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{\bf Question}

If $T:z\to\frac{z-i}{z+i}$ compute $T^{-1}$ and hence or otherwise
show that $T$ maps the upper half plane $\{z|{\rm Im}(z)>0\}$ onto
the unit disc $D=\{z|\,|z|<1\}$.  Show further that $T$ maps the
positive real axis to the lower half of the unit circle and the
negative real axis to the upper half of the unit circle.

Show that $z\to e^{\pi z}$ maps the strip $\{0<{\rm Im}(z)<1\}$
conformally onto the upper half plane and find the images of the
lines Im$(z)=0$ and Im$(z)=1$ under this transformation.

Hence find a conformal transformation which maps $D$ onto this
strip and takes the lower half circle to the real axis and the
upper half circle to the line Im$(z)=1$.



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{\bf Answer}

$\ds w=\frac{z-i}{z+i} \hspace{0.3in} z=i\frac{1+w}{1-w}$

If $z$ is real then $\ds w=\frac{x-i}{x+i} \hspace{0.3in}
\bar{w}=\frac{x+i}{x-i}$, so $w\bar{w}=1$

so $w$ is on the unit circle.

If $w=e^{i\theta}$, then $\ds
z=i\frac{(1+e^{i\theta})}{1-e^{i\theta}}=
i\frac{e^{-i\frac{\theta}{2}}+e^{i\frac{\theta}{2}}}
{e^{-i\frac{\theta}{2}}-e^{i\frac{\theta}{2}}}=
-\cot\frac{1}{2}\theta$ - real

so the real axis maps to $|z|=1$ and vice versa.

$z=i\to w=0$ so $U$ maps to $D$.

Now $-\pi<\theta<0 \Leftrightarrow -\cot\frac{1}{2}\theta>0$

And $0<\theta<\pi \Leftrightarrow -\cot\frac{1}{2}\theta<0$

So the lower half of the circle $\rightarrow$ positive real axis.

and the upper half of the circle $\rightarrow$ negative real axis.

$w=e^{\pi z}$ - analytic

$z=x\Rightarrow w=e^{\pi x}$ positive real axis

$z=x+i\Rightarrow w=e^{\pi x+i}=-e^{\pi x}$ negative real axis

$z=x+iy_0\Rightarrow w=e^{\pi x}e^{i\pi y_0}$ which is a ray from
0 at angle $\pi y_0$, which goes from 0 to $\pi$ as $y_0$ goes
from 0 to 1.  So the strip maps conformally to the upper half
plane.

$T^{-1}$ maps $D$ to $U$.

$\ds z=\frac{1}{\pi}\log w$ maps $U$ to $S$

$\ds z=\frac{1}{\pi}\log i\frac{1+w}{1-w}$ maps $D$ to $S$

and maps the halves of the circle in the required fashion.


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