I choose (a)
@fresh_42, which is the original problem I posed.
Ok, but then be aware of the fact that [imath] y [/imath] is our unknown, and [imath] x [/imath] is a given, known real number.
@fresh_42
So because \(\displaystyle \sqrt [6] x\) is a 6th degree polynomial, according to the fundamental theorem of algebra, it had 6 roots i.e. \(\displaystyle (x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5)(x - r_6) = 0\), is thatwhat you're saying? Is it a 6th degree polynomial though?
No. You have made another choice.
The polynomial is thus [imath] y^6-x=(y-c_1)\cdot (y-c_2)\cdot(y-c_3)\cdot(y-c_4)\cdot(y-c_5)\cdot(y-c_6) [/imath] by the fundamental theorem of algebra, yes. The complex numbers [imath] c_k\in \mathbb{C} [/imath] are called zeros or roots of the polynomial.
If you divide the circle into 6 parts, each angle will be 60 degrees, yep. Are we using symmetry to distribute the roots?
The other way around. The zeros [imath] c_k [/imath] turn out to be distributed that way. But let's proceed step by step.
We have [imath] y^6=x. [/imath] This can be written as [imath] 1=\dfrac{y^6}{x}= \left(\dfrac{y}{\sqrt[6]{x}}\right)^6=z^6 [/imath] with a new variable [imath] z=\dfrac{y}{\sqrt[6]{x}}. [/imath] Our equation is now [imath] z^6=1. [/imath] This is qualitatively the same problem as before, but I will not have to carry that nasty real radius [imath] \sqrt[6]{x} [/imath] with me all the time. We now have a radius one, or in your language
[math] z^6=1 \text{ AND } z\in \mathbb{R}\Longrightarrow z=\pm \sqrt[6]{1}=\pm 1. [/math]But what about the other, complex roots? We have
[math]\begin{array}{lll}
1&=z^6=(z-c_1)\cdot (z-c_2)\cdot(z-c_3)\cdot(z-c_4)\cdot(z-c_5)\cdot(z-c_6) \\
&=z^6 - z^5\cdot (c_1 + c_2 + c_3 + c_4 + c_5 + c_6)\\
&\phantom{=} + z^4\cdot (c_1 c_2 + c_1 c_3 + c_2 c_3 + c_1 c_4 + c_2 c_4 + c_3 c_4 + c_1 c_5 + c_2 c_5 + c_3 c_5 + c_4 c_5 + c_1 c_6 + c_2 c_6 + c_3 c_6 + c_4 c_6 + c_5 c_6 )\\
&\phantom{=} - z^3\cdot( c_1 c_2 c_3 + c_1 c_2 c_4 + c_1 c_3 c_4 +c_2 c_3 c_4 + c_1 c_2 c_5 + c_1 c_3 c_5 + c_2 c_3 c_5 + c_1 c_4 c_5 + c_2 c_4 c_5 + c_3 c_4 c_5 + c_1 c_2 c_6 + c_1 c_3 c_6 + c_2 c_3 c_6 + c_1 c_4 c_6 + c_2 c_4 c_6 + c_3 c_4 c_6 + c_1 c_5 c_6 + c_2 c_5 c_6 + c_3 c_5 c_6 + c_4 c_5 c_6 )\\
&\phantom{=} + z^2\cdot (c_1 c_2 c_3 c_4 + c_1 c_2 c_3 c_5 + c_1 c_2 c_4 c_5 + c_1 c_3 c_4 c_5 + c_2 c_3 c_4 c_5 + c_1 c_2 c_3 c_6 + c_1 c_2 c_4 c_6 + c_1 c_3 c_4 c_6 + c_2 c_3 c_4 c_6 + c_1 c_2 c_5 c_6 + c_1 c_3 c_5 c_6 + c_2 c_3 c_5 c_6 + c_1 c_4 c_5 c_6 + c_2 c_4 c_5 c_6 + c_3 c_4 c_5 c_6 )\\
&\phantom{=} - z^1\cdot ( c_1 c_2 c_3 c_4 c_5 + c_1 c_2 c_3 c_4 c_6+ c_1 c_2 c_3 c_5 c_6 + c_1 c_2 c_4 c_5 c_6 + c_1 c_3 c_4 c_5 c_6 + c_2 c_3 c_4 c_5 c_6)\\
&\phantom{=} + z^0\cdot+ c_1 c_2 c_3 c_4 c_5 c_6 \\
&=z^6-p_5(c_1,c_2,c_3,c_4,c_5,c_6)z^5+p_4(c_1,c_2,c_3,c_4,c_5,c_6)z^4-p_3(c_1,c_2,c_3,c_4,c_5,c_6)z^3+p_2(c_1,c_2,c_3,c_4,c_5,c_6)z^2-p_1(c_1,c_2,c_3,c_4,c_5,c_6)z+p_0(c_1,c_2,c_3,c_4,c_5,c_6)
\end{array}[/math]
We can try and solve this equation system for the roots [imath] c_k [/imath] now by comparing the coefficients of the polynomials. This gives
[math]\begin{array}{lll}
p_5(c_1,c_2,c_3,c_4,c_5,c_6)&=0\\
p_4(c_1,c_2,c_3,c_4,c_5,c_6)&=0\\
p_3(c_1,c_2,c_3,c_4,c_5,c_6)&=0\\
p_2(c_1,c_2,c_3,c_4,c_5,c_6)&=0\\
p_1(c_1,c_2,c_3,c_4,c_5,c_6)&=0
\end{array}[/math]and [imath] p_0(c_1,c_2,c_3,c_4,c_5,c_6)=c_1\cdot c_2\cdot c_3\cdot c_4\cdot c_5\cdot c_6 =1 .[/imath] This is nasty.
Your symmetry argument is faster. We can enumerate the roots however we want. Every permutation of the [imath] c_k [/imath] results in the exact same system of equations [imath] p_k(c_1,c_2,c_3,c_4,c_5,c_6)=0\; (k=1,\ldots,5). [/imath] There is no way to distinguish the roots from here. We also know that [imath] c_1=1 [/imath] is a valid root. It is a point on the unit circle, the circle of radius one and center in the origin of a Cartesian coordinate system. This means that all roots have the length one, since otherwise, we could distinguish them from [imath] c_1 [/imath] which we can't. So all roots are on the unit circle. They also have to be distributed uniformly, since otherwise, we could distinguish them from [imath] c_1 [/imath] which we can't. All multivariate polynomials [imath] p_k(c_1,c_2,c_3,c_4,c_5,c_6) (k<6) [/imath] are zero and the product of all roots is one.
The coordinate system with the unit circle lives in the plane with a real coordinate and a pure imaginary coordinate because our roots [imath] c_k [/imath] are complex numbers. (I cannot name the axis since you occupied [imath] x,y [/imath] and any other letters are misleading. So we have a real axis and a purely imaginary axis and six roots uniformly distributed on the unit circle, in this case by a distance of [imath] \dfrac{360°}{6}=60° [/imath] by counting from [imath] c_1=1 [/imath] on the real axis, i.e. at [imath] 0°. [/imath]
Euler's formula gives us all six roots [imath] (k=1,2,3,4,5,6) [/imath]
[math]
c_k= \mathfrak{Re}(c_k)+ i \cdot \mathfrak{Im}(c_k) =\cos \left(\dfrac{k}{n}\cdot 360°\right)+i \cdot \sin \left(\dfrac{k}{n}\cdot 360°\right)=e^{\frac{360° i \cdot k}{n}}=e^{ \frac{2\pi i k}{n}}
[/math]
