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I want to simplify $|a+b|^2 + |a-b|^2$ where $a, b \in \mathbb{C}$. I've used Wolfram Alpha to get $$ |a+b|^2 + |a-b|^2 = 2\left(|a|^2 + |b|^2\right) $$ I'm trying to understand the steps involved in arriving at this result: $$\begin{eqnarray*} |a+b|^2 + |a-b|^2 &=& |(a+b)^2| + |(a-b)^2| \\ &=& | a^2 + 2ab + b^2 | + | a^2 - 2ab + b^2 | \end{eqnarray*} $$ But I'm at a loss as to how to continue from here; I find it hard to work symbolically with absolute values.

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    I'm confused. What exactly are you trying to do? You say you are "trying to deduce the derivation" of a certain equation, but seem to have done so. Is there a step you do not understand?2012-08-08
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    You can write $(a+b)^2$ instead of $|a+b|^2$. (For any real number $x$ the equality $x^2=(-x)^2=|x|^2$ holds.)2012-08-08
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    No no, I am trying to understand the details of how you go from A to B.2012-08-08
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    @MartinSleziak but can you do the same when $x$ is complex?2012-08-08
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    @Martin, these numbers are not necessarily real, according to the question.2012-08-08
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    This stems directly from the identity $|z|^2=z\bar z$ applied to $z=a\pm b$. One gets $|a\pm b|^2=(a\pm b)(\bar a\pm\bar b)=|a|^2\pm a\bar b\pm\bar ab+|b|^2$. Add the $+$ case and the $-$ case.2012-08-08
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    Oh, I did not notice that we work with complex numbers. (I should have read the post more thoroughly.) Since adding complex numbers is the same as adding vectors, this is basically the [Parallelogram law](http://en.wikipedia.org/wiki/Parallelogram_law).2012-08-08

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$|z|^2=zz'$ where $z'$ stands for the complex conjugate of $z$.

$$|a+b|^2+|a-b|^2=(a+b)(a'+b')+(a-b)(a'-b')=aa'+ab'+ba'+bb'+aa'-ab'-ba'+bb'=2aa'+2bb'$$ and you're pretty much there.