1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

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

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted