Why $|e^{itx}| = 1$?












2












$begingroup$


I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?






probability-theory characteristic-functions







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  • 7




    $begingroup$
    ... because $cos (tx)^2+sin(tx)^2=1$
    $endgroup$
    – Fakemistake
    Dec 29 '18 at 12:21


















2












$begingroup$


I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?






probability-theory characteristic-functions







share|cite|improve this question









$endgroup$








  • 7




    $begingroup$
    ... because $cos (tx)^2+sin(tx)^2=1$
    $endgroup$
    – Fakemistake
    Dec 29 '18 at 12:21
















2












2








2





$begingroup$


I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?






probability-theory characteristic-functions







share|cite|improve this question









$endgroup$




I'm studying some demonstrations of properties of characteristic function in which I have to use that $|e^{itx}| = 1$ but I don't understand it at all. Could you give a clue to demonstrate it?






probability-theory characteristic-functions




probability-theory characteristic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 12:19









Miguel AnguitaMiguel Anguita

485




485








  • 7




    $begingroup$
    ... because $cos (tx)^2+sin(tx)^2=1$
    $endgroup$
    – Fakemistake
    Dec 29 '18 at 12:21
















  • 7




    $begingroup$
    ... because $cos (tx)^2+sin(tx)^2=1$
    $endgroup$
    – Fakemistake
    Dec 29 '18 at 12:21










7




7




$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21






$begingroup$
... because $cos (tx)^2+sin(tx)^2=1$
$endgroup$
– Fakemistake
Dec 29 '18 at 12:21












2 Answers
2






active

oldest

votes


















2












$begingroup$

A key property of the exponential map is



$$e^ze^{z^prime} = e^{z+z^prime}$$



This can be proven using Cauchy product.



Based on that, you get for $y in mathbb R$



$$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much to you and all who answered!
    $endgroup$
    – Miguel Anguita
    Dec 29 '18 at 12:53



















3












$begingroup$

Hint: $$e^{itx}=cos(tx)+isin(tx)$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    A key property of the exponential map is



    $$e^ze^{z^prime} = e^{z+z^prime}$$



    This can be proven using Cauchy product.



    Based on that, you get for $y in mathbb R$



    $$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much to you and all who answered!
      $endgroup$
      – Miguel Anguita
      Dec 29 '18 at 12:53
















    2












    $begingroup$

    A key property of the exponential map is



    $$e^ze^{z^prime} = e^{z+z^prime}$$



    This can be proven using Cauchy product.



    Based on that, you get for $y in mathbb R$



    $$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much to you and all who answered!
      $endgroup$
      – Miguel Anguita
      Dec 29 '18 at 12:53














    2












    2








    2





    $begingroup$

    A key property of the exponential map is



    $$e^ze^{z^prime} = e^{z+z^prime}$$



    This can be proven using Cauchy product.



    Based on that, you get for $y in mathbb R$



    $$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$






    share|cite|improve this answer









    $endgroup$



    A key property of the exponential map is



    $$e^ze^{z^prime} = e^{z+z^prime}$$



    This can be proven using Cauchy product.



    Based on that, you get for $y in mathbb R$



    $$vert e^{iy}vert^2= e^{iy} overline{e^{iy}}=e^{iy}e^{-iy}=e^{iy-iy}=1$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 29 '18 at 12:28









    mathcounterexamples.netmathcounterexamples.net

    26.9k22158




    26.9k22158












    • $begingroup$
      Thank you very much to you and all who answered!
      $endgroup$
      – Miguel Anguita
      Dec 29 '18 at 12:53


















    • $begingroup$
      Thank you very much to you and all who answered!
      $endgroup$
      – Miguel Anguita
      Dec 29 '18 at 12:53
















    $begingroup$
    Thank you very much to you and all who answered!
    $endgroup$
    – Miguel Anguita
    Dec 29 '18 at 12:53




    $begingroup$
    Thank you very much to you and all who answered!
    $endgroup$
    – Miguel Anguita
    Dec 29 '18 at 12:53











    3












    $begingroup$

    Hint: $$e^{itx}=cos(tx)+isin(tx)$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hint: $$e^{itx}=cos(tx)+isin(tx)$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hint: $$e^{itx}=cos(tx)+isin(tx)$$






        share|cite|improve this answer









        $endgroup$



        Hint: $$e^{itx}=cos(tx)+isin(tx)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 12:23









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        79k42867




        79k42867






























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