Property of distance of a point from centre and focus of a rectangular hyperbola using complex numbers 
If P is a point on the hyperbola whose axis are equal, prove that $S_2 P×S_1P=CP^2$ where $C$ is centre where $S_1$ and $S_2$ are the focii
Advanced problems in JEE , Vikas Sharma aka Pink Book

I thought of approaching this question using complex numbers, I took my hyperbola equation as $x^2 - y^2 = a^2$, then it must be that the focii are at $(a\sqrt{2},0)$ and $(-a\sqrt{2},0)$, hence we want to evaluate:
$$ d= | z-a \sqrt{2}| | z+a\sqrt{2}|= |z^2 - 2a^2| = |(x^2 -y^2-2a^2) + 2xiy|= |(x^2 -y^2) - 2xiy|$$
Now, the problem is I have to prove that the final expression is equal to $|x+iy|$, but I can't see any direct algebra way to do it. Maybe I can show it by infusing geometry but I can't see how.

An observation I made, we see that $xy$ is equal to twice the area of triangle with vertices as centre, the point, and foot of perpendicular at x axis.
 A: I figured it out after I made the triangle observation, let $d$ be the distance from origin to the point, then $x= d \cos \theta$ and $y = d \sin \theta$, upon substitution into the length expression:
$$ |x^2 -y^2 - 2xiy | = d | \cos 2 \theta - i \sin 2 \theta| = d^2$$

Or, even better, follow dxiv's comment and notice that the expression is equal to $|x-iy|= |x+iy|$, this completes the proof.
A: The problem is invariant to translations, so we can assume WLOG that the center is at the origin and the hypervbola is given by $\big||z+a|-|z-a|\big|=2c\,$. Then, squaring:
$$
|z+a|^2 + |z-a|^2 - 2 \,|z+a|\,|z-a| = 4c^2
$$
It follows that:
$$
\require{cancel}
\begin{align}
2 \,|z+a|\,|z-a| &= |z+a|^2 + |z-a|^2 - 4c^2
\\ &= (z+a)(\bar z + \bar a) +  (z-a)(\bar z - \bar a) - 4c^2
\\ &= |z|^2+\cancel{z\bar a}+\bcancel{\bar z a}+|a|^2+|z|^2-\cancel{z \bar a} - \bcancel{\bar z a} +|a|^2 -4c^2
\\ \iff \;\;\;\; |z+a|\,|z-a| &= |z|^2 + |a|^2 - 2c^2
\end{align}
$$
The result follows for $\,|a|^2 = 2c^2\,$.
