Algebraic expansion with complex variable... How do I expand the following 
$$ \big||y| - x \big|^2 $$ 
note that $|.|$ means magnitude of complex variable and $x$ is complex variable..
Is the solution
$$ \big||y| - x \big|^2 = \big(|y| - x \big)^2= |y|^2+|x|^2-2 |y|x$$
My other question is the following inequality true?
$$ x \geq |x| $$ always?
Thanks 

Inequalities do not make sense for complex numbers (what would it mean to say $i \geq |i| = 1$?) Further, this isn't even true for real numbers: $-1 \leq |-1|$ (in fact the opposite is true for real numbers: $r \leq |r|$).
With regard to your first question, no that is not a solution. The critical error is that, for a complex number $|x|^2 \neq x^2$ in general. What is true is that
$$
|x|^2 = x\overline x,
$$
i.e. if $x = u + iv$ then 
$$|x|^2 = (u + iv)(u-iv) = u^2 + iuv - iuv - i^2v^2 = u^2 + v^2.
$$
So, to carefully write out your expression (assuming $y$ is real, and $x = u + iv$ is complex, and let $c = |y|$ to save me writing absolute values over and over):
$$
||y| -x|^2 = |c - u - iv|^2 = ((c - u) - iv)\overline{((c - u) - iv)} \\
= ((c - u) - iv)((c - u) + iv) = (c - u)^2 + v^2 = c^2 -2uc + u^2 + v^2 \\
= |y|^2 - 2|y|\textrm{Re}(x) + |x|^2.
$$
If $y$ is meant to be complex, write $y = s + it$, note $|y| = \sqrt{s^2 + t^2}$ and then proceed carefully.
