$1 + {alpha}^2 + 2 alpha cos(omega)$ as a square of distance












0












$begingroup$


How to prove that



$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$



starting from



$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$



(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)






complex-analysis complex-numbers







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$endgroup$








  • 2




    $begingroup$
    Look for the roots of this second degree polynomial in $alpha$.
    $endgroup$
    – Stefan Lafon
    Jan 18 at 15:35








  • 4




    $begingroup$
    Just do the 'backward' proof in reverse.
    $endgroup$
    – lightxbulb
    Jan 18 at 15:39
















0












$begingroup$


How to prove that



$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$



starting from



$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$



(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)






complex-analysis complex-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Look for the roots of this second degree polynomial in $alpha$.
    $endgroup$
    – Stefan Lafon
    Jan 18 at 15:35








  • 4




    $begingroup$
    Just do the 'backward' proof in reverse.
    $endgroup$
    – lightxbulb
    Jan 18 at 15:39














0












0








0





$begingroup$


How to prove that



$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$



starting from



$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$



(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)






complex-analysis complex-numbers







share|cite|improve this question











$endgroup$




How to prove that



$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$



starting from



$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$



(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)






complex-analysis complex-numbers




complex-analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 15:43









KM101

6,0351525




6,0351525










asked Jan 18 at 15:35









DiscreteMathDiscreteMath

414




414








  • 2




    $begingroup$
    Look for the roots of this second degree polynomial in $alpha$.
    $endgroup$
    – Stefan Lafon
    Jan 18 at 15:35








  • 4




    $begingroup$
    Just do the 'backward' proof in reverse.
    $endgroup$
    – lightxbulb
    Jan 18 at 15:39














  • 2




    $begingroup$
    Look for the roots of this second degree polynomial in $alpha$.
    $endgroup$
    – Stefan Lafon
    Jan 18 at 15:35








  • 4




    $begingroup$
    Just do the 'backward' proof in reverse.
    $endgroup$
    – lightxbulb
    Jan 18 at 15:39








2




2




$begingroup$
Look for the roots of this second degree polynomial in $alpha$.
$endgroup$
– Stefan Lafon
Jan 18 at 15:35






$begingroup$
Look for the roots of this second degree polynomial in $alpha$.
$endgroup$
– Stefan Lafon
Jan 18 at 15:35






4




4




$begingroup$
Just do the 'backward' proof in reverse.
$endgroup$
– lightxbulb
Jan 18 at 15:39




$begingroup$
Just do the 'backward' proof in reverse.
$endgroup$
– lightxbulb
Jan 18 at 15:39










2 Answers
2






active

oldest

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2












$begingroup$

assuming $alpha$ is a real number (not complex):



$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint



    Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      assuming $alpha$ is a real number (not complex):



      $$1-2 alpha cos(omega)+|alpha|^2 $$
      $$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
      $$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
      $$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
      $$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
      $$|1-alpha e^{jomega}|^2$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        assuming $alpha$ is a real number (not complex):



        $$1-2 alpha cos(omega)+|alpha|^2 $$
        $$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
        $$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
        $$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
        $$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
        $$|1-alpha e^{jomega}|^2$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          assuming $alpha$ is a real number (not complex):



          $$1-2 alpha cos(omega)+|alpha|^2 $$
          $$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
          $$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
          $$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
          $$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
          $$|1-alpha e^{jomega}|^2$$






          share|cite|improve this answer











          $endgroup$



          assuming $alpha$ is a real number (not complex):



          $$1-2 alpha cos(omega)+|alpha|^2 $$
          $$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
          $$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
          $$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
          $$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
          $$|1-alpha e^{jomega}|^2$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 16:31

























          answered Jan 18 at 16:19









          Bill MooreBill Moore

          1376




          1376























              0












              $begingroup$

              Hint



              Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint



                Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint



                  Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.






                  share|cite|improve this answer









                  $endgroup$



                  Hint



                  Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 15:41









                  Mostafa AyazMostafa Ayaz

                  15.6k3939




                  15.6k3939






























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