Good treatment of Euler’s formula

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I need help because I'm a bit confused about the correct way to treat Euler's formula.



If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.



Note: I will appreciate a link to a good explanation about that kind of things.



Thank you







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  • 2




    It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
    – Lord Shark the Unknown
    Aug 24 at 8:21














up vote
1
down vote

favorite
1












I need help because I'm a bit confused about the correct way to treat Euler's formula.



If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.



Note: I will appreciate a link to a good explanation about that kind of things.



Thank you







share|cite|improve this question


















  • 2




    It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
    – Lord Shark the Unknown
    Aug 24 at 8:21












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I need help because I'm a bit confused about the correct way to treat Euler's formula.



If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.



Note: I will appreciate a link to a good explanation about that kind of things.



Thank you







share|cite|improve this question














I need help because I'm a bit confused about the correct way to treat Euler's formula.



If I ask Wolfram Alpha (WA) about $(i^i)^4$ the answer is $e^(-2 pi)$, therefore, I thought, ok this must be equal to $(e^i 2 pi)^i$ but, this is not true, WA says that is equal 1...but, it says that its general form is $e^-2pi n$ but that can not be ever 1...so my head explodes here.



Note: I will appreciate a link to a good explanation about that kind of things.



Thank you









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 8:30









InertialObserver

37719




37719










asked Aug 24 at 8:20









Perico Cruel

245




245







  • 2




    It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
    – Lord Shark the Unknown
    Aug 24 at 8:21












  • 2




    It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
    – Lord Shark the Unknown
    Aug 24 at 8:21







2




2




It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
– Lord Shark the Unknown
Aug 24 at 8:21




It's dangerous to write $a^b$ unless either $a$ is a positive real or $b$ is an integer.
– Lord Shark the Unknown
Aug 24 at 8:21










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










The result is obtained from



$$i=e^fracipi2+2 k pi i$$



then



$$i^i=e^-fracpi2-2k pi$$



and



$$(i^i)^4=e^-2pi-8k pi$$






share|cite|improve this answer


















  • 1




    You mean $i=e^fracipi2+2kpi i$ etc.
    – Lord Shark the Unknown
    Aug 24 at 8:36










  • @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
    – gimusi
    Aug 24 at 8:39










  • But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
    – Perico Cruel
    Aug 24 at 9:24











  • What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
    – Perico Cruel
    Aug 24 at 9:30











  • @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
    – gimusi
    Aug 24 at 10:29

















up vote
2
down vote













The complex exponential does not satisfy $a^bc=(a^b)^c$.



Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).



Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers



More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    The result is obtained from



    $$i=e^fracipi2+2 k pi i$$



    then



    $$i^i=e^-fracpi2-2k pi$$



    and



    $$(i^i)^4=e^-2pi-8k pi$$






    share|cite|improve this answer


















    • 1




      You mean $i=e^fracipi2+2kpi i$ etc.
      – Lord Shark the Unknown
      Aug 24 at 8:36










    • @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
      – gimusi
      Aug 24 at 8:39










    • But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
      – Perico Cruel
      Aug 24 at 9:24











    • What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
      – Perico Cruel
      Aug 24 at 9:30











    • @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
      – gimusi
      Aug 24 at 10:29














    up vote
    2
    down vote



    accepted










    The result is obtained from



    $$i=e^fracipi2+2 k pi i$$



    then



    $$i^i=e^-fracpi2-2k pi$$



    and



    $$(i^i)^4=e^-2pi-8k pi$$






    share|cite|improve this answer


















    • 1




      You mean $i=e^fracipi2+2kpi i$ etc.
      – Lord Shark the Unknown
      Aug 24 at 8:36










    • @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
      – gimusi
      Aug 24 at 8:39










    • But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
      – Perico Cruel
      Aug 24 at 9:24











    • What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
      – Perico Cruel
      Aug 24 at 9:30











    • @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
      – gimusi
      Aug 24 at 10:29












    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    The result is obtained from



    $$i=e^fracipi2+2 k pi i$$



    then



    $$i^i=e^-fracpi2-2k pi$$



    and



    $$(i^i)^4=e^-2pi-8k pi$$






    share|cite|improve this answer














    The result is obtained from



    $$i=e^fracipi2+2 k pi i$$



    then



    $$i^i=e^-fracpi2-2k pi$$



    and



    $$(i^i)^4=e^-2pi-8k pi$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 24 at 8:41

























    answered Aug 24 at 8:26









    gimusi

    69.7k73686




    69.7k73686







    • 1




      You mean $i=e^fracipi2+2kpi i$ etc.
      – Lord Shark the Unknown
      Aug 24 at 8:36










    • @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
      – gimusi
      Aug 24 at 8:39










    • But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
      – Perico Cruel
      Aug 24 at 9:24











    • What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
      – Perico Cruel
      Aug 24 at 9:30











    • @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
      – gimusi
      Aug 24 at 10:29












    • 1




      You mean $i=e^fracipi2+2kpi i$ etc.
      – Lord Shark the Unknown
      Aug 24 at 8:36










    • @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
      – gimusi
      Aug 24 at 8:39










    • But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
      – Perico Cruel
      Aug 24 at 9:24











    • What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
      – Perico Cruel
      Aug 24 at 9:30











    • @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
      – gimusi
      Aug 24 at 10:29







    1




    1




    You mean $i=e^fracipi2+2kpi i$ etc.
    – Lord Shark the Unknown
    Aug 24 at 8:36




    You mean $i=e^fracipi2+2kpi i$ etc.
    – Lord Shark the Unknown
    Aug 24 at 8:36












    @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
    – gimusi
    Aug 24 at 8:39




    @LordSharktheUnknown Opssss yes of course! I fix that. Thanks
    – gimusi
    Aug 24 at 8:39












    But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
    – Perico Cruel
    Aug 24 at 9:24





    But, if I do $i=e^fracipi2+2 k pi i$ then $i^4=e^2pi i-8k pi i$ and $(i^4)^i=e^-2pi+8k pi$...but is not true, because $(i^i)^4 != (i^4)^i$
    – Perico Cruel
    Aug 24 at 9:24













    What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
    – Perico Cruel
    Aug 24 at 9:30





    What I mean is: Is that true: $((e^(i*(2/pi) + 2*k*pi*i)) ^i)^4 = ((e^(i*(2/pi) + 2*k*pi*i)) ^4)^i$ ?
    – Perico Cruel
    Aug 24 at 9:30













    @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
    – gimusi
    Aug 24 at 10:29




    @PericoCruel If we consider the multivalued $i^4=e^ileft(fracpi2+2kpiright)$ then $(i^4)^i=e^-2pi-8k pi=(i^i)^4$.
    – gimusi
    Aug 24 at 10:29










    up vote
    2
    down vote













    The complex exponential does not satisfy $a^bc=(a^b)^c$.



    Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).



    Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers



    More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.






    share|cite|improve this answer
























      up vote
      2
      down vote













      The complex exponential does not satisfy $a^bc=(a^b)^c$.



      Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).



      Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers



      More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        The complex exponential does not satisfy $a^bc=(a^b)^c$.



        Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).



        Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers



        More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.






        share|cite|improve this answer












        The complex exponential does not satisfy $a^bc=(a^b)^c$.



        Of course the question is how you define the complex exponential in the first place. In gimusi's answer, it is a multi-valued function. Personally, I prefer to define it as single-valued (which means it is not defined or discontinuous somewhere).



        Some information is on Wikipedia at https://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers



        More application-oriented textbooks (such as Wunsch, Complex Variables with Applications or Brown-Churchill, Complex Variables and Applications) define $a^b$ as multi-valued. More theory-oriented textbooks (Lang, Complex Analysis or Stein-Shakarchi, Complex Analysis) tend to make $a^b$ single-valued, but only defined on a subset of the complex numbers, and dependent on a choice of a branch of the logarithm.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 24 at 8:43









        Kusma

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