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A real number may have a complex $n^{\text{th}}$ root. Can a non-real complex number have a real $n^{\text{th}}$ root?

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    $\begingroup$ By 'non real complex number' do you mean an imaginary number? Also, whilst I think I understand what you're trying to say, the question as you've presented it makes no sense $\endgroup$
    – mrnovice
    Commented Feb 25, 2017 at 6:47
  • $\begingroup$ @mrnovice how does the question make no sense if you think you've understood what they're trying to say? $\endgroup$
    – Mike Pierce
    Commented Feb 25, 2017 at 6:51
  • $\begingroup$ That's a fair point $\endgroup$
    – mrnovice
    Commented Feb 25, 2017 at 7:04

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Can a nonreal complex number have a real nth root?

Hint:  can a real number raised to the $n^{th}$ power be a non-real complex number?

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