Why does $|e^ix|^2 = 1$?

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Why does $|e^ix|^2 = 1$?




The book said $e^ix = cos x + isin x$, and square it, then $|e^ix|^2 = cos^2x + sin^2x = 1$.



But, when I calculated it, $ |e^ix|^2 = left|cos x + isin xright|^2 = cos^2x - sin^2x + 2isin xcos x$.



I can't make it to be equal $1.$ How can I do it?




complex-numbers absolute-value




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  • 7




    $|z|^2 = zbar z$, not $z^2$.
    – Ted Shifrin
    Aug 3 at 4:01














up vote
5
down vote

favorite













Why does $|e^ix|^2 = 1$?




The book said $e^ix = cos x + isin x$, and square it, then $|e^ix|^2 = cos^2x + sin^2x = 1$.



But, when I calculated it, $ |e^ix|^2 = left|cos x + isin xright|^2 = cos^2x - sin^2x + 2isin xcos x$.



I can't make it to be equal $1.$ How can I do it?




complex-numbers absolute-value




share|cite|improve this question

















  • 7




    $|z|^2 = zbar z$, not $z^2$.
    – Ted Shifrin
    Aug 3 at 4:01












up vote
5
down vote

favorite









up vote
5
down vote

favorite












Why does $|e^ix|^2 = 1$?




The book said $e^ix = cos x + isin x$, and square it, then $|e^ix|^2 = cos^2x + sin^2x = 1$.



But, when I calculated it, $ |e^ix|^2 = left|cos x + isin xright|^2 = cos^2x - sin^2x + 2isin xcos x$.



I can't make it to be equal $1.$ How can I do it?




complex-numbers absolute-value




share|cite|improve this question














Why does $|e^ix|^2 = 1$?




The book said $e^ix = cos x + isin x$, and square it, then $|e^ix|^2 = cos^2x + sin^2x = 1$.



But, when I calculated it, $ |e^ix|^2 = left|cos x + isin xright|^2 = cos^2x - sin^2x + 2isin xcos x$.



I can't make it to be equal $1.$ How can I do it?





complex-numbers absolute-value





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 4:14









Math Lover

12.2k21132




12.2k21132









asked Aug 3 at 3:59









uninopkn

353




353







  • 7




    $|z|^2 = zbar z$, not $z^2$.
    – Ted Shifrin
    Aug 3 at 4:01












  • 7




    $|z|^2 = zbar z$, not $z^2$.
    – Ted Shifrin
    Aug 3 at 4:01







7




7




$|z|^2 = zbar z$, not $z^2$.
– Ted Shifrin
Aug 3 at 4:01




$|z|^2 = zbar z$, not $z^2$.
– Ted Shifrin
Aug 3 at 4:01










4 Answers
4






active

oldest

votes

















up vote
4
down vote



accepted










If $a,b$ are real then $displaystyle left| a+bi right| = sqrta^2+b^2,, = sqrt(a+bi)(a-bi),.$






share|cite|improve this answer





















  • Thanks.. I missed some point
    – uninopkn
    Aug 3 at 5:04

















up vote
4
down vote













If $x in mathbbR$,
$$cos^2x - sin^2 x= cos(2x)$$



$$2 sin x cos x =sin(2x)$$



$$|(cos x + i sin x)^2|=cos^2(2x)+sin^2(2x)=1$$






share|cite|improve this answer





















  • up to the point .. +1
    – Ahmad Bazzi
    Aug 3 at 4:12

















up vote
1
down vote













you can first apply the modulus $$|e^ix|=sqrtcos^2 x + sin^2 x$$ then square it whole you will get $$|e^ix|^2=cos^2 x + sin^2 x=1$$






share|cite|improve this answer




























    up vote
    1
    down vote













    $|z|^2=zbarz$



    But the complex conjugate of $e^xi$ is $e^-xi$.



    $|e^xi|^2=e^xie^-xi=1$






    share|cite|improve this answer





















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      If $a,b$ are real then $displaystyle left| a+bi right| = sqrta^2+b^2,, = sqrt(a+bi)(a-bi),.$






      share|cite|improve this answer





















      • Thanks.. I missed some point
        – uninopkn
        Aug 3 at 5:04














      up vote
      4
      down vote



      accepted










      If $a,b$ are real then $displaystyle left| a+bi right| = sqrta^2+b^2,, = sqrt(a+bi)(a-bi),.$






      share|cite|improve this answer





















      • Thanks.. I missed some point
        – uninopkn
        Aug 3 at 5:04












      up vote
      4
      down vote



      accepted







      up vote
      4
      down vote



      accepted






      If $a,b$ are real then $displaystyle left| a+bi right| = sqrta^2+b^2,, = sqrt(a+bi)(a-bi),.$






      share|cite|improve this answer













      If $a,b$ are real then $displaystyle left| a+bi right| = sqrta^2+b^2,, = sqrt(a+bi)(a-bi),.$







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Aug 3 at 4:05









      Michael Hardy

      204k23185460




      204k23185460











      • Thanks.. I missed some point
        – uninopkn
        Aug 3 at 5:04
















      • Thanks.. I missed some point
        – uninopkn
        Aug 3 at 5:04















      Thanks.. I missed some point
      – uninopkn
      Aug 3 at 5:04




      Thanks.. I missed some point
      – uninopkn
      Aug 3 at 5:04










      up vote
      4
      down vote













      If $x in mathbbR$,
      $$cos^2x - sin^2 x= cos(2x)$$



      $$2 sin x cos x =sin(2x)$$



      $$|(cos x + i sin x)^2|=cos^2(2x)+sin^2(2x)=1$$






      share|cite|improve this answer





















      • up to the point .. +1
        – Ahmad Bazzi
        Aug 3 at 4:12














      up vote
      4
      down vote













      If $x in mathbbR$,
      $$cos^2x - sin^2 x= cos(2x)$$



      $$2 sin x cos x =sin(2x)$$



      $$|(cos x + i sin x)^2|=cos^2(2x)+sin^2(2x)=1$$






      share|cite|improve this answer





















      • up to the point .. +1
        – Ahmad Bazzi
        Aug 3 at 4:12












      up vote
      4
      down vote










      up vote
      4
      down vote









      If $x in mathbbR$,
      $$cos^2x - sin^2 x= cos(2x)$$



      $$2 sin x cos x =sin(2x)$$



      $$|(cos x + i sin x)^2|=cos^2(2x)+sin^2(2x)=1$$






      share|cite|improve this answer













      If $x in mathbbR$,
      $$cos^2x - sin^2 x= cos(2x)$$



      $$2 sin x cos x =sin(2x)$$



      $$|(cos x + i sin x)^2|=cos^2(2x)+sin^2(2x)=1$$







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Aug 3 at 4:02









      Siong Thye Goh

      76.6k134794




      76.6k134794











      • up to the point .. +1
        – Ahmad Bazzi
        Aug 3 at 4:12
















      • up to the point .. +1
        – Ahmad Bazzi
        Aug 3 at 4:12















      up to the point .. +1
      – Ahmad Bazzi
      Aug 3 at 4:12




      up to the point .. +1
      – Ahmad Bazzi
      Aug 3 at 4:12










      up vote
      1
      down vote













      you can first apply the modulus $$|e^ix|=sqrtcos^2 x + sin^2 x$$ then square it whole you will get $$|e^ix|^2=cos^2 x + sin^2 x=1$$






      share|cite|improve this answer

























        up vote
        1
        down vote













        you can first apply the modulus $$|e^ix|=sqrtcos^2 x + sin^2 x$$ then square it whole you will get $$|e^ix|^2=cos^2 x + sin^2 x=1$$






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          you can first apply the modulus $$|e^ix|=sqrtcos^2 x + sin^2 x$$ then square it whole you will get $$|e^ix|^2=cos^2 x + sin^2 x=1$$






          share|cite|improve this answer













          you can first apply the modulus $$|e^ix|=sqrtcos^2 x + sin^2 x$$ then square it whole you will get $$|e^ix|^2=cos^2 x + sin^2 x=1$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 3 at 4:08









          Nebo Alex

          1,079824




          1,079824




















              up vote
              1
              down vote













              $|z|^2=zbarz$



              But the complex conjugate of $e^xi$ is $e^-xi$.



              $|e^xi|^2=e^xie^-xi=1$






              share|cite|improve this answer

























                up vote
                1
                down vote













                $|z|^2=zbarz$



                But the complex conjugate of $e^xi$ is $e^-xi$.



                $|e^xi|^2=e^xie^-xi=1$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $|z|^2=zbarz$



                  But the complex conjugate of $e^xi$ is $e^-xi$.



                  $|e^xi|^2=e^xie^-xi=1$






                  share|cite|improve this answer













                  $|z|^2=zbarz$



                  But the complex conjugate of $e^xi$ is $e^-xi$.



                  $|e^xi|^2=e^xie^-xi=1$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 3 at 5:21









                  Mason

                  1,1271223




                  1,1271223






















                       

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