Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. I know it is supposed to be $\Re z <{1\over 2}$, but unfortunately I've been trying to show this for too long and my sanity along with my concentration drops monotonically. I would really appreciate any help with this.

Let $z=x+iy$ with $x,y\in\mathbb{R}$. We have
$$\frac{z}{z-1}=\frac{x+\mathrm{i}y}{(x-1)+\mathrm{i}y}=\frac{x(x-1)+y^2}{(x-1)^2+y^2}-\mathrm{i}\frac{y}{(x-1)^2+y^2}$$
so
\begin{align}
\Re\frac{z}{z-1} & =\frac{x(x-1)+y^2}{(x-1)^2+y^2}\\
& =\frac{x^2+y^2-x}{x^2+y^2-2x+1}\\
& < \frac{1-x}{2-2x}\\
& =\frac{1}{2}\frac{1-x}{1-x}\\
& =\frac{1}{2}
\end{align}
whence $f\left(D\right)\subset\{z\in\mathbb{C}\mid \Re z<\frac{1}{2}\}$.
Now take $\omega=\left(\frac{1}{2}-\varepsilon\right)+\mathrm{i}y$ with $\varepsilon>0$ and $y\in\mathbb{R}$. We can check that $f$ is an involution, that is $f$ is bijective and $f^{-1}=f$, so we see that we can choose $z=\frac{\omega}{\omega-1}$ to have $f(z)=\omega$. Moreover, this choice is licit since
\begin{align}
|z| & =\left|\frac{\omega}{\omega-1}\right|\\
& =\frac{\left|\left(\frac{1}{2}-\varepsilon\right)+\mathrm{i}y\right|}{\left|-\left(\frac{1}{2}+\varepsilon\right)+\mathrm{i}y\right|}\\
& = \frac{\left(\frac{1}{2}-\varepsilon\right)^2+y^2}{\left(\frac{1}{2}+\varepsilon\right)^2+y^2}\\
& = \frac{\frac{1}{4}+\varepsilon^2+y^2-\varepsilon}{\frac{1}{4}+\varepsilon^2+y^2+\varepsilon}\\
& <1.
\end{align}
Hence, we have $\{z\in\mathbb{C}\mid \Re z<\frac{1}{2}\}\subset f\left(D\right)$.
