How do you prove that the image of the real axis lies on a circle? The question is, Consider the map $$f(z) = (z + i)/(z - i)$$
Show that the image of the real
axis under z lies on a circle centred at the origin of the complex plane. 
Basically, I've gotten up to this point: 
$$=((x^2 - 1) + 2xi)/(x^2 + 1)$$
And I'm stuck here. 
Thanks for any help! 
 A: Hint:


*

*any set $Z\subseteq \mathbb C$ lies on a circle centered at the origin if $\|z\|$ is constant for all $z\in\mathbb Z$.

*$\|z\|$ is constant if and only if $\|z\|^2$ is constant

A: Well, you arrived at
$$\frac{x^2-1}{x^2+1}+\frac{2x}{x^2+1}i$$
If we identify this with a generic point on the plane $\;\Bbb R^2\;$ , we get
$$\left(\frac{x^2-1}{x^2+1}\,,\,\,\frac{2x}{x^2+1}\right)=:(a,b)$$
observe now that
$$a^2+b^2=\frac{x^4-2x^2+1+4x^2}{(x^2+1)^2}=\frac{(x^2+1)^2}{(x^2+1)^2}=1$$
and there you have your canonical circle: the unit one. You were very close indeed.
A: First note that if $z$ is real, then $|z+i| = |z-i|$ (this is most easily seen geometrically), so $|f(z)|= 1$. Thus the image of every real number lies on the unit circle.
To show that it covers (almost) all of the unit circle, note that if $\theta$ is the angle of $z+i$, then $z-i$ has the angle $-\theta$, and therefore the angle of $f(z)$ is $2\theta$. As $z$ moves along the real axis, $\theta$ can get arbitrarily close to $\pi$ from below, and arbitrarily close to $0$ from above. In other words, $\theta \in (0, \pi)$, which means that the angle of $f(z)$ can be anywhere in $(0,2\pi)$, so the only value on the unit circle that isn't attained by $f$ is $1$.
