Confusion in conformal mapping

Question

If we have some region $\mathbb{C}-[-1,1]$ in the complex plane, define a map $R(z) =\frac{1}{z}$ which takes the region $\mathbb{C}-[-1,1] \rightarrow \mathbb{C}-(-\infty,-1]\cup[1,\infty)$ denoted by $\Omega$. For some function $\frac{z+1}{z-1}$ on $\mathbb{C}-[-1,1]$, the new function on the region $\Omega$ is given by $\frac{1/z+1}{1/z-1}$. Why is this the case? Shouldnt it be just $\frac{1}{\bigg(\frac{z+1}{z-1}\bigg)}$??

Answer 1

If $z \in \mathbb C - (-\infty,-1] \cup [1,\infty)$ you first map it to a point of $\mathbb C - [-1,1]$ using the inverse of $\frac 1 z$ (which is $\frac 1 z$ itself!) and then apply the given function on $\mathbb C - [-1,1]$. So what you get $\frac {(\frac 1 z )+1} {(\frac 1 z )-1}$.