The Complexity is Real

Let $z = a+bi$ , with $a^2 + b^2 = 1$ , and $\mathrm{abs}(a) \neq 1$ , or equivalently, $b \neq 0$ .

Evaluating the given expression, we have

$i \bigg( \frac{ (a-1) + bi}{(a+1) +bi} \bigg)$

$i \bigg( \frac{ (a-1) + bi}{(a+1) +bi} \cdot \frac{ (a+1) - bi}{(a+1) - bi} \bigg)$

$i \bigg( \frac{ a^2-1 + b^2 + (a+1)bi - (a-1)bi}{(a+1)^2+b^2} \bigg)$

$i \bigg( \frac{ \color{#D61F06}a^2 + b^2 \color{#333333} - 1 + 2bi}{\color{#D61F06} a^2+b^2 \color{#333333} + 2a + 1} \bigg)$

$i \bigg( \frac{ \color{#D61F06} 1 \color{#333333} - 1 + 2bi}{\color{#D61F06} 1 \color{#333333} + 2a + 1} \bigg)$

$i \bigg( \frac{2bi}{2(a+1)} \bigg)$

$\bigg( \frac{-b}{a+1} \bigg)$

And there we have it. That resulting fraction will always be real, given the conditions.

$|z|=1\to z=e^{2ix}\to i\dfrac{e^{2ix}-1}{e^{2ix}+1}= \dfrac{i(e^{ix}-e^{-ix})}{e^{ix}-e^{-ix}}=\dfrac{-\sin(x)}{\cos(x)}=-\tan(x)$ $|Re(e^{2ix})|\neq 1 \to x\neq \dfrac{\pi*n}{2} \to -tan(x) \in R$