Let z, w $\in\mathbb{C}$ such that z$\ne\overline{w}$. Prove that if zw isn't purely imaginary, then $\lvert \frac{(z+\overline{w})}{(z-\overline{w})}\rvert \ne 1$.
I've written out z=a+bi and w=c+di, then w=c-di and expanded all of it. I've just been getting lost on how to make sure it's not actually equal to 1. Any help would be appreciated.
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Let $z, w \in \mathbb{C}$ be such that $z \neq \bar{w}$. As $\bar{z} \neq w$, we can multiply both sides of the fraction by $(\bar{z} -w)$. We get :
$$\frac{(z+\bar{w})}{(z-\bar{w})} = \frac{(z+\bar{w})(\bar{z} -w)}{(z-\bar{w})(\bar{z} -w)}$$
The denominator is :
$$(z-\bar{w})(\bar{z} -w) = z\bar{z} - w\bar{w}-zw - \underbrace{\bar{z}\bar{w}}_{= \overline{zw}} = |z|^2 - |w|^2 - 2 \mathcal{Re}(zw)$$
Let's develop the numerator:
$$(z+\bar{w})(\bar{z} -w) = z\bar{z} - w\bar{w}-zw + \underbrace{\bar{z}\bar{w}}_{= \overline{zw}} = |z|^2 - |w|^2 - 2 \mathcal{Im}(zw) i$$
Suppose now that $zw$ isn't purely imaginary (ie $\mathcal{Re}(zw) \neq 0$). We have to check that the module of the numerator isn't equal to the module of the denominator, but this now amounts to show that $(\mathcal{Im}(zw)i)^2 \neq \mathcal{Re}(zw)^2$, which is obvious since $\mathcal{Re}(zw) \neq 0$ : the left hand side is negative, while the right hand side is strictly positive.
From the thesis, we have: $$|z+\overline{w}|\neq|z-\overline{w}|$$ Now, let $z=a+ib \land w=c+id$, we obtain: $$\sqrt{(a+c)^2+(b-d)^2}\neq\sqrt{(a-c)^2+(b-d)^2} \leftrightarrow 2ac-2bd\neq-2ac+2bd \leftrightarrow ac\neq bd \leftrightarrow ac-bd\neq 0$$ From the hypotesis, we have: $$(a+ib)(c+id)\neq ik, k \in R \leftrightarrow (ac-bd)+i(bc+ad)\neq ik \leftrightarrow ac-bd \neq 0$$ The two condition are the same, so your thesis is correct.