Finding the transformation

Question

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

Answer 1

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 $$(u-2)^2+v^2=1$$

Answer 2

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 $1$.