# Sum of roots of a number

#### GeoC

Is there a theorem related to the sum of the roots of a positive number?

For example, for any integer n=1,2,3.....N, and positive number $$\displaystyle \lambda$$, such that $$\displaystyle \lambda _k=\left | \lambda \right |^{1/n}exp^{2\pi ik/n} ...... k=0,1,2....n$$ are the nth roots of $$\displaystyle \lambda$$, what is the sum of the roots,

i.e. $$\displaystyle \sum_{k=0}^{n-1}\lambda _k = ?$$.

In other words, is there a closed form, and general solution for the sum of roots, i.e. $$\displaystyle \left | \lambda \right |^{1/n}\sum_{k=0}^{n-1}exp^{2\pi ik/n}$$

#### Bruno J.

MHF Hall of Honor
Yes, it's $$\displaystyle 0$$, unless $$\displaystyle n=1$$!

Hint : the sum of the roots of a polynomial $$\displaystyle f(z)$$ is (up to sign) the coefficient of $$\displaystyle z$$, and the roots of a complex number $$\displaystyle \lambda$$ are the roots of $$\displaystyle f(z)=z^n-\lambda$$.

• GeoC

#### GeoC

So, are you saying the sum of the roots of $$\displaystyle \lambda^{1/n}$$ is 0 for n not equal 1, and, for n=1, the sum must be simply $$\displaystyle \lambda$$? Thanks.. looking at the $$\displaystyle \sum$$, I guess that makes sense.. I had long since forgotten, if I ever even knew it, that the sum of roots equals coefficient of Z in the polynomial.

#### Bruno J.

MHF Hall of Honor
So, are you saying the sum of the roots of $$\displaystyle \lambda^{1/n}$$ is 0 for n not equal 1, and, for n=1, the sum must be simply $$\displaystyle \lambda$$?
Yes!

Here's another way to prove it : if $$\displaystyle n>1$$, let $$\displaystyle \omega = e^{2\pi i / n}$$ and $$\displaystyle S=1+\omega + \dots + \omega^{n-1}$$. Then, since $$\displaystyle \omega^n=1$$, we have $$\displaystyle \omega S = \omega + \dots + \omega^n = S$$. Since $$\displaystyle \omega \neq 1$$, we must have $$\displaystyle S=0$$.

• HallsofIvy

#### wonderboy1953

Comment for Bruno J

#### chisigma

MHF Hall of Honor
In general if we have a polynomial of degree $$\displaystyle n$$ in $$\displaystyle z$$ it can be written as...

$$\displaystyle p(z) = z^{n} + a_{n-1} z^{n-1} + \dots + a_{1} z + a_{0} = \prod_{k=0}^{n-1} (z-z_{k})$$ (1)

... where the $$\displaystyle z_{k}$$, $$\displaystyle k=0,1,\dots , n-1$$ are the roots of the polynomial. If the roots are all distinct then from (1) is easy to derive that is...

$$\displaystyle \sum_{k=0}^{n-1} z_{k} = - a_{n-1}$$ (2)

If $$\displaystyle p(z) = z^{n} - \lambda$$ with $$\displaystyle \lambda>0$$ is...

$$\displaystyle \sum_{k=0}^{n-1} z_{k} = \sum_{k=0}^{n-1} \lambda^{\frac{1}{n}} e^{2 \pi i \frac{k}{n}} =0$$ (3)

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### Bruno J.

MHF Hall of Honor