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I Definition of the root of -1 for different roots

  1. Mar 15, 2017 #1
    How does the value of ##\displaystyle \sqrt[a]{-1}## vary as ##a## varies as any real number? When is this value complex and when is it real? For example, we know that when a = 2 it is complex, but when a = 3 it is real. What about when ##a = \pi##, for example?
     
  2. jcsd
  3. Mar 15, 2017 #2
  4. Mar 15, 2017 #3

    Mark44

    Staff: Mentor

    As far as I know, there is no such thing as the "π-th root" of a number. The index of a radical is a positive integer that is 2 or larger. You can however raise a number to an arbitrary power. For example, ##2^{1/\pi} = (e^{\ln 2})^{1/\pi} = e^{\frac{\ln 2}{\pi}}##, but see the link that @SlowThinker posted.
     
  5. Mar 15, 2017 #4
    if ##x^{1/a} = e^{\frac{\ln x}{a}}##, and we know that ##(-1)^{1/3} = -1##, does that mean that ##(-1)^{1/3} = e^{\frac{\ln (-1)}{3}} = -1##?
     
  6. Mar 15, 2017 #5

    Mark44

    Staff: Mentor

    Not if ln means the usual natural logarithm function whose domain is positive real numbers.
     
  7. Mar 15, 2017 #6
    So in general when would ##(-1)^{1/a}## be complex and when would it be real?
     
  8. Mar 15, 2017 #7
    ##(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}=-1## this is correct but you dont consider the other results,
    ##\sqrt4=2## is correct but not complete, becuase ##\sqrt4=\pm 2##
    in the same way are other results for ##(-1)^{\frac{1}{3}}=e^\frac{ln(-1)}{3}##

    look the post https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
     
  9. Mar 15, 2017 #8

    Mark44

    Staff: Mentor

    No. ##\sqrt 4## is generally accepted to mean the principal square root of 4, a positive number.
     
  10. Mar 15, 2017 #9

    Stephen Tashi

    User Avatar
    Science Advisor

    That definition only applies when ##x## is positive.

    (In the domain of real numbers, there is no standard definition for raising a negative number to an irrational exponent. )
     
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