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Let $f(z)=\frac{z+1}{z-1}$. I shown that $f:\mathbb C\backslash \{1\}\longrightarrow \mathbb C$ is bijective and it's inverse if $f$ him self. I want to calculate $f(i\mathbb R)$. I have the impression that I'm doing the reverse work, could you check it and tell me what's wrong.

1) Let $z\in f(i\mathbb R)$. Then $z=f(it)$ for a certain $t\in\mathbb R$. Therefore $it=\frac{z+1}{z-1}$ and thus $\Re\left(\frac{z+1}{z-1}\right)=0$. Then, $$\frac{1}{2}\left(\frac{z+1}{z-1}+\frac{\bar z+1}{\bar z-1}\right)=0\implies |z|=1,$$ therefore $f(i\mathbb R)\subset S_1(0)\backslash \{1\}$ the circle of center $0$ and radius $1$.

2) Conversely, let $w\in S_1(0)\backslash \{1\}$. Then $w\bar w=1$. $$w=f(z)\implies z=f(w)\implies z=\frac{w+1}{w-1}=\frac{\frac{1}{\bar w}+1}{\frac{1}{\bar w-1}}=-\frac{\bar w+1}{\bar w-1}=-z$$ and thus $z\in i\mathbb R$. Therefore $w\in f(i\mathbb R)$.

Strangely, in both case (particularly in 2)), I have the impression that I'm doing the reverse work since I suppose que $w=f(z)$ and I show that $z\in i\mathbb R$... it's not correct, right ?

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1) is correct. 2) is indeed strange for the redaction. So you do what you did in your draft. Then you do as following. Let $\omega \in \mathcal S_1(0)\backslash \{1\}$. In particular, if $z=\frac{\omega +1}{\omega -1}$ then $\omega =f(z)$. Moreover, $\bar z=-z$ and thus $z\in i\mathbb R$. Therefore, $\mathcal S_1(0)\subset f(i\mathbb R)$.