# Proof that the sum of complex roots are 0

1. Nov 10, 2013

### Seydlitz

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. Nov 10, 2013

### Curious3141

Why not try an exponential notation approach? It becomes a geometric series that sums very easily.

3. Nov 10, 2013

### D H

Staff Emeritus
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.

4. Nov 10, 2013

### bcrowell

Staff Emeritus
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.

5. Nov 10, 2013

### Hyrum

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 $ax^3 + bx^2 + cx + d = 0$ with roots $z_0, z_1, z_2$, 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 $x^n - 1 = 0$, 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.

6. Nov 10, 2013

### Seydlitz

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.