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Math Help - Roots of Unity represented on a regular polygon - lenght of side.

  1. #1
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    Roots of Unity represented on a regular polygon - lenght of side.

    Hello,

    I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.

    I did all of that , but I have a critical question ,

    is there a relation ship between the power n and the length of the polygon side?

    I mean , for the euqation Z^3 - 1 = 0 for example, is there a formula relating the power to the lenght of one side of the polygon ?
    Thanks.
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  2. #2
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    For each value of n, how is the length of the side of a polygon related to the central angle subtended by the side? (The side is a chord of the unit circle.)
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Quote Originally Posted by Hyunqul View Post
    I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.
    There are n nth roots of unity.
    \rho_k  = \exp \left( {\frac{{2\pi ki}}{n}} \right),~k=0,1,\cdots,n-1.
    The length of the side of the polygon is |\rho_{k+1}-\rho_{k}|.
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Thank you very much for your response,

    But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Quote Originally Posted by Hyunqul View Post
    Thank you very much for your response,
    But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?
    O.K. Let \theta_n=\frac{2\pi}{n} then
    \exp(\theta_nk\mathbf{i})=\cos(\theta_nk)+\mathbf{  i}\sin(\theta_nk)
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Thanks alot sir , I highly appreciate you help.

    If I could ask one more question ... is there a proof for this rule ? The one relating roots of unity to the length of one side.
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  7. #7
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    1 can be represented in polar form in the obvious way: 1e^{i(0)}. The nth roots are given by 1^{1/n}e^{\frac{0+ 2k\pi}{n}}= e^{\frac{2k\pi}{n}} for k= 0 to n- 1. That is, 1 has n nth roots, equally spaced around the unit circle. If you draw lines from each root to the origin, you have n isosceles triangles and the angle at the vertex (the origin) is \frac{2\pi}{n}. If you draw a perpendicular from the origin to one side, it divides that isosceles triangle into two right triangles having hypotenuse 1 and angle \frac{\pi}{n}. The "opposite side" to that angle is given by [itex]sin(\frac{\pi}{n})[/tex] and is 1/2 the length of one side of the polygon. The length of one side, then, is 2sin\left(\frac{\pi}{n}\right).

    As a check note that if n= 4, the polygon is a square, having diagonals of length 2. s= 2sin(\frac{\pi}{4})= 2\left(\frac{\sqrt{2}}{2}\right)= \sqrt{2} while, by the Pythagorean theorem a^2+ a^2= 2^2 so a^2= 2, a= \sqrt{2}.

    Similarly, if n= 6, the polygon is a hexagon having diagonals of length 2. s= 2 sin(\frac{\pi}{6})= 2\left(\frac{1}{2}\right)= 1.
    Last edited by HallsofIvy; September 2nd 2011 at 02:01 PM.
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Thanks for your generous contribution sir ,

    In the same context , I was asked to obtain solutions for the equation Z^n-i = 0 for three cases where n = 1,2,3

    I did that for all of them , for example when n = 3 the solutions set was :
    i^(1/3) = 1^(1/3) e^[i(π/2 + 2kπ)]/3 = 1 e^[i(π/6 + 2kπ/3)] , where k = 0, 1, and 2 ... and I followed the same approach for n = 4 and 5.
    Then I had to plot each of those roots on an argand diagram , and I did that as well.

    After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Quote Originally Posted by Hyunqul View Post
    After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks
    First let \theta=\text{Arg}(a+b\mathbf{i}).

    Then z=\exp\left(\frac{\theta +2\pi k}{n}\mathbf{i}\right) where k=0,1,\cdots,n-1

    Recall that \exp(\phi \mathbf{i})=\cos(\phi)+\mathbf{i}\sin(\phi)
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Yes , sir..thanks but I would like to calrify the question more :

    First , it is required to generalize and prove the resuls for z when z^n = cos (x) + isin (X)

    Secondly . it is required to generalize for Z when z^n = cos (Kx) + isin (Kx)

    Thanks and sorry for disturbance.
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Sorry I made a mistake ...

    First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)

    Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 )

    Sorry again.
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    Re: Roots of Unity represented on a regular polygon - lenght of side.

    Quote Originally Posted by Hyunqul View Post
    First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)
    Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 )
    Why don't you just read about this topic?
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