Recent content by ribbon

  1. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Ah yes so forget the negative ninth pi, was 17pi/9 correct? And 19pi/9?
  2. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Cosine is an even function so throw in -pi/9 as well. As for the third how about 17∏/9?
  3. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    cos(pi/3) = cos(3θ) cant I take cos inverse of both sides yielding pi/3 = 3θ? then θ=1/9 pi of course times some scalar k?
  4. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Sure so r = 1 and θ=1pi/9
  5. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Ok we have cosine of an angle corresponding to 1/2 and the sin corresponding to (sqrt3)/2 so the θ in question should be ∏/3 (60 degrees). The r = (1/2)^2 + (sqrt3/4)^2 r = 1/4 + 3/4 r = 1
  6. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    That kind of looks like the set up to DeMoivre theorem? I know if it is r(cosθ + isinθ)^n = r^n(cos nθ + isin nθ) But how does this help us?
  7. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    No my book never mentioned anything about this, but it makes sense that we have a positive and a negative version of the root. That would give me four if I included the conjugates to the one I wrote, what would the 3rd set be??
  8. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Oh no, do you get something different when you take the cube root?
  9. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    However we would have 6 roots to this polynomial would we not, by the fundamental theorem of algebra?
  10. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    OMG how silly of me yes you'd take the cube root of the positive and negative answers each so, z = (-1 + i3^1/6)/(2^1/3) z = (-1 - i3^1/6)/(2^1/3)
  11. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Wicked alright let u = z^3 We have u^2 + u + 1 = 0 then u = -1/2 +- [(√3)/2]i Now given that we took u = z^3, can I just cube the results to get the answers for z??
  12. R

    Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

    Homework Statement Find all complex solutions of z^6 + z^3 + 1 (z^3 + 1)/(z^3 - 1) = i Homework Equations The Attempt at a Solution I am going crazy with trial and error with these, there must be some systematic method or tricks that I am oblivious of. For the second question I...
  13. R

    Cardinality of a subset of [0,1]

    Hmm, ok well could the sequence be 0.1, 0.101, 0.010101... till I hit that n (n being the finite number of non zero digits?
  14. R

    Cardinality of a subset of [0,1]

    It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality. But it seems to be otherwise now.
  15. R

    Cardinality of a subset of [0,1]

    Could I go, f(1) = 0.11 f(2) = 0.12 f(3) = 0.13 ... and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?