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Proving/Creating a conjecture on the roots of complex numbers

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Formulate a conjecture for the equation (z^3)-1=0, (z^4)-1=0 (z^5)-1=0
    and prove it.

    2. Relevant equations
    r^n(cosnθ + isinnθ)


    3. The attempt at a solution

    Well my conjecture is that 2pi/n and 2pi/n + pi are possible values. I'm a bit iffy on how to word it. don't know which way I should go for a proof
     
  2. jcsd
  3. Oct 22, 2012 #2

    tiny-tim

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    Hi Daaniyaal! :smile:

    (try using the X2 button just above the Reply box :wink:)
    Start "if rn(cosnθ + isinnθ) = 1, then …" :smile:
     
  4. Oct 23, 2012 #3
    if rn(cosnθ + isinnθ)=1 then possible values for n are 2pi/n and 2pi/n+pi.
     
  5. Oct 23, 2012 #4

    tiny-tim

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    i was thinking, more like …

    if rn(cosnθ + isinnθ) = 1,

    then r = 1, and cosnθ = 1,

    so nθ = … ? :smile:
     
  6. Oct 23, 2012 #5
    0 most definitely
     
  7. Oct 23, 2012 #6

    Mark44

    Staff: Mentor

    Aren't there other values for which cos(nθ) = 1?
     
  8. Oct 23, 2012 #7
  9. Oct 23, 2012 #8
    By the way how would I put capital pi notation into this?

    My conjecture is sqrt(2-2coskpi/n)) in pi notation: n-1, k=1
     
  10. Oct 23, 2012 #9

    Mark44

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    ???
    How did you get that?
     
  11. Oct 23, 2012 #10

    Mark44

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    What do you mean by "capital pi notation?"
     
  12. Oct 24, 2012 #11

    tiny-tim

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    (just got up :zzz:)
    Come on, Daaniyaal …

    what are all the solutions to cosψ = 1 ?​

    (draw a graph of cos)
     
  13. Oct 24, 2012 #12
  14. Oct 24, 2012 #13
    I think an alternate form of writing this would be e^(i x) = 1
     
  15. Oct 24, 2012 #14
    and I can't get the superscript button to work :(
     
  16. Oct 24, 2012 #15

    tiny-tim

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    correct … ψ = 2π times n , for any value of n

    but since we're already using n in the question, let's write that as …

    ψ = 2π times k , for any value of k

    sooo … what are all the solutions (for θ) of cos = 1 ?
    yes :smile:
    do you have javascript turned off?

    [noparse]alternatively, you can just type before and after[/noparse] :wink:
     
  17. Oct 24, 2012 #16
    2,4,6,8,10,12 and so on?
     
  18. Oct 24, 2012 #17
    All even numbers basically
     
  19. Oct 24, 2012 #18

    tiny-tim

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    θ = 2,4,6,8,10,12 … are the solutions to cosnθ = 1 ?
     
  20. Oct 24, 2012 #19

    Mark44

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  21. Oct 24, 2012 #20

    Mark44

    Staff: Mentor

    Daaniyaal, it would be helpful if you replied with complete statements. tiny-tim is trying to get you to do that with what he says below.

     
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