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}
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Hint:
Consider the case $y=-x \ne 0$.
Emilio Novati
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@Cali Don't assume when you can check yourself. Insert and calculate, see what you get. – Arthur Nov 04 '16 at 16:23
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@Cali: if you substitute $y=-x$ in your equation you can easily see that it's wrong. – Emilio Novati Nov 04 '16 at 16:26
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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.
yurnero
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$$ \begin{align} |x+y|^2 & = (x+y)(\bar x + \bar y) \\ & = x \bar x + x \bar y + \bar x y + y \bar y \\ & = |x|^2 + |y|^2 + 2 \operatorname{Re} (x \bar y) \\ & = |x|^2 + |y|^2 + 2 |x y| \cos \theta \\ & \ne |x|^2 + |y|^2 + 2 |x y| \quad \text{unless the angle between} \;x,y \;\text{is}\;\theta=0 \end{align} $$
dxiv
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that means if $y=\bar{x}$, i.e. y is conjugated complex version of x, in that case my initial equality would hold? – Cali Nov 04 '16 at 16:45
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@Cali I fixed the post above, that was meant to be the angle between $x,y$. If $y = \lambda x$ with $\lambda \in \mathbb{R}^+$ then the equality holds. – dxiv Nov 04 '16 at 16:56
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So in the case of conjugate complex if would essentially have that |$x$+$\bar{x}$|^2=($x$+$\bar{x}$)($\bar{x}$+${x}$)=($x$+$\bar{x}$)^2? or I am wrong again? – Cali Nov 04 '16 at 17:27
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@Cali $x+\bar x \in \mathbb{R}$ and $|r|^2 = r^2$ for $\forall r \in \mathbb{R}$. – dxiv Nov 04 '16 at 17:30