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

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...

• Do you know the formula for $|x|$ if $x=a+bi$? – kccu Nov 2 '16 at 3:05
• I feel like I should know...but I forgot...is it $\sqrt{a^2+b^2}$? – suomynonA Nov 2 '16 at 3:06
• $\sqrt{a^2+b^2}$, yes. Plug $-x$ into that formula. – Sophie Nov 2 '16 at 3:07
• Oh, thanks. Why do I always miss this kind of obvious stuff? – suomynonA Nov 2 '16 at 3:08
• Now repeat for $$|x| = \left|\left(\cos t + i \sin t\right)x\right|,$$ with $t$ a any real number. – Sean Lake Nov 2 '16 at 3:10

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

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

Hence $|x|=|-x|$

• I think you want to write $|-x|=\sqrt{(-a)^2+(-b)^2}$. – kccu Nov 2 '16 at 3:09
• Yes, thank you! – Alpp Nov 2 '16 at 3:10

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

• What does $\overline{x}$ mean? I only know it as "not $x$"... – suomynonA Nov 2 '16 at 3:21
• @suomynonA That's a common notation for the complex conjugate. – dxiv Nov 2 '16 at 3:22
• Didn't know that, thanks – suomynonA Nov 2 '16 at 3:24