I'll need your help to find the image of the area $C_1 \;$and $\,C_2$ under a Möbius transformation
$$C_1 = \left\{x>0 \, , y>0\right\}\;$$(the first quadrant)
of the Möbius transformation :
$$ z \longmapsto \frac{z-i}{z+i}$$
And
$$C_2 = \left\{|z|< 1 \, , Im(z) >0\right\}$$(semicircle)
of the Möbius transformation :
$$ z \longmapsto \frac{2z-i}{iz+2}$$
Thanks in advance for your help.
On
For the first transformation, let $w=\frac{z-i}{z+i}$. Then, after rearranging, we get $$z=i\frac{-w-1}{w-1}$$ Therefore, to find the image of $C_1$ under the map $z\mapsto w$, we have to solve \begin{align} \operatorname{Re}\left(i\frac{-w-1}{w-1}\right)&=\operatorname{Re}z>0\text{ and}\\ \operatorname{Im}\left(i\frac{-w-1}{w-1}\right)&=\operatorname{Im}z>0. \end{align} Multiply both the numerator and the denominator by the complex conjugate of $w-1$ to obtain \begin{align}\frac{-w-1}{w-1}\cdot\frac{\overline{w-1}}{\overline{w-1}}&=\frac{1}{|w-1|^2}(-w-1)(\overline w - 1)=\frac{1}{|w-1|^2}(-|w|^2+w-\overline w+1)\\ &=\frac{-|w|^2+1}{|w-1|^2}+2i\frac{\operatorname{Im}w}{|w-1|^2} \end{align} Because the whole thing is multiplied by $i$, we get that \begin{align} \operatorname{Re}\left(i\frac{-w-1}{w-1}\right)&=\color{red}{2}\frac{\operatorname{Im}w}{|w-1|^2}>0\\ \operatorname{Im}\left(i\frac{-w-1}{w-1}\right)&=\frac{-|w|^2+1}{|w-1|^2}>0. \end{align} This simplifies to $\operatorname{Im}w>0$ and $|w|^2<1$. The second inequality simplifies further to $|w|<1$, therefore, the image is precisely $C_2$.
There is a tiny mistake above: the red 2 should be -2, implying $\operatorname {Im} w<0$, and not greater than, meaning that the image of $C_1$ is the complex conjugate of $C_2$. Fortunately, this doesn't change much from below, so I'll leave it as is, with a couple of notes.
Using the same approach with the second transformation leads to a tricky system of inequalites (try it yourself). However, it is very convenient that the image of $C_1$ under the first transformation $z\mapsto w$ is exacty the region $C_2$ which we want to consider in the second transformation.
*(Note: as mentioned above, the image is not $C_2$ but its complex conjugate $\overline{C_2}$, but let's continue.)
So let $w\mapsto u$ denote the second transformation, i.e. $$u=\frac{2i-w}{iw+2}\iff w=\frac{-2u-i}{iu-2}.$$ We want to figure out the image of $C_2$ under $w\mapsto u$, which is going to be the image of $C_1$ under the composition $z\mapsto w\mapsto u$.
*(Note: what we actually want is $z\mapsto\overline w\mapsto u$, because $z\mapsto\overline w$ maps $C_1$ onto $C_2$, so we should be substituting $\overline w = \frac{z-i}{z+i} \iff w = \frac{\overline{z-i}}{\overline{z+i}}$ below instead)
So we want to express $u$ in terms of $z$:
$$u=\frac{2i-\left(\frac{z-i}{z+i}\right)}{i\left(\frac{z-i}{z+i}\right)+2}=\frac{(2-i)z+(1-2i)}{(2+i)z+(1+2i)}\iff z=\frac{-u(2i+1)+(1-2i)}{u(2+i)+(-2+i)}$$ and find the real and imaginary part of the right-hand side by multiplying the numerator and the denominator with the complex conjugate of the denominator. Because we will be solving the inequalities $\operatorname{Re}w>0$ and $\operatorname{Im}w>0$, we only care about the numerator, so we have to consider \begin{align}(-u(2i+1)+(1-2i))(\overline u(2-i)+(-2-i))&=-(4+3i)|u|^2+5i(u-\overline u)-4+3i\\ &=-(4+3i)|u|^2+5i(2i\operatorname{Im}u)-4+3i\\ &=-(4+3i)|u|^2-10\operatorname{Im}u-4+3i\\ &=(-4|u|^2-10\operatorname{Im}u-4)+i(-3|u|^2+3) .\end{align}
Finally, the inequalities that we have to solve are \begin{align} -4|u|^2-10\operatorname{Im}u-4>0\\ -3|u|^2+3>0. \end{align}
The second inequality gives $|u|^2<1$, whichi is equivalent to $|u|<1$. The first inequality we can rewrite as $$\operatorname {Re}(u)^2+\operatorname {Im}(u)^2+\frac52\operatorname {Im}(u)+1<0$$ and we can complete the square for $\operatorname {Im}(u)$ to obtain $$\operatorname {Re}(u)^2+\left(\operatorname {Im}(u)+\frac54\right)^2-\left(\frac54\right)^2+1<0$$ equivalent to $$\operatorname {Re}(u)^2+\left(\operatorname {Im}(u)+\frac54\right)^2<\left(\frac34\right)^2$$ Therefore, the image of $C_2$ is the intersection of the open unit disk, and the open disk centered at $-i\frac54$ with radius $\frac34$. This intersection is clearly nonempty.
$G$ is where the image should be , under the x-axis and inside the circle
$G$ is where the image should be , at the intersection