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?
 A: 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.
A: To simplify, you could note that you only really need to compute the image of the upper half-circle and of the real segment [-1, 1], because your transformation is continuous. Other than that, the expression $x^2 + y^2 = 1$ (no need to take square roots there) will magically simplify : see also the wikipedia article on Möbius transformations. It's a general fact that these send circles to circles and lines to lines.
A: You have errors in your calculations of $T(-1), \space T(0)$
First verify that you can get $$T(-1)=-\frac{3}{5}-\frac{4}{5}i \quad T(0)=-\frac{i}{2}\quad T(1)=\frac{3}{5}-\frac{4}{5}i\quad T(i)=i$$
Next since you're looking for the mapping of the area between the real axis and the upper half unit disk, you first need to find the mapping of the real axis and the unit disk.  
Starting with the unit disk, observe that $T(1) = \frac{3}{5}-\frac{4}{5}i, \space T(i) = i, \space T(-1)=-\frac{3}{5}-\frac{4}{5}i\space$ These are points on the unit circle so the unit circle is mapped to itself. We went counter-clockwise from $1$ to $i$ to $-1$ and got to the points $\frac{3}{5}-\frac{4}{5}i,\space i,\space -\frac{3}{5}-\frac{4}{5}i\space$ also counter-clockwise. So the interior is mapped to itself. Alternatively to this "direction" approach you can take one interior (or exterior) point and check where it's mapped to. We know that $T(0)=-\frac{i}{2}$ so an interior point is mapped to an interior point which means that the interior area is mapped to the interior area.  
To check where the real axis is mapped to we look at $-1, 0, 1$ which are mapped to $-\frac{3}{5}-\frac{4}{5}i, \space -\frac{i}{2},\space \frac{3}{5}-\frac{4}{5}i$ respectively. So as we go over the real axis "from left to right" we get points on the circle in clock-wise direction. This means that points to the left of the real axis (the upper half plane) are mapped to points to the left of the circle (outside the circle).
So now you know that your image set is all the points inside the unit circle but outside the other circle.
To find that other circle plug in $z=t \in \mathbb{R}$ and represent the curve $h(t)$ with $x-y$ coordinates, take the derivative $h'(t)$ and check when the $y$ coordinate is $0$ which should be at $t=0, \infty$. Then you get $2$ points on the circle with distance $2r$. You should get the circle $|z+\frac{5}{4}i|=\frac{3}{4}$.  
Update
For $z=t\in \mathbb{R}$ we have $w=\frac{2t-i}{2+it} = \frac{3t}{4+t^2}-i\frac{2+2t^2}{4+t^2}$ so the image curve is given by $h(t)=(\frac{3t}{4+t^2},-\frac{2+2t^2}{4+t^2})$
Now you can compute $h'(t)$ and check where the second coordinate is zero which will give you $2$ points where the tangent to the circle is parallel to the real axis meaning these points are at distance $2r$ from each other where $r$ is the radius.
