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Size:  73.7 KB if this is true for when Z^n=1 then why, for z^10=1, does k= +/-1,+/-2,+/-3, +/-4 and 5? I have the general solution as cos kpi/5+isin kpi/5. I would be grateful for any explanation😅


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    (Original post by Mr Pussyfoot)
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Size:  73.7 KB if this is true for when Z^n=1 then why, for z^10=1, does k= +/-1,+/-2,+/-3, +/-4 and 5? I have general solution as cos kpi/5+isin kpi/5. I would be grateful for any explanation😅


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    Those values of k are the principal solutions
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    (Original post by Protoxylic)
    Those values of k are the principal solutions
    So wouldn't the solutions be all the numbers -1<k<10?


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    (Original post by Mr Pussyfoot)
    So wouldn't the solutions be all the numbers -1<k<10?


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    You can either have all the values of k for which the the solutions have argument (0,2pi) or (-pi,pi) these tend to be the principal domains. Technically any domain with period 2pi, but the principal tends to be the above two.
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    (Original post by Mr Pussyfoot)
    So wouldn't the solutions be all the numbers -1<k<10?


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    They are, remember that the -1 and n-1 solutions are the same as they differ by 2*pi. With -2 and n-2, -3 and n-3 being the same as well etc.
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    By principal do you mean between 0 and 2pi? Sorry I hadn't read the previous posts. Ignore this😂


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    (Original post by Protoxylic)
    You can either have all the values of k for which the the solutions have argument (0,2pi) or (-pi,pi) these tend to be the principal domains.
    All right cool I understand now


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    (Original post by CharlieGEM)
    They are, remember that the -1 and n-1 solutions are the same as they differ by 2*pi. With -2 and n-2, -3 and n-3 being the same as well etc.
    Thanks for the reminder, I hadn't spotted the relation at first.



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