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I know what the roots are - according to this page: How to solve $x^3 = 1$?

But i don't understanding why the roots are: $e^{\frac23i\pi}$ and $e^{\frac43i\pi}$.

Community
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user11001
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  • What does mean the expression "equivalent roots"? Two numbers can't be equivalent but they can be equal (and here, it's not the case). – paf Aug 20 '16 at 15:22
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    Easy non-explanation answer: because they work, try them. You can follow the algebra in the link you posted....but also, there is a picture that goes with this. Once you see the picture it becomes much clearer, though the analysts will disagree, why they are the same and why there are only three(although the algebraist will disagree). – ReverseFlowControl Aug 20 '16 at 15:22
  • How much do you understand complex analysis. If you know the basics it's ... basic. If you don't it's imponderable. Every non-zero complex number a + bi may be written as $r(\cos x + i \sin x)$ where $r = |a+ bi| = \sqrt{a^2 + b^2}$ and cos x = a/r and sin y = b/r. For notation (and other reasons) we define $r(\cos x + i \sin x) = e^{ix}$. If you experiment it's easy to show if $z = re^{ix}$ and $w = se^{iy}$ then $zw = (rw)e^{i(x + y)}$. Try it, it works. So $z^3 = r^3{e^{ix}}^3 = 1= r*e^{i0}$ means $r = \sqrt[3]{1}=1$ and $3x = 0 = 2\pi=4\pi$ so $x = 0,2\pi/3, 4\pi/3$. – fleablood Aug 21 '16 at 19:06

2 Answers2

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Whenever you want to solve an equation, you need to be precise about the field in which you are looking for the roots. In general, unless you are in $\mathbb{C}$ where any polynomial has as many roots as its degree, an equation may have no root (example $x^2-2$ over $\Bbb{Q}$) one root (example $x^3-1$ over $\Bbb{R}$)...

In your case when looking to complex roots of the form $x=\rho\cdot e^{i\theta}$ one has $x^3=1=\rho^3\cdot e^{3i\theta}$ and this means besides $\rho=1$

$$3i\theta=2k\pi\,\,\,(k=0,1,2)$$

And this gives three distinct solutions $1,e^{2i\pi/3},e^{4i\pi/3}$

marwalix
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Because of $e^{i2\pi}=1$ and therefore also $e^{i4\pi}=1$ the roots are solving the equation. Think about $(-1)\cdot(-1)=1\cdot 1$, that's the same type of problem.

user90369
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