Conformal mapping of a semi-circle and a finite line Can I map a semi-circle and a finite line separated by a distance $h$ to two parallel lines?
Since I am new to con-formal mapping, I used the $w=atan(z$) con formal function but I guess this is for infinite line case, and the circle formed is not exactly a semi-circle. Can please someone help in this regards.What other con-formal function can be used or normalization added to $atan(z)$ function. THANKS.

 A: The semicircle can be transform into a line segment by a composition of linear fractional transformations.
The first goal is to transform your semicircle into a circle of the form $$x^2+(y-1)^2=1$$ with $-1\le x \le 1$ and $1 \le y \le 2$. This can be accomplished by translating your semicircle's center to the origin, rescaling the semicircle to a radius of 1, rotating the semicircle so that it is opening downwards and then translating again so that the center is at $(0,1)$.
From here you can perform an inversion. This will make a  line segment. Since $$\frac{1}{z} = \frac{\bar z}{|z|^2} = \frac{x-iy}{x^2+y^2} = \frac{x-iy}{x^2+(y-1)^2+2y-1} = \frac{x-iy}{2y} = \frac{x}{2y} - i\frac{1}{2}.$$
Now we have a line segment along $a-(0.5)i$, which is parallel to the real line. The bounds on $a$ can be determined by examining the bounds on the semicircle.

This procedure works provided you have only a semicircle to convert to a line segment.
For your problem in particular, you can arrange to have the line transformed to be along the real axis. The semi circle wont be the upper half of the circle we described above, but it will still be part of that circle. The semi-circle will still be a transformed to a line segment parallel to the real axis after inversion. Note that the line (after rotated and translated to lie along the real axis) will be unchange by inversion.
However for this to work as described the radius of the semicircle must be $H$ as well.
