1. Originally Posted by ThePerfectHacker

Here is another hint: after you get the half-disk map the half-disk by $\displaystyle (1-z)/(1+z)$.
$\displaystyle f_{1}(z) = z^{1/2}$ to get the half disc map
$\displaystyle f_{2}(z) = \frac{1-z}{1+z}$ this maps the half disc onto the 4th quadrant.
$\displaystyle f_{3}(z) = z^2$ maps the 4th quadrant onto $\displaystyle \Im(z) < 0$
$\displaystyle f_{4}(z) = iz$ to rotate the region onto $\displaystyle \Re(z) > 0$
$\displaystyle f_{5}(z) = \frac{1-z}{1+z}$ maps the region onto the unit disc.

$\displaystyle f_5 \circ f_4 \circ f_3 \circ f_2 \circ f_1 = \frac{1 - i \left( \frac{1 - \sqrt{z}}{1+\sqrt{z}} \right)^2}{1 + i \left( \frac{1 - \sqrt{z}}{1+\sqrt{z}} \right)^2}$ I'm sure that can be simplified is it any good TPH?

Bobak

2. Originally Posted by bobak
I'm sure that can be simplified is it any good TPH?
That looks correct.

I have ran out of lecture material, so we would have to end it here.
Hopefully you now know some stuff about conformall mappings.

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