<s>[INST] 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.
 [/INST] So let’s evaluate the expression by plugging in $x = a + bi$. $\frac{x-1}{x+1} = \frac{(a-1)+bi}{(a+1)+bi}$.
Multiply by the conjugate of the denominator, which is $(a+1)-bi$, to get $\dfrac{((a-1)+bi)((a+1)-bi)}{((a+1)+bi)((a+1)-bi)}$.
We know that the resulting denominator must be real and equal to $(a+1)^2+b^2$. We can expand the numerator to $(a-1)(a+1)+(a+1)bi-(a-1)bi+b^2$. Discard the imaginary middle terms, and we're left with $(a-1)(a+1)+b^2 = a^2 - 1 + b^2$. We know that $a^2+b^2=1$, and $1-1=0$. Thus, the real part of the numerator is zero, and since the denominator is also real, we know that the real part of the fraction is zero.
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