- #1
Seydlitz
- 263
- 4
Homework Statement
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.
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