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!

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.
