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Math Help - Complex numbers Factorization

  1. #1
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    Complex numbers Factorization

    Hello,

    I want to factorize the equation Z^n=1 over complex numbers

    I reached (x-1)(X^4+X^3+X^2+x+1) andd then got stucked
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  2. #2
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    Re: Complex numbers Factorization

    Hmm
    X
    x
    Z
    n (=5)??
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  3. #3
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    Re: Complex numbers Factorization

    Quote Originally Posted by Hyunqul View Post
    Hello,

    I want to factorize the equation Z^n=1 over complex numbers

    I reached (x-1)(X^4+X^3+X^2+x+1) andd then got stucked
    First of all, I don't know where x and X have come from, since your equation is in terms of z...

    Anyway, let \displaystyle z = r\,e^{i\theta}, then

    \displaystyle \begin{align*} z^n &= 1 \\ \left(r\,e^{i\theta}\right)^n &= e^{2\pi i} \\ r^ne^{i\theta n} &= e^{2 \pi i} \\ r^n = 1\textrm{ and }e^{i\theta n} &= e^{2 \pi i} \\ r = 1 \textrm{ and } i\theta n &= 2 \pi i \\ r = 1 \textrm{ and } \theta &= \frac{2 \pi}{n} \end{align*}

    Therefore \displaystyle z = e^{\frac{2 \pi i}{n}} = \cos{\left(\frac{2 \pi}{n}\right)} + i\sin{\left(\frac{2 \pi }{n}\right)}

    The solutions are all evenly spaced about a circle, so all have the same modulus and are separated by an angle of \displaystyle \frac{2\pi}{n}, so the solutions are

    \displaystyle \begin{align*} z_1 &= \cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)} \\ z_2 &= \cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)} \\ z_3 &= \cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)} \\ &\vdots \\ z_n &= \cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)} \\ &= \cos{(2\pi)} + i\sin{(2\pi)} \\ &= 1 + 0i \\ &= 1 \end{align*}

    Each solution \displaystyle z_i implies a factor of \displaystyle \left(z - z_i\right), so that means the factorised form (finally) is

    \displaystyle \begin{align*} z^n &= 1 \\ z^n - 1 &= 0 \\ \left\{ z - \left[\cos{\left(\frac{1\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{1 \cdot 2\pi}{n}\right)}\right]\right\} \cdot  \left\{z - \left[\cos{\left(\frac{2\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{2\cdot 2\pi}{n}\right)}\right]\right\} \cdot \left\{z - \left[\cos{\left(\frac{3\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{3\cdot 2\pi}{n}\right)}\right]\right\} \cdot \cdot \cdot \left\{ z - \left[\cos{\left(\frac{n\cdot 2\pi}{n}\right)} + i\sin{\left(\frac{n\cdot 2\pi}{n}\right)}\right] \right\} &=  0 \end{align*}
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  4. #4
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    Re: Complex numbers Factorization

    Here is a compact way to factor z^n-1=0.
    Let \xi  = \exp \left( {\frac{{2\pi i}}{n}} \right) then write \prod\limits_{k = 0}^{n - 1} {\left( {z - \xi ^k } \right)}  = 0
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  5. #5
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    Re: Complex numbers Factorization

    Quote Originally Posted by Plato View Post
    Here is a compact way to factor z^n-1=0.
    Let \xi  = \exp \left( {\frac{{2\pi i}}{n}} \right) then write \prod\limits_{k = 0}^{n - 1} {\left( {z - \xi ^k } \right)}  = 0
    Which is what we got in Post #3
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  6. #6
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    Re: Complex numbers Factorization

    The nth roots of unity, z such that z^n= 1 are given by z= re^{i\theta} such that z^n= r^ne^{ni\theta}= 1= (1)e^{i0} so we must have r= 1, n\theta= 0 which is, of course, equivalent to n\theta= 2\pi. That is, geometrically, the n nth roots of unity lie on the unit circle in the complex plane, equally space around the circle. The five fifth roots of unity, in particular, form a "pentagon" on the unit circle.
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