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Proof that the sum of complex roots are 0

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello guys,

    I need to prove that the sum of complex roots are 0.

    In the Boas book, it is actually written 'show that the sum of the n nth roots of any complex number is 0.' I believe it's equivalent.

    3. The attempt at a solution
    I have managed to obtain this summation. It is in fact a big series.

    $$
    \sum_{k=0}^{n} ( \cos (\frac{\theta + 2 \pi k}{n}) + i \sin (\frac{\theta + 2 \pi k}{n}) )$$

    Note that there should be the radius value, but because it's constant I just omit it from the series.

    My problem is currently showing that the cosine and sine function sums up to 0. If ##n=2## it is simple to see because then we can have two cosines with opposing signs and the sum will be 0. (Because of ##\pi## shift in the periodicity). The same goes to the sine function.

    However if n is arbitrary, i don't see direct way to get 0. Is there any useful identity or method with summing trigonometric function?

    I realized that this problem is actually very simple if we are to use algebraic approach, because there's no ##z## so the roots must sum to 0. But I've spent several hours on this approach, I really want to finish it if possible.

    Thank You
     
  2. jcsd
  3. Nov 10, 2013 #2

    Curious3141

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    Why not try an exponential notation approach? It becomes a geometric series that sums very easily.
     
  4. Nov 10, 2013 #3

    D H

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    Your limits are incorrect. For example, this sums over three roots of z1/2, and in general over n+1 roots of z1/n.

    Even after correcting the limits, this is the hard way to approach your proof. Try following Curious3141's advice.
     
  5. Nov 10, 2013 #4

    bcrowell

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    If you add 2π to θ, your sum is unchanged. But in the complex plane, adding 2π to θ corresponds to rotating each term in the sum by 2π/n about the origin. So for n>1 you have a geometrical figure (set of points) whose center of mass is unchanged when you rotate the figure about the origin in a nontrivial way.
     
  6. Nov 10, 2013 #5
    All the nth complex roots of any number z are z1/n times the complex roots of 1.

    All of the complex roots of 1 are solutions to this equation:

    $$x^n - 1 = 0$$

    Recall that given a polynomial equation such as [itex]ax^3 + bx^2 + cx + d = 0[/itex] with roots [itex]z_0, z_1, z_2[/itex], then
    $$z_0z_1z_2 = \frac{d}{a}$$
    $$z_0z_1 + z_0z_2 + z_1z_2 = \frac{c}{a}$$
    $$z_0 + z_1 + z_2 = \frac{b}{a}$$

    And so on for other degree polynomials.

    Returning to our original equation [itex]x^n - 1 = 0[/itex], we see that the product of its roots is -1 but the sum of its roots and also the sum of combinations of products following the rule shown above is 0.

    Hope it helps. I wanted to do it in a slightly more elegant method than to sum them all up using trigonometry and calculus concepts.
     
  7. Nov 10, 2013 #6
    I almost cried. Yes, I did try it once you tell me and I got the proof in just like 30 seconds. Thanks. I got so fixated with the trigonometrical function that I forget almost everything else.

    I realized that but then the third root would be ##2\pi## anyway.

    I followed his advice finally.

    I understand the figures, and I think it's quite straightforward. I just can't find the way to relate its geometric property into finding the proof until now.

    Thanks for your help hyrum but I already stated in op that I wanted to try the trigonometric approach instead. Nevertheless I think to know how the root is related to the original equation is very useful, I'll brush up my knowledge on that so that I don't get caught off guard like this in the future.
     
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