Let $u,v \in \mathbb C, u\ne\pm v, |u| = |v| = 1$. Show $$ \frac {uv}{(u+v)^2} \in \mathbb R, \quad \frac {u+v}{u-v}i \in \mathbb R.$$ Source: Art of Problem Solving.
My proof is below. I request verification and critique on both the proofs and exposition as well as other approaches.
Proof 1: Since $u\bar u = 1 = v \bar v$, then $$uv(\bar u + \bar v)^2 = u \bar v + v \bar u + 2 = \bar u \bar v (u + v)^2$$ then $$\frac {uv} {(u + v)^2} = \overline{\frac{uv}{(u+v)^2}}$$ and is real. Likewise, since $$(u+v)(\bar u - \bar v) = v \bar u - u \bar v = -(u-v)(\bar u + \bar v)$$ then $$\frac {u+v}{u-v} = -\overline{\frac{u+v}{u-v}}$$ and is pure imaginary.
Proof 2: Let $w = u+v$. Quadrilateral $O,U,V,W$ is a parallelogram with $OU \cong OV$ and therefore a rhombus. Therefore, its diagonal $OW$ bisects $\angle UOV$. Then $$\arg(u+ v)= \arg w = \frac 1 2 (\arg u + \arg v) = \frac 1 2 \arg(uv)$$ and $$\arg (u+v)^2 = \arg(uv).$$ Consequently, their ratio $\frac {uv}{(u+v)^2}$ is real.
Likewise, $u+v$ and $u-v$ are diagonals of this rhombus and consequently mutually perpendicular. Their ratio is therefore pure imaginary, completing the proof.
Draw on argand plane. $u$ and $v$ lies on the unit circle centered on the origin. $u-v$ is parallel to the secant line passing through $u$ and $v$. $u+v$ is parallel to line passing through the origin and the midpoint of $uv$. Consequently, $u+v\perp u-v$ or
$$ \frac{u-v}{u+v}\in \mathbb{I} $$
For the second part,
$$ \frac{uv}{(u+v)^{2}} = \frac{1}{4}\left( 1 + \left( \frac{u-v}{u+v} \right)^{2} \right) $$
Square of pure imaginary number is real, proven.