How can u prove that the sum of the roots of a complex number is zero?

Question

How can u prove that the sum of the roots of a complex number is zero?

Answer 1

It seems that the previous answerers are focused on an analytic proof. For perhaps another perspective, here's an algebraic proof.



For any monic polynomial P(z) = z^n + a_(n-1)*z^(n-1) + ... + a_1*z + a_0, -a_(n-1) is the sum of the roots of P; this can be seen by writing P(z) = (z - r_1)...(z - r_n), where r_1, ..., r_n are the roots of P, and multipyling out this product (in the general setting, we need to take an algebraic closure of the field in which we're working, but in the complex numbers, this is moot, as the complex numbers are already algebraically closed).



For your specific problem, then, we are considering P(z) = z^n + a_0, where n > 1 -- that is, -a_(n-1) = 0.

Answer 2

The idea is this:



Factorise your equation with real roots as much as you can. You will be left with a polynomial so maybe something like (x-a)(x-b)(polynomial) where a,b, are real roots



This polynomial WILL be of an even power by the intermediate value theorem which says "a odd power polynomial will have atleast one real root" i.e. a or b



Break this polynomial into x^2 polynomial say x^2+cx+d where there is no real solutions. Recall the quadratic formula, x = [ -b +- sqrt(b^2 - 4ac) ] /2a



Let's look at the sqrt. We know there is no real solution to the sqrt, i.e. it is imaginary and it is also constant. So eg. b^2-4ac = -4 then sqrt(-4) = +2i and -2i.



What I am getting to is that what is inside the sqrt is a CONSTANT and will produce an IMAGINARY where one is positive and negative thus when adding them up will equal zero.



Check out: Complex conjugate root theorem

http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem



I know the above isn't a proof proof, it just the idea. Hopefully by reading what I wrote and the link, you could prove it yourself?



Thanks



PS: With the De Moive "proof", it doesn't work in general as it is assuming it is a "root of unity" i.e. in the form z^n = 1 (or any constant). My version is for a more general polynomial.

Answer 3

You would have to start with the complex roots theorem, which gives an expression for the nth roots of a complex number in polar form. It's simple to come up with discrete examples (square roots of 4, cube roots of i, and so on), but doing the necessary expansion of the complex roots theorem (for the cases k=1, 2, 3, ......n) is not something I can accomplish here on Yahoo Answers. Start by writing out the general case, expand, group real and imaginary components, use properties of sine and cosine and see if you can get to the end. Should be a lot of work.

Answer 4

A Complex number does not have any root. Some seems to be misguiding you. Be careful.

Answer 5

I think you mean to say, "How can one prove that the sum of the complex roots of any number is zero?" For example, the roots of the following equation:



x^5 - 1 = 0



are 1, -(-1)^(1/5), (-1)^(2/5), -(-1)^(3/5), (-1)*(4/5), the sum of which comes to 0. The easiest way to do show this is to use the de Moivre's formula (see wiki) to locate all the roots on the complex plane, which form a regular n-gon, where n is the power in the polynomial, x^n - A = 0, and A being the number. If one imagines the complex roots to be vectors, then the vectors add up to zero. Think of n vectors all pulling with equal force from a point at equal angles apart.



Rigorously speaking, what you have to work out is that for any n and θ,



∑(k = 1 to n) Cos( (2π/n)k + θ ) = 0



∑(k = 1 to n) iSin( (2π/n)k + θ ) = 0



Of course, what you can also do is to expand (x - a)(x - b)(x - c)(x - d)(x - e), where a, b, c, d, e are the roots, and since the coefficient for the x^4 has to be zero, the sum a + b + c + d + e must also be zero. Up to you which way you want to go.