ds w=\frac{1+e^z}{1-e^z}$ maps $\Omega \rightarrow$ first
quadrant.

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