# Sum of complex vectors on unit circle

Problem Statement

Let $$j=0,...,N-1$$. I want to evaluate

$$S_N=\sum_{j=0}^{N-1}e^{i2\pi\frac{j}{N}}$$

I think that $$S_N=0$$ for all $$N\ge 2$$.

Attempt 1: Series method

I tried using the formula for a finite geometric series, namely $$S_N=a\left(\frac{1-r^{N}}{1-r}\right)$$ where $$a=1$$ and $$r=e^{i2\pi\frac{1}{N}}$$.

$$S_N=\frac{1-e^{i2\pi\frac{N}{N}}}{1-e^{i2\pi\frac{1}{N}}} = 0$$

To be honest this seems a little too easy, so I'm looking for a sanity check on my application of this formula.

Attempt 2: Trigonometry method

I also tried expanding the sum. I know that all the sines from $$e^{i\theta}=\cos\theta+i\sin\theta$$ will cancel nicely, by symmetry of the following figure (figure drawn for $$N=7$$)

unit circle figure

Like the figure, assume that $$N$$ is odd. Cancelling the sines, we have

\begin{align} S_N&=\sum_{j=0}^{N-1}e^{i2\pi\frac{j}{N}}\\ &=\sum_{j=0}^{N-1}\cos\left(2\pi\frac{j}{N}\right)\\ &=1+2\sum_{j=1}^{\frac{N-1}{2}}\cos\left(2\pi\frac{j}{N}\right) \end{align}

where the last equality follows from $$\cos(\theta)=\cos(2\pi-\theta)$$. But this is where I get stuck.

Questions

1. Is my series method attempt correct? It seems too easy.
2. Is there a way to complete my trigonometry method?

Related question:

Why is such sum of cosines always zero?

• This is where I got the idea to try the geometric series formula.
• Your method is correct. Another way to verify is to factor the polynomial $x^N-1$. Commented Jun 5, 2022 at 4:32

Your series method is valid. Observe that $$\{e^{i2\pi \frac{j}{N}}\}_{j=0}^{N-1}$$ are the $$N$$-ths root of unity, so they are the roots of the polynomial $$z^N-1$$. Then:

$$z^N-1 = (z-1)(z-e^{i2\pi \frac{1}{N}})\dots (z-e^{i2\pi \frac{N-1}{N}})$$

Now, the sum identity you use follows from the fact that $$z^N - 1 = (z-1)(z^{N-1} + \dots + 1)$$. Substituting and cancelling the term $$(z-1)$$ we have that for all $$z\neq 1$$:

$$z^{N-1} + \dots + 1 = (z-e^{i2\pi \frac{1}{N}})\dots (z-e^{i2\pi \frac{N-1}{N}})$$

Evaluating at $$z=e^{i2\pi \frac{1}{N}}$$ we get the equality you wanted to prove.

I want to post another proof (similar to Bruno's) and also inspired by this webpage.

The Nth roots of unity are roots of the polynomial $$f(z)=z^N-1$$. Denote these roots as

$$\Omega_N=\{\omega_j\}_{j=0}^{N-1}=\left\{e^{i2\pi \frac{j}{N}}\right\}_{j=0}^{N-1}$$

Consider $$f(z)=z^N-1$$ and $$g(z)=(z-\omega_0)(z-\omega_1)\cdots(z-\omega_{N-1})$$. We will show that these polynomials are equal and that the coefficient on the $$z^{N-1}$$ term is both equal to zero and the sum of the Nth roots of unity.

It must be the case that $$f(z)=g(z)$$, because the interpolating polynomial of degree at most $$N$$ for $$N$$ distinct points is unique. We can prove this as follows. Consider $$H(z)=f(z)-g(z)$$. Then $$H(\omega_i)=0$$ for $$0\leq i \leq N-1$$, so $$H(z)$$ has $$N$$ distinct roots. However, we can deduce that $$H(z)$$ is degree at most $$N-1$$, because the $$z^N$$ from $$f(z)$$ will cancel with the $$z^N$$ from $$g(z)$$ (if you were to multiply everything out). Thus $$H(z)$$ is a $$N-1$$ degree polynomial with $$N$$ roots. This implies $$H(z)=0$$ identically, and thus $$f(z)=g(z)$$.

Now, notice that the coefficient for the $$N-1$$ degree term in $$f(z)$$ is zero. I claim that the coefficient for the $$N-1$$ degree term in $$g(z)$$ is $$-\omega_0-\omega_1-\cdots -\omega_{N-1}$$, which implies that $$\sum_{j=0}^{N-1}\omega_j =0$$.

We can prove this by induction. For the base case $$N=1$$, we have $$g_1(z)=(z-\omega_0)$$. The coefficient on $$z^{N-1}=z^0$$ is $$-\omega_0$$. Now assume that the result holds for $$N\leq k$$. We want to show that it holds for $$N=k+1$$.

\begin{align*} g_{k+1}(z)&=g_{k}(z)(z-\omega_k)\\ &=(z-\omega_0)(z-\omega_1)\cdots(z-\omega_{k-1})(z-\omega_k)\\ &=\left(z^k+\left(-\sum_{j=0}^{k-1}\omega_j\right)z^{k-1} + p_{k-2}(z)\right)(z-\omega_k)\\ &=\left(z^{k+1}-\omega_k z^k+ \left(-\sum_{j=0}^{k-1}\omega_j\right)z^{k}+p_{k-1}(z)\right)\\ &=\left(z^{k+1}+\left(-\sum_{j=0}^{k}\omega_j\right)z^{k}+p_{k-1}(z)\right) \end{align*}

where $$p_{k-2}(z),p_{k-1}(z)$$ are polynomials of degree $$k-2$$ and $$k-1$$, respectively.

Thus $$\sum_{j=0}^{N-1}\omega_j =0$$.