Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$ 
Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$

I have tried to express $x$ in terms of $a$ and $b$, namely: $x = a+bi, |x|= \sqrt{a^2+b^2}$ and to evaluate this but this simply led me nowhere. I guess there must be a clever way to solve that I don't see. I'd be thankful if you could give me some hints (not the actual solution) on how to proceed with this problem.

Hint 1: $\overline{x} + 1 = \overline{x+1}$.

 Let $x=a+ib$.  Then
 $$ \overline{x} + 1 = \overline{a + ib} + 1 = a - ib + 1 = (a+1) - ib, \tag{1}$$
 and
 $$\overline{x+1} = \overline{a+ib+1} = \overline{(a+1) + ib} = (a+1) - ib. \tag{2}$$
 Since (1) and (2) agree, we conclude that $\overline{x} + 1 = \overline{x+1}$.

Hint 2: $(x+1)(\overline{x+1}) = |x+1|^2$.

 This is an application of the results that $z\overline{z} = |z|^2$ for any $z\in \mathbb{C}$.  To see this, recall that if $z = a+ib$, then
 $$|z| := \sqrt{a^2 + b^2}. $$
 But
 $$ z\overline{z} = (a+ib)(a-ib) = a^2 - aib + iba - i^2 b^2 = a^2 + b^2 = |z|^2, $$
 which is the desired identity.

Hint 3: $(x-1)(\overline{x}+1) = |x|^2 + 2i\Im(x) - 1$.

 Let $x = a+ib$.  Then
 \begin{align} (x-1)(\overline{x}+1) &= x\overline{x} + x - \overline{x} - 1 \\&= |x|^2 + (a+ib) - (a+ib) - 1 \\&= |x|^2 + 2ib - 1 \\&= |x|^2 + 2i \Im(x) - 1. \end{align}

If you can show each of these and put them together in the right order, they can be used to deduce a complete solution to your problem.

 Putting the pieces together, we have
 \begin{align}\frac{x-1}{x+1}&= \frac{x-1}{x+1} \cdot \frac{\overline{x+1}}{\overline{x+1}} \tag{"rationalize" the denominator}\\&= \frac{(x-1)(\overline{x}+1)}{(x+1)(\overline{x+1})} \tag{Hint 1} \\ &= \frac{|x|^2 + 2i\Im(x) - 1}{|x+1|^2}. \tag{Hints 2 and 3}\end{align}
 Then the real part is given by
 $$ \Re\left( \frac{|x|^2 + 2i\Im(x) - 1}{|x+1|^2} \right) = \frac{|x|^2 - 1}{|x+1|^2}. $$
 But this is zero by the assumption that $|x|=1$.

