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For what values $a$, the equation $x^n+a^n=0$ has $n$ different solutions? what are the solutions? (the question refers to complex solutions).

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If your equation were $x^n=a^n$ then the solutions would be $x=a\zeta_n^k$ for $0 \le k < n$, where $\zeta_n$ is a primitive $n$th root of unity. These are all distinct, save for when $a=0$. How might you modify this argument for your question? –  Clive Newstead Aug 11 '12 at 10:32

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By the Fundamental theorem of algebra, since for every value of $n>0$ this is a polynomial of degree $n>0$ there are exactly $n$ roots (counting multiplicy), in particular there is always $z\in\mathbb{C}$ that satisfies $z^n+a^n=0$.

In order to get all solutions you can use de Moivre's formula

Edit: For $a=0$ we have $x^n=0$ hence $x=0$ is the only solution, for $a\neq 0$ there are $n$ different solutions.

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i meant, different solutions.. –  Tomer Galanti Aug 11 '12 at 10:32
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@TomerGalanti - all the solutions are indeed different in your case, see Clive comment –  Belgi Aug 11 '12 at 10:34
    
Provided $a \neq 0$, all $n$ solutions are different. If $a=0$ then the only solution is $x=0$. –  user22805 Aug 11 '12 at 11:07
    
@DavidWallace - thank you, I edited –  Belgi Aug 11 '12 at 11:13

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