Find the image of the $x$ and $y$ axes under $f(z)=\frac{z+1}{z-1}$ 
Find the image of the $x$ and $y$ axes under $f(z)=\frac{z+1}{z-1}$

Atemppt
Notice that the real axis is given by $x+0i$ for $x\in \mathbb{R}$, then the image are
$f(x)=\frac{x+1}{x-1}$ which is a hyperbola in the plane.
Now for the $y$ axis, we should have $0+iy$ but then $f(y)=\frac{iy+1}{iy-1}$ from here I can´t obtain a expression of in the plane, I tryed multiply $f$ by $\frac{iy-1}{iy-1}$ but it doesn´t work.
Too I tryed calculate $|f(y)|^2=1$, I think that it is a unit circle, but I don´t now if it least is sufficient.
Any hint or comment was useful.
 A: If $x\in\Bbb R$, then $f(x)=\frac{x+1}{x-1}$ as a point in the complex plane, so the image under $f$ of the real axis is
$$\left\{\frac{x+1}{x-1}:x\in\Bbb R\right\}=\left\{1+\frac2{x-1}:x\in\Bbb R\right\}\,.$$
Plainly this is a subset of the real axis. Try to show that it is all of the real line except one point and identify the missing point.
For the image of the imaginary axis we want
$$\left\{\frac{yi+1}{yi-1}:y\in\Bbb R\right\}\,,$$
again as a subset of $\Bbb C$, the complex plane. Note that
$$\begin{align*}
\frac{yi+1}{yi-1}&=\frac{yi+1}{yi-1}\cdot\frac{yi+1}{yi+1}\\
&=\frac{(yi+1)^2}{-y^2-1}\\
&=-\frac{1-y^2+2yi}{1+y^2}\\
&=\frac{y^2-1-2yi}{y^2+1}\,.
\end{align*}$$
Now compute $\left|\frac{y^2-1-2yi}{y^2+1}\right|$ and verify that all of these points lie on the unit circle. Finally, show that they include all of the unit circle except one point, and identify the missing point.
A: $0\mapsto-1$,
$1\mapsto \infty$, and
$-1\mapsto0$
From that we know the image of the  $x $-axis is itself.
$i\mapsto \dfrac {i+1}{i-1} $, and
$-i\mapsto\dfrac {i-1}{1+i} $.  Thus, since $0,i,-i $ map to three different points on the unit circle,  the image of the  $y $-axis is the unit circle.
In both cases the standard fact about Möbius transformations was used: namely they map generalized circles to generalized circles.
