2

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.
nwsteg
  • 329

2 Answers2

1

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.

Bruno Aceves
  • 163
0

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

nwsteg
  • 329