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    So when calculating roots of complex numbers e.g. if z^n = r[cos(x) + isin(x)] then z = r^1/n[cos(x/n + 2pik/n) + isin(x/n + 2pik/n)] but why is it that you choose values up to k=n-1?

    Also why is it that in some problems you can do k=-1, k=-2, etc?
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    (Original post by thekidwhogames)
    So when calculating roots of complex numbers e.g. if z^n = r[cos(x) + isin(x)] then z = r^1/n[cos(x/n + 2pik/n) + isin(x/n + 2pik/n)] but why is it that you choose values up to k=n-1?

    Also why is it that in some problems you can do k=-1, k=-2, etc?
    You know that the n^th roots of z, say \omega_i have to satisfy \omega_i^n = z. This is a polynomial equation of degree n, so can have at most n solutions.

    In your premise, taking k=0,1,2,3,... n-1 generates n different solutions (since cos and sine have least period 2pi) and so that must be all of the roots. To demonstrate this you have with k=0 the root z = r^{1/n}(\cos(x/n) + i \sin(x/n)) = r^{1/n}(\cos(x/n + 2\pi n / n) + i \sin(x/n + 2\pi n/n) which is the same as the root you get with k=0, since cos(y + 2pi) = cos y and sin(y+2pi) = sin y and 2pi * n/n = 2pi.

    Note that taking k=0,1,2,3,...n-1 is just convenient sometimes. In reality, taking k = {any sequence of n consecutive integers} will achieve the same thing, you will get n different roots and so that must be all the roots. In some problems, it is easier to take k = -n/2, -n/2 + 1,...,0,1,2,...,n/2 (for n even) but is equally valid.
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    The above boils down to: i) there are at most n nth roots of a complex number (for n an integer) and ii) cos and sin have least period 2pi, so cos(y + 2pi k/n) differs from cos(y+2pi m/n) whenever k is not 0 modulo m.
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    (Original post by Zacken)
    You know that the n^th roots of z, say \omega_i have to satisfy \omega_i^n = z. This is a polynomial equation of degree n, so can have at most n solutions.

    In your premise, taking k=0,1,2,3,... n-1 generates n different solutions (since cos and sine have least period 2pi) and so that must be all of the roots. To demonstrate this you have with k=0 the root z = r^{1/n}(\cos(x/n) + i \sin(x/n)) = r^{1/n}(\cos(x/n + 2\pi n / n) + i \sin(x/n + 2\pi n/n) which is the same as the root you get with k=0, since cos(y + 2pi) = cos y and sin(y+2pi) = sin y and 2pi * n/n = 2pi.

    Note that taking k=0,1,2,3,...n-1 is just convenient sometimes. In reality, taking k = {any sequence of n consecutive integers} will achieve the same thing, you will get n different roots and so that must be all the roots. In some problems, it is easier to take k = -n/2, -n/2 + 1,...,0,1,2,...,n/2 (for n even) but is equally valid.
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
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    (Original post by thekidwhogames)
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
    As long as your choice of k's work, yes.
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    (Original post by Zacken)
    As long as your choice of k's work, yes.
    Alright thanks!
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    No problem
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    (Original post by thekidwhogames)
    Ahh, that makes more sense, thanks! So markers in the exam will give us the mark if we chose other ks that are convenient?
    You should take care to notice the allowed domain. Sometimes the question might specify a complex number with argument -\pi \leq \theta < \pi, so there you would need to choose solutions as appropriate to fit that domain. I tend to do it by finding roots 1, 2, 3, ... n-1 and then subtracting 2\pi from solutions out of the domain.
 
 
 
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