Mobius image of a line I've been looking at this question;
Let gamma be the mobius transformation defined by:$$\gamma(z) = \frac{2iz-2}{z+1}, z\ne -1$$
Show that $\gamma$ maps the line $\{z: Im z = Re z - 1\}$ into the circle $C(i,1)$.
I've gone about this in a similar method to an example in my notes however I'm not sure if I have the correct answer or not, and if I have I can't really see what I am doing if you understand what I mean.
So I have;
$\gamma(z) \in C(i,1) \Leftrightarrow |\gamma(z)-i|=1$ 
$\\\Leftrightarrow|\frac{2iz-2}{z+1} -i|=1 \\\Leftrightarrow |iz-2-i|^2=|z+1|^2 \\\Leftrightarrow x^2-2x+y^2+4y+5=x^2+2x+y^2+1 \\\Leftrightarrow x=1$
Writing this out I just have no idea what this is supposed to be... Thanks for any help!
 A: Let $$w=\frac{2iz-2}{z+1}\implies z=-\frac{w+2}{w-2i}$$
Now write $z=x+iy$ and $w=u+iv$
Then after a couple of lines of algebra, we get $$x=-\frac{u^2+v^2-2v+2u}{u^2+(v-2)^2}$$ and $$y=\frac{-2u+2v-4}{u^2+(v-2)^2}$$
Now substitute these into the given line equation $y=x-1$ and after some simplification we get $$u^2+(v-1)^2=1$$ which is the circle you are given.
A: If you are allowed to use that these transformations keep clines (they map (circles or lines) into (circles or lines)), then it's enough to check the image of 3 points from the line, as 3 points (the images of the ones you just picked) uniquely determine a cline. Pick easy ones, if you can.
For example -i, 1, 2 + i come to mind.
Their images are, in order: $0, -1 + i, \frac{2i(2+i) - 2 }{3+i} = \frac{4i - 4}{3+i}$.
Check the distance of these points from $i$. 
$$|0-i| = 1,\quad |-1 + i - i| = 1,\quad  | \frac{4i - 4 - i(3+i)}{3+i} | = |\frac{3+i}{3+i}| = 1 $$
So as you can see, the image of $3$ points from the line are on the unit circle around $i$. Since the image of the line is a cline, you know that it can only be a circle. And if two circles meet in 3 points, then they are the same.
For the slightly more numbery approach:
$\\...\\\Leftrightarrow|\frac{2iz-2}{z+1} -i|=1 \\\Leftrightarrow |iz-2-i|^2=|z+1|^2$ 
Up to this point you did well. Now comes the step where you write $\mathbb C \ni z = x + iy$.
Thus $x = \Re (z), y = \Im (z)$. So...
$|i(x+iy) - 2 - i| = | x + iy + 1|$ 
Then group by real and imaginary parts...
$|(-2 -y) + i(x-1)| = | (x+1) + i y|$ 
Now realize that these are lengths of vectors...
$(-2-y)^2 + (x-1)^2 = (x+1)^2 + y^2$
Simplify...
$(2+y)^2 - y^2 + (x-1)^2 - (x+1)^2 = 0 \Leftrightarrow 2(2+2y) + (-2)(2x) = 0 \Leftrightarrow (1+y) - x = 0$
And finish up...
$y = x-1 \quad \text{or equivalently } \Im (z) = \Re (z) -1$
