Squared magnitude of two complex numbers I believe one easy questions for maths experts. If I have two complex numbers x=a+ib and y=c+di is the squared magnitude of their sum equal to:
\begin{equation}
|x+y|^2=|x|^2+2|xy|+|y|^2
\end{equation}

Your LHS equals
$$
(a+c)^2+(b+d)^2=(a^2+b^2)+2(ac+bd)+(c^2+d^2)=|x|^2+2(ac+bd)+|y|^2
$$
so it's almost the RHS, except for the fact that
\begin{aligned}
2|xy|&=2|(ab-cd)+i(ac+bd)|=2\sqrt{(ab-cd)^2+(ac+bd)^2}\\
&=2\sqrt{a^2b^2+c^2d^2+a^2c^2+b^2d^2}\\
&\neq 2(ac+bd).
\end{aligned}
In fact,
$$
(ac+bd)^2=a^2c^2+b^2d^2+2(ab)\cdot (cd)\leq a^2b^2+c^2d^2+a^2c^2+b^2d^2.
$$
So in fact you have LHS $\leq$ RHS, not LHS $=$ RHS.
