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!

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...
$y = x-1 \quad \text{or equivalently } \Im (z) = \Re (z) -1$
