How can I know the mapping of the x and y axes in this analytic function?

Question

I am given the function (z+1)/(z-1)

It asks me what the mappings of the x and y axes would be. The answer is the x axis maps to the x axis excluding 1, and the y axis maps to the unit circle excluding 1. Can someone explain how to do this? I'm very unclear.

BTW, if it makes a difference, this is in a complex analysis book.

Answer 1

1. Define $f: D:=\mathbb R \setminus \{1\} \to \mathbb R$ by $f(x)=\frac{x+1}{x-1}$.

Then it is easy to show that $1 \notin f(D)$ and that to each $u \in D$ there is a unique $x \in D$ such that $f(x)=u$. This is your turn !

2. If $y \in \mathbb R$, then show that the equation $\frac{iy+1}{iy-1}=1$ has no solution. Furthemore show that $|\frac{iy+1}{iy-1}|=1$ .

If $w \in \mathbb C$ and $w \ne 1$, then show that there is a unique $y \in \mathbb R$ with $\frac{iy+1}{iy-1}=w$.

Answer 2

Let $f$ be your function.

1. If $x\in\mathbb R$, $f(x)=\frac{x+1}{x-1}=1+\frac2{x-1}$. So, it's just a matter of noticing that the image of the map$$\begin{array}{ccc}\mathbb{R}\setminus\{1\}&\longrightarrow&\mathbb R\\x&\mapsto&1+\frac2{x-1}\end{array}$$is $\mathbb{R}\setminus\{1\}$.
2. If $y\in\mathbb R$, then$$\bigl|f(yi)\bigr|=\left|\frac{yi+1}{yi-1}\right|=\frac{\sqrt{1+y^2}}{\sqrt{1+y^2}}=1.$$Therefore, $f(yi)$ belongs to the circle centered at $0$ with radius $1$. But $f(yi)$ cannot be $1$, because$$f(yi)=1\iff yi+1=yi-1.$$On the other hand,$$f(yi)=\frac{1-y^2}{1+y^2}+\frac{2y}{1+y^2}i.$$Can you take from here?