0
$\begingroup$

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...

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

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

$\endgroup$
2
$\begingroup$

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

Hence $|x|=|-x|$

$\endgroup$
  • $\begingroup$ I think you want to write $|-x|=\sqrt{(-a)^2+(-b)^2}$. $\endgroup$ – kccu Nov 2 '16 at 3:09
  • $\begingroup$ Yes, thank you! $\endgroup$ – Alpp Nov 2 '16 at 3:10
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.