# Sum of nth roots of unity

Question: If $c\neq 1$ is an $n^{th}$ root of unity then, $1+c+...+c^{n-1} = 0$

Attempt: So I have established that I need to show that $$\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}}=\frac{1-e^{\frac{ik2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}} =0$$ by use of the sum of geometric series'. My issue is proceeding further to show that this is indeed true.

The right hand side should be

$$\frac{1-e^{\frac{in2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}}=0$$

since $e^{\frac{in2\pi}{n}}=e^{i2\pi}=1$.

• That actually solved it for me. Thank you for noticing that. – TfwBear Jan 27 '15 at 17:37

If a finite set of complex numbers is symmetric about a line passing through the origin, then its sum must lie on that line; if it is symmetric about two different lines through the origin, then its sum must be zero. The $n^{\text{th}}$ roots of unity are the vertices of a regular $n$-gon centered at the origin, which has $n$ lines of symmetry. Hence, for $n\ge2,$ the sum is zero.

Are you sure you are supposed to use complex analysis to do this? If so what I am saying is irrelevant but here is another 'easy' method.

If $c$ is an nth root of unity then $c$ satisfies the equation $c^n-1$. Factorise this and you should be able to get to the answer.

• Yes, I'm taking a complex analysis course and he wasn't very explicit about how we had to solve it. – TfwBear Jan 27 '15 at 17:36

Since $c \neq 1$ we have $1+c+\cdots+c^{n-1} = { 1-c^n \over 1-c}$. Then $c^n=1$ finishes.

A variation of the answer of bof: the $n$-th roots of unity are the vertices of a regular $n$-gon centered at the origin and the product by $e^{2\pi i/n}$ is a rotation of angle $2\pi/n$ that leaves invariant the set of $n$-th roots. This means that $$e^{2\pi i/n}\sum^{n-1}_{k=0} e^{2k\pi i/n} = \sum^{n-1}_{k=0} e^{2k\pi i/n}$$ i.e., $$(e^{2\pi i/n}-1)\sum^{n-1}_{k=0} e^{2k\pi i/n} = 0.$$

Yet another answer, with more general applicablity:

For any polynomial $p(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_1 x +a_0$, with roots $x_1,x_2,\ldots,x_n$, the sum of the roots is given by $x_1+x_2+\ldots+x_n=-\frac{a_{n-1}}{a_n}$.

This is because $p(x)=a_n(x-x_1)(x-x_2)\cdots(x-x_n)$, and you can work out from there that the coefficient of $x^{n-1}$ is minus the sum of the roots, times the $a_n$ prefactor. (Notice it's also easy to get the product of the roots, as well as the sum of products of any fixed number of terms from $1$ to $n$).

In your case you're looking at the root of the polynomial $p(x)=x^n-1$, so $a_n=1,a_0=-1$ and $a_k=0$ for other $k$. In particular, unless $k=1$, you have $a_{n-1}=0$ so the sum of the roots is zero.