<s>[INST] Finding the transformation I am trying to find the transformation of $x = 2$ under $T(z) = \frac{z+1}{z-1}$
My approach was as follows,
I put $u +iv = \frac{z+1}{z-1} $ and then rationalized the RHS.
I get $u = \frac{3-y^2}{1+y^2}$ and $v = \frac{4y}{1+y^2}$ 
I was trying to substitute $y = tan(\theta)$ and prove its a circle.
Thanks
Anupam
 [/INST] Note that if $\Re(z)=2$,
then $\forall y\in \mathbb{R}$,
$z=2+iy$
$$T(z)=\frac{2+iy+1}{2+iy-1}=\frac{3+iy}{1+iy}=\frac{(1-iy)(3+iy)}{1+y^2}=\frac{3+y^2-2iy}{1+y^2}=2+\frac{1-y^2-2iy}{1+y^2}$$
Now notice that:
$$|T(z)-2|^2=\frac {1}{(1+y^2)^2} ((1-y^2)^2+2^2y^2)=\frac{1+2y^2+y^4}{(1+y^2)^2}=1$$
Which means that the distance between $T(z)$ and the point $2$ doesn't depend on the value of $y$, so it is the circle centered at $2$ and has radius $1$.
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