Prove (∀z∈ℂ\{1,-1} : |z|=1)(∃x∈ℝ) where z=(x+i)/(x-i)

Question

I am having trouble proving the next problem:

Prove that (∀$z$∈ℂ \ {1,-1} : |$z$|=1)(∃x∈ℝ) where $z=\frac{x+i}{x-i}$

What have I done:

• I observed complex number $z$ as dots on a circle with radius of 1, because |$z$|=1=$r$ (but $z$ cannot be 1 or -1)

• I changed the form of $z$ to $z=\frac{(x^2-1)}{x^2+1}$+$\frac{(2x)}{x^2+1}i$

• After that, I tried to calculate modul of $z$, but it gave me no new information: $|z|=\left(\sqrt\frac{(x^2-1)^2+(2x)^2}{(x^2+1)^2}\right)=1$

It only proved that, yes, |$z$|=1, which is already stated. What am I supposed to conclude based on given information? What am I not 'seeing'? Any help would be greatly appreciated.

Answer 1

Imagine that this problem is a game. I think of some $z$ in $\mathbb{C} \text{ omit } \{1,-1\} |z|=1$. You give me an $x$ such that $z=\frac{x+i}{x-i}$. What do information do you need to show you are representing $z$?

I think you were making some real progress with changing the form.

You calculated the modulus of $z$. What else could you need to calculate to pin down the value of $z$?

Updating:

We know that $\tan \phi = \frac{2x}{x^2-1}$ We can therefore rearrange this and solve for x to get: $x = \tan \frac{\phi}{2}$ or $x = \cot \frac{\phi}{2}$. We now know that if you give us some $z$ we can calculate $\phi$ and from there we can find the appropriate $x$.

Answer 2

The map is a Möbius map. It maps lines (here will be real axis) to circles (here will be the unit circle $|z|=1$.) Note that exclusion of $+1$ and $-1$ corresponds to the fact that they are achieved at infinities.

However, I can't see right away, why the real line would map to the unit circle. Maybe you should see if you can show that the inverse map, which will again be Mobius, maps the circle to the real line.