Image of a unit disk under a Möbius transformation I have to find the image of the disk $|z| < 1$ under the Möbius Transformation $w(z) = \frac{iz-i}{z+1}$. This is how I approached it:
Since every Möbius function is determined uniquely by its action on three points. I took three points on the boundary of the disk (i.e. unit circle) $i, 1$ and $-i$ whose images were $-i, 0$ and $1$ respectively. 
Since Möbius transformations are conformal and the unit disk was on the right hand side of the three points, the image must be on the right hand side of the three images. This implies that the image must be the 4th quadrant, excluding the positive real axis and negative imaginary axis. 
Is this a valid solution? I was trying to attempt a lot of exercises like this during self-learning complex functions. Is my approach rigorous and valid enough?
 A: You want to find the image of $\vert z\vert<1$ under $T(z)=\frac{iz-i}{z+1}=\frac{i(z-1)}{z+1}$.
Take 3 boundary points: $1,i,-1$ and 
note that 


*

*$T(1)=0$

*$T(i)=-1$

*$T(-1)=\infty$
Also observe the direction of the points, first is $1$ then $i$ and then $-1;$ and the interior is in the LHS, thus following the same direction of the points $0,-1,\infty$, the interior should be also in the LHS.
Alternatively you could take an arbitrary point of the interior of $|z|<1,$ say $0$, and $T(0)=-i$ tells you where the image is.
Thus the image of $|z|<1$ is the lower half-plane.
A: Every Möbius transformation turns circles in the Riemann Sphere $\overline{\Bbb C}=\Bbb C\cup\{\infty\}$ to circles in the same space.
Now: how does a circle in $\overline{\Bbb C}$ look like? Distinguish between those who touch the point $\infty$ and those who don't: looking at these two types of circles just in the complex plane $\Bbb C$ (so making the choice of a chart... but just don't care of this sentence, it's more important to get the intuitive idea) they will look like straight lines in the former case and usual (euclidean) circles in the latter.
Now the image of the unit circle $C:=\{|z|=1\}$ under $z\mapsto T(z):=i\frac{z-1}{z+1}$ is a circle in $\overline{\Bbb C}$ touching $\infty$, so it is a straight line in $\Bbb C$. So it's enough to compute two points of the $T(C)$, e.g. $T(1)=0$ and $T(-i)=1$. Hence the image of $C$ is the real line and since we are interested in the inner part of $C$ we can just compute $T(0)=-i$ which belongs to $\{|z|<0\}$; so all the points of the disk $\Delta:=\{|z|<1\}$ are such: $T(\Delta)\subseteq\{|z|<0\}$. The inclusion $\supseteq$ follows by repeating the same argument for the inverse of $T$, which is still a Möbius transformation, inverting the role of the real line and of the unit circle $C$.
