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

[Siya]

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

Hello,

I really hope someone could help me with solving the next problem:

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

What have I done so far:
- I observed z, a complex number, as dots on a circle in a complex plane (exluding 1 and -1), with r=1=|z|
- Hoping it will lead me somewhere, I changed the form of z, so it has the form

. . .\(\displaystyle \dfrac{x^2\, -\, 1}{x^2\, +\, 1}\, +\, \dfrac{2\, x\, i}{x^2\, +\, 1}\)

Now, using |z|=sqrt(Re(z)^2 + Im(z)^2) only proved that |z|=1 and nothing more. That is all I have. What information am I not seeing to help me prove x∈ℝ?

Any help would be greatly appreciated. Also, one more thing: If someone can suggest where could I find a book with these types of problems, I would be greatful. Book that I have, and that has been given to us, doesn't offer much help nor does it have these types of problems, yet are being given to us on exams.

Thank you.

 

[Siya]

This is what I have:

Since z equals to

. . .\(\displaystyle \dfrac{x^2\, -\, 1}{x^2\, +\, 1}\, +\, \dfrac{2\, x\, i}{x^2\, +\, 1}\)

we look at the angle, which is

. . .\(\displaystyle \tan(f)\, =\, \dfrac{2\,x}{x^2\, -\, 1}\)

Now, using

. . .\(\displaystyle \tan(2\,a)\, =\, \dfrac{2\, \tan(a)}{1\, -\, \tan^2(a)}\)

(where a=f/2) we have

. . .\(\displaystyle \tan(2\,a)\, =\, \dfrac{2\,x}{x^2\, -\, 1}\)

Now, we would have

. . .\(\displaystyle 2\, \times\, \dfrac{\tan(a)}{1\, -\, \tan^2(a)}\, =\, \dfrac{2\,x}{x^2\, -\, 1}\)

When I replace a=f/2, I get that x=-tan(f/2) which belongs to field of real numbers, thus for any complex number z, I can find an angle and calculate x.

Am I right to assume this is the correct answer?