Describe the image of the set $\{z:|z|<1, Im(z)>0\}$ under the mapping $w =\frac{2z-i}{2+iz}$ Describe the image of the set $\{z:|z|<1, Im(z)>0\}$  under the
mapping $w =\frac{2z-i}{2+iz}$
First I need to find the inverse which is $z=\frac{2w+i}{2-iw}$.
Now let $w=u+iv$, we have 
$$z=\frac{2w+i}{2-iw}=\frac{2u+2iv+i}{2+v-iu}$$
From this I get $x= \frac{3u}{(2+v)^2 +u^2}$ and $y=\frac{5v+2u^2 +2v^2+2}{(2+v)^2 +u^2 }$
Since $Im(z)>0$, $5v+2u^2 +2v^2+2>0$ and $|z|<1$ so $ \sqrt{x^2+y^2} <1$.
$$x^2 +y^2<1$$
$$(\frac{3u}{(2+v)^2 +u^2})^2 +(\frac{5v+2u^2 +2v^2+2}{(2+v)^2 +u^2 })^2 <1$$
$$3u^4+3v^4+20u^3+8v^3+6u^2v^2+8u^2v+4v^2-3v-12<0$$
Now I'm stuck, so I tried Mr. Blatter method and got$T(-1)=\frac{-2-i}{2+i}$, $T(0)=1$, $T(1)=\frac{2-i}{2+i}$, $T(i)=i$. So is this telling me that the image is the left side?

The Moebius transformation
$$T:\quad z\mapsto w:={2z-i\over 2+iz}$$
maps "circles" (i.e., circles or lines) in $\bar{\Bbb C}$ onto "circles". Since "circles" are determined by three points on them it is sufficient to compute the image points of three points on the unit circle and of three points on the real axis.  Computing the four complex numbers $T(-1)$, $T(0)$, $T(1)$, and $T(i)$ will do the job.  Furthermore $T(0)$, resp. $T(i)$, will tell you on which side of the two image "circles" the upper half of the unit disk will be mapped.
One computes
$$T(-1)={-3-4i\over5},\quad T(0)=-{i\over2},\quad T(1)={3-4i\over5},\quad T(i)=i\ .$$
Since $T(-1)$, $T(1)$, and $T(i)$ are lying on the unit circle $\partial D$ we know that $T$ maps $\partial D$ onto $\partial D$. Since both $0$ and $T(0)$ lie in the interior of $\partial D$ we can conclude that $T$ maps the unit disk $D$ onto $D$.
$T$ maps the real axis onto the circle $\gamma$ through the three points $T(-1)$, $T(0)$, $T(1)$ in the lower half plane. The point $i$ is in the upper half plane, and $T(i)$ is in the exterior $E$ of $\gamma$. This implies that $T$ maps the upper half plane $H$ onto $E$.
The upper semidisk in question is the intersection $D\cap H$. Its image is then the intersection  $D\cap E$  – an Apple-logo turned ${\pi\over2}$ clockwise.
