<s>[INST] Arithmetic proof of absolute value function of complex numbers I am looking for the arithmetic proof that: 
$ |z| = \sqrt{(x^2 + y^2)} $    where  $ z = x + i y $
Previously I assumed squaring a function then square rooting it would be analogous to the absolute value function (modulus) but it seems not to be the case in the complex domain. Consider the following simple counter example:
Let $ z = \cos(x) + i \times \sin (x) $
$|z| = |\cos(x) + i \times \sin(x)| = \sqrt{((\cos(x) + i \times \sin(x))^2} $
$ = \sqrt{2 \times i \times \cos(x) \times \sin(x) + \cos^2(x) +i^2 \times \sin^2(x) }$
$ = \sqrt{2i \times \cos(x)\sin(x) + \cos^2(x) - \sin^2(x) } \neq \cos^2(x) + \sin^2(x)$
Why is that?
 [/INST] First,
the absolute value 
of a complex number 
is a real number.
The standard definition is
$|z|^2
=z\, \bar{z}
$.
From this,
if $z = x+iy$,
$|z|^2
=z\, \bar{z}
=(x+iy)(x-iy)
=x^2+y^2
$.
</s>