I"ve noticed the roots of $$(x^k-1 =0) = (1-x^k=0)$$ This is simple enough since changing +/- signs gives you the additive inverse which is equal at zero. However this property can be also be thought of as the two permutations of the coefficents with the 2nd roots of unity e.g.
$ [(1)a+(-1)b=0] = [(-1)a +(1)b=0]$
Was wondering if this property holds for other roots of unity and under what conditions i.e.
Let $\omega =$ 3rd root of unity. $$(\omega^1)a+(\omega^2)b+(\omega^3)c = 0$$
Do the six (3!) permutations of this equation have the same roots? Perhaps only if there are three distinct roots? What about composite roots of unity?
Any hints or further reading suggestions welcome. Thank you.
Edit: To be more clear. If we take a random 3rd degree trinomial such as: $3x^3+5x+14$ and multiply each coeffecient by one of the three 3rd roots of unity ($\omega^3 =1$ of course) and set the equation equal to zero we get: $$\omega^1 3x^3+\omega^2 5x+14=0$$ This can be permutated in 6 ways by changing which of the three roots each coeffecient gets paired with: $$ 1.\omega^1 3x^3+\omega^2 5x+14 =0$$ $$2.\omega^1 5x+\omega^2 3x^3+14 =0$$ $$3.\omega^1 14+\omega^2 5x+3x^3 =0$$ $$4.\omega^1 14+\omega^2 3x^3+5x =0$$ $$5.\omega^1 3x^3+\omega^2 14+5x =0$$ $$6.\omega^1 5x+\omega^2 14+3x^3 =0$$
Per wolfram the roots of these equations are indeed identical.
The deeper idea here is that just as an additive inverse will be exactly the same at =0 as it is just a reflection. The same can hold for all nth roots of unity which can be seen as reflectional/rotational inverses of the nth degree and will be therefore be exactly equivalent at 0.
Edit: #2
After some more analysis and feedback from comments I think we can confirm this hypothesis is likely true only when:
1.n is prime.
2.The equation is of degree n.
Edit: #3 Upon closer inspection the six equations don't have identical roots. Instead there are two sets of complex roots that differ in which of their parts are positive/negative. Each set is only valid for 3 equations. So while perhaps still interesting, the hypothesis is false.
Two equations of the same degree can have same set of roots (with same multiplicity) iff one is obtained from the other by uniformly multiplying the coefficients by the same number.
For higher degree ($>2$) your modification ("permutation of coefficients" in your words) will amount to each coefficient getting multiplied by different elements and hence will not have the same roots.