Simplifying $|a+b|^2 + |a-b|^2$

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? –  Alex Becker Aug 8 '12 at 5:58
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.) –  Martin Sleziak Aug 8 '12 at 5:59
No no, I am trying to understand the details of how you go from A to B. –  Ezequiel Muns Aug 8 '12 at 6:00
@MartinSleziak but can you do the same when $x$ is complex? –  Ezequiel Muns Aug 8 '12 at 6:02
@Martin, these numbers are not necessarily real, according to the question. –  Gerry Myerson Aug 8 '12 at 6:03
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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.