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).
What are the solutions of $x^n+a^n=0$?
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polynomials
complex-numbers
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4If 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? – 2012-08-11
1 Answers
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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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0@DavidWallace - thank you, I edited – 2012-08-11