Help understanding the image of complex transformation So in this example, I have a semi-circle of radius 1 given by |z| <= 1 and Im Z >= 0
and I need to apply two transformations: 
a) $\frac{1}{z+1}$
b) $\frac{i}{z+1}$
the solutions for this can be seen in the image below: 
https://imgur.com/a/6kHzNNO
I understand the first step where z + 1 is applied, but I don't undertstand how is it applied the final step of the $\frac{1}{z+1}$. Can someone explain me this? 
Thank you 
 A: We want to find the image of the half-circle $S_+=\{|z-1|\leq1\}\cap\{Im(z)\geq0\}$, under the map $z\to 1/z$. This is a Moubius transformation, and we know that it transforms circles and lines nicley, and preserves angles. So it's a good idea to look at the image of the whole circle, $S={|z-1|\leq 1}$ and then think what happens to each side. Note further, that it is enough to see where the boundary goes (and where one point from the interior goes). Now, because the boundary $\partial S=\{|z-1|=1 \}$ contains the point $0$ which goes to infinity in $\hat{\mathbb{C}}$, we know that it must be mapped to a line. This line must be orthogonal to the image of the real axis, which gets mapped to itslef. As the point $2$ is also on $\partial S$, we know that this line contains the point $1/2$; and so it is the line $l=\{z|z=x+iy,x=1/2\}$.
Now, Let's what where $S_+$ goes. The boundary of $S_+$ is composed of the arc $A_+=\partial S\cap\{Re(z)\geq 0\}$  and the segment $[0,2]$. Let's see where each of them goes:
$A_+$ is a connected components of $\partial S$ and so it must go to a connected component of the image of $\partial S$, $l$. moreover it contains the point $2$ which goes to $1/2$, and the point $0$ which goes to infinity; So it must go to one of the rays $l_+=\{z|z=x+iy,x=1/2,y\geq0\}$ or $l_-=\{z|z=x+iy,x=1/2,y\leq 0\}$. Picking any point would show that it is the later line.
$[0,2]$ is much easier to handle: simple calculus shows that it is mapped to $[-1/2,\infty]$.
To sum up: the boundary of $S_+$ goes to the boundary of the domain which is specified in the image you sent. Now we know that $S_+$ goes to a connected component of $\mathbb{C}$, so it must goes to the gray area, or to the white area. But the point $1$ goes to $1$, so it is the first option.
I hope everything makes sense to you - if I used any statement which you are not familiar with I'll be happy to provide further info :)
