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Math Help - Roots of unity

  1. #1
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    Roots of unity

    In solving  z^n = 1 we get the distinct solutions  z = e^{i(2k \pi/n)} . We get this as follows:

    Let  z = re^{i \theta} . Then  (re^{i \theta})^{n} = 1 or  r^{n}e^{i n \theta} = 1e^{i0} . Thus   r = 1 and  \theta = 2 k \pi/n .

    How do we draw the solutions? Why do we write  \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n} ? Like the three cube roots of  1 form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    In solving  z^n = 1 we get the distinct solutions  z = e^{i(2k \pi/n)} . We get this as follows:

    Let  z = re^{i \theta} . Then  (re^{i \theta})^{n} = 1 or  r^{n}e^{i n \theta} = 1e^{i0} . Thus   r = 1 and  \theta = 2 k \pi/n .

    How do we draw the solutions? Why do we write  \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n} ? Like the three cube roots of  1 form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?
    On the complex plane, the number   z = e^{i \theta} is defined as a rotation of \theta radians from the point 1+0i.

    So let n = 1,2,3,4,5, and find out what  \theta is at these points. Then simply rotate that amount about the unit circle from the point 1+0i.

    Unless you're talking about drawing the general case... which is pretty meaningless.
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  3. #3
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    Complex roots of unity

    Hello particlejohn
    Quote Originally Posted by particlejohn View Post
    In solving  z^n = 1 we get the distinct solutions  z = e^{i(2k \pi/n)} . We get this as follows:

    Let  z = re^{i \theta} . Then  (re^{i \theta})^{n} = 1 or  r^{n}e^{i n \theta} = 1e^{i0} . Thus   r = 1 and  \theta = 2 k \pi/n .

    How do we draw the solutions? Why do we write  \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n} ? Like the three cube roots of  1 form an equilateral triangle inscribed in the unit circle. But does it matter how we draw it?
    Your working is OK up to and including

      r = 1 and  \theta = 2 k \pi/n

    But then you've written

     \omega_{n} = \cos \frac{2 \pi}{n} + i \sin \frac{2 \pi}{n}

    Which isn't quite right: you've missed the k out. So let's put it right. Putting the values of r and \theta into z = re^{i \theta} we get:

    z = 1e^{i \theta}=\cos \theta + i\sin\theta

    =\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}

    Now note two things:

    • because r = 1, the points representing the values of z lie on a circle, radius 1, centre (0,0)
    • distinct points are obtained by values of k between 0 and n-1 inclusive. When k = 2n, 2n+1, \dots the points are repeated.

    So, where do these points lie?

    When k=0, z is wholly real. This is simply the value z=1, and is represented by the point (1,0).

    The remaining values of k divide the whole circle into n equal parts, corresponding to angles at the centre of \frac{\pi}{n}, \frac{2\pi}{n},\frac{3\pi}{n}, \dots The points representing these values of z, then, lie on the vertices of a regular n-sided polygon, with centre (0,0), one vertex at (1,0) and each side subtending an angle at the centre of \frac{2\pi}{n}.

    I hope that clears it up.

    Grandad
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  4. #4
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    Actually, isn't my  \omega_n correct? Because then all the roots can be expressed as  1, \omega, \omega^{2}, \ldots, \omega^{n-1} ?
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  5. #5
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    Quote Originally Posted by particlejohn View Post
    Actually, isn't my  \omega_n correct? Because then all the roots can be expressed as  1, \omega, \omega^{2}, \ldots, \omega^{n-1} ?
    The other roots can indeed be constructed from \omega_n:

     1, \omega_{\color{red}n}, \omega_{\color{red}n}^{2}, \ldots, \omega_{\color{red}n}^{n-1}

    and you get the same values as Grandad.

    But you never said this is what you meant in your original post (and your notation is not quite right).
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