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?
Best Answer
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 $.