I was finding the solutions to this equation:
$x(x)^{1/3} = 81 $
and got $27,-27,27i$ and $-27i$ as solutions but apparently $27$ is the only solution according to wolfram. Why is that?
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Tina
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7What is your definition of $x^\frac{1}{3}$, when $x$ is complex? – zetko Aug 28 '24 at 20:22
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1Cf. this question – J. W. Tanner Aug 28 '24 at 20:43
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1Please youse mathjax to increase the readability of your posts. Here is a link for a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Tina Aug 28 '24 at 20:46
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@TimonKolt The question already uses MathJax. – Ben Steffan Aug 28 '24 at 20:47
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1@BenSteffan yes but only because i edited it – Tina Aug 28 '24 at 20:48
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1@BenSteffan You can see who edited the question and the original version clicking in "edited x mins ago". When a comment seems to make no sense it's a good idea to check the timeline of the edits before disqualifying a comment. – jjagmath Aug 28 '24 at 21:01
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In Mathematics there are two usual conventions for the meaning of $z^{1/n}$.
One is consider it as a multivaluated expression, so $z^{1/3}$ would actually represent three complex numbers. This brings several problems when considering things like $z^{1/3} + w^{1/2}$ having six possible values, or when it's used in an equation raises the question which value are you considering for the equation.
The other convention is to define more generally $$z^w = e^{w\ \text{Log}(z)}$$ where only the principal branch of $\text{Log}$ is used.
That's the definition used by Wolfram. Using this definition you have $$z^{1/3} = e^{\tfrac{1}{3}\text{Log}(z)}=e^{\tfrac{1}{3}(\log(|z|)+i\arg(z))}=e^{\tfrac{1}{3}\log(|z|)+\tfrac{1}{3}i\arg(z)}=|z|^{1/3}\left(\cos(\tfrac{1}{3}\arg(z))+i\sin(\tfrac{1}{3}\arg(z))\right)$$
You can verify that with this definition $27$ is the only solution.
jjagmath
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