$-1$ to the power of a irrational number












3












$begingroup$


According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.



But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.



So why is $(-1)^pi$ a complex number, if there is no even denominator?



*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.






complex-numbers exponentiation pi







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Do you know how $(-1)^pi$ is defined?
    $endgroup$
    – James
    Mar 9 at 14:13












  • $begingroup$
    In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
    $endgroup$
    – Oscar Lanzi
    Mar 9 at 14:15
















3












$begingroup$


According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.



But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.



So why is $(-1)^pi$ a complex number, if there is no even denominator?



*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.






complex-numbers exponentiation pi







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Do you know how $(-1)^pi$ is defined?
    $endgroup$
    – James
    Mar 9 at 14:13












  • $begingroup$
    In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
    $endgroup$
    – Oscar Lanzi
    Mar 9 at 14:15














3












3








3





$begingroup$


According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.



But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.



So why is $(-1)^pi$ a complex number, if there is no even denominator?



*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.






complex-numbers exponentiation pi







share|cite|improve this question











$endgroup$




According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.



But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.



So why is $(-1)^pi$ a complex number, if there is no even denominator?



*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.






complex-numbers exponentiation pi




complex-numbers exponentiation pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 10 at 3:18









YuiTo Cheng

2,0532637




2,0532637










asked Mar 9 at 14:12









EnzoEnzo

662




662








  • 4




    $begingroup$
    Do you know how $(-1)^pi$ is defined?
    $endgroup$
    – James
    Mar 9 at 14:13












  • $begingroup$
    In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
    $endgroup$
    – Oscar Lanzi
    Mar 9 at 14:15














  • 4




    $begingroup$
    Do you know how $(-1)^pi$ is defined?
    $endgroup$
    – James
    Mar 9 at 14:13












  • $begingroup$
    In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
    $endgroup$
    – Oscar Lanzi
    Mar 9 at 14:15








4




4




$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13






$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13














$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15




$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15










2 Answers
2






active

oldest

votes


















13












$begingroup$

This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.



However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
    $endgroup$
    – drhab
    Mar 9 at 14:23












  • $begingroup$
    @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
    $endgroup$
    – saulspatz
    Mar 9 at 14:25






  • 2




    $begingroup$
    You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
    $endgroup$
    – user1551
    Mar 9 at 22:14






  • 2




    $begingroup$
    @user1551 Even when $z$ is real?
    $endgroup$
    – saulspatz
    Mar 9 at 22:47










  • $begingroup$
    The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
    $endgroup$
    – user1551
    Mar 10 at 5:36



















8












$begingroup$

For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.



Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.






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









    13












    $begingroup$

    This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
    which is the value given by WolframAlpha.



    However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
      $endgroup$
      – drhab
      Mar 9 at 14:23












    • $begingroup$
      @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
      $endgroup$
      – saulspatz
      Mar 9 at 14:25






    • 2




      $begingroup$
      You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
      $endgroup$
      – user1551
      Mar 9 at 22:14






    • 2




      $begingroup$
      @user1551 Even when $z$ is real?
      $endgroup$
      – saulspatz
      Mar 9 at 22:47










    • $begingroup$
      The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
      $endgroup$
      – user1551
      Mar 10 at 5:36
















    13












    $begingroup$

    This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
    which is the value given by WolframAlpha.



    However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
      $endgroup$
      – drhab
      Mar 9 at 14:23












    • $begingroup$
      @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
      $endgroup$
      – saulspatz
      Mar 9 at 14:25






    • 2




      $begingroup$
      You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
      $endgroup$
      – user1551
      Mar 9 at 22:14






    • 2




      $begingroup$
      @user1551 Even when $z$ is real?
      $endgroup$
      – saulspatz
      Mar 9 at 22:47










    • $begingroup$
      The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
      $endgroup$
      – user1551
      Mar 10 at 5:36














    13












    13








    13





    $begingroup$

    This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
    which is the value given by WolframAlpha.



    However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.






    share|cite|improve this answer











    $endgroup$



    This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
    which is the value given by WolframAlpha.



    However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 10 at 1:59









    J. W. Tanner

    3,3351320




    3,3351320










    answered Mar 9 at 14:20









    saulspatzsaulspatz

    17k31435




    17k31435












    • $begingroup$
      You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
      $endgroup$
      – drhab
      Mar 9 at 14:23












    • $begingroup$
      @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
      $endgroup$
      – saulspatz
      Mar 9 at 14:25






    • 2




      $begingroup$
      You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
      $endgroup$
      – user1551
      Mar 9 at 22:14






    • 2




      $begingroup$
      @user1551 Even when $z$ is real?
      $endgroup$
      – saulspatz
      Mar 9 at 22:47










    • $begingroup$
      The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
      $endgroup$
      – user1551
      Mar 10 at 5:36


















    • $begingroup$
      You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
      $endgroup$
      – drhab
      Mar 9 at 14:23












    • $begingroup$
      @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
      $endgroup$
      – saulspatz
      Mar 9 at 14:25






    • 2




      $begingroup$
      You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
      $endgroup$
      – user1551
      Mar 9 at 22:14






    • 2




      $begingroup$
      @user1551 Even when $z$ is real?
      $endgroup$
      – saulspatz
      Mar 9 at 22:47










    • $begingroup$
      The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
      $endgroup$
      – user1551
      Mar 10 at 5:36
















    $begingroup$
    You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
    $endgroup$
    – drhab
    Mar 9 at 14:23






    $begingroup$
    You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
    $endgroup$
    – drhab
    Mar 9 at 14:23














    $begingroup$
    @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
    $endgroup$
    – saulspatz
    Mar 9 at 14:25




    $begingroup$
    @drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
    $endgroup$
    – saulspatz
    Mar 9 at 14:25




    2




    2




    $begingroup$
    You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
    $endgroup$
    – user1551
    Mar 9 at 22:14




    $begingroup$
    You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
    $endgroup$
    – user1551
    Mar 9 at 22:14




    2




    2




    $begingroup$
    @user1551 Even when $z$ is real?
    $endgroup$
    – saulspatz
    Mar 9 at 22:47




    $begingroup$
    @user1551 Even when $z$ is real?
    $endgroup$
    – saulspatz
    Mar 9 at 22:47












    $begingroup$
    The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
    $endgroup$
    – user1551
    Mar 10 at 5:36




    $begingroup$
    The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
    $endgroup$
    – user1551
    Mar 10 at 5:36











    8












    $begingroup$

    For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.



    Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
    exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$    .






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.



      Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
      exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$      .






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.



        Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
        exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$        .






        share|cite|improve this answer









        $endgroup$



        For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.



        Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
        exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$        .







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 9 at 14:25









        GEdgarGEdgar

        63.1k267171




        63.1k267171






























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