Prove that for any complex number $|x|=|-x|$

Question

Prove that for any complex number $|x|=|-x|$

So we can substitute $a+bi$ for $x$, so the equation becomes $$|a+bi|=|-a-bi|$$

I don't know how to continue; sorry if it is really obvious and I missed it...

Answer 1

$$ |x|= |a+ib| = \sqrt{ a^2 + b^2} = \sqrt{ (-a)^2 + (-b)^2 } = |-a-bi|= |-x| $$

Answer 2

Just $|x|=\sqrt{a^2+b^2}$ and $|-x|=\sqrt{(-a)^2+(-b)^2}=\sqrt{a^2+b^2}$.

Hence $|x|=|-x|$

Answer 3

Hint: $\;\;|x|^2=x \cdot \overline{x}=(-x)(-\overline{x}) = (-x)\overline{(-x)} = |-x|^2$