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

Let $$w=u+iv=\frac{z+1}{z-1}\implies z=\frac{w+1}{w-1}$$

Hence $$z=\frac{1+u+iv}{u-1+iv}\cdot\frac{u-1-iv}{u-1-iv}$$

The real part is $$\frac{u^2+v^2-1}{(u-1)^2+v^2}=2$$

And this simplifies to $$(u-2)^2+v^2=1$$