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Is {(-2)^5}^(1/5) a complex number ?

  1. Aug 7, 2013 #1
    Last edited by a moderator: Aug 7, 2013
  2. jcsd
  3. Aug 7, 2013 #2

    pwsnafu

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    Science Advisor

    Notice how the alpha page says
    If you select the real valued option you'll get -2.
     
  4. Aug 7, 2013 #3

    lurflurf

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    You have a typo
    E = {(-2)^5}^(1/5)=(-32)^(1/5)
    There are five numbers such that x^5=-32
    (-32)^(1/5) should be one of them
    they are
    1.61803398874989+1.17557050458495 i
    -0.61803398874989+1.90211303259031 i
    -2
    -0.61803398874989-1.90211303259031 i
    1.61803398874989-1.17557050458495 i

    We chose 1.61803398874989+1.17557050458495 i as it is first on the list and the most reasonable choice. Some people inexplicably pick -2.
     
  5. Aug 7, 2013 #4

    D H

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    There are *five* solutions to x5=(-2)5. One of them is the real number -2. The other four are complex numbers. Wolfram Alpha gives you the principal root by default. You can force it to yield the real-valued root by clicking on "Use the real‐valued root instead".

    You can see all five solutions if you instead ask WA to solve x5=(-2)5.
     
  6. Aug 7, 2013 #5
    got it. Thanks everyone!
     
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