Let \(P(z)\) be a polynomial of degree less than \(k\) with complex coefficients, and let \[ \omega_1,\ldots,\omega_k \] be the \(k\)th roots of unity in \(\mathbb C\). Prove that \[ \frac1k\sum_{i=1}^k P(\omega_i)=P(0). \]
Solution
Write \[ P(z)=a_0+a_1z+\cdots+a_{k-1}z^{k-1}, \] where some of the final coefficients may be zero.
1. For every integer \(j\) with \(1\le j<k\),\[ \sum_{i=1}^k\omega_i^j=0. \]
Let \(\zeta\) be a primitive \(k\)th root of unity. Multiplication by \(\zeta\) permutes the set of all \(k\)th roots of unity. Hence \[ \sum_{i=1}^k\omega_i^j = \sum_{i=1}^k(\zeta\omega_i)^j = \zeta^j\sum_{i=1}^k\omega_i^j. \] Since \(1\le j<k\) and \(\zeta\) has order \(k\), one has \(\zeta^j\neq1\). Therefore \[ (1-\zeta^j)\sum_{i=1}^k\omega_i^j=0, \] which forces the displayed sum to vanish.
2. One has\[ \sum_{i=1}^k P(\omega_i)=ka_0. \]
Expanding the polynomial term by term gives \[ \begin{aligned} \sum_{i=1}^kP(\omega_i) &= \sum_{i=1}^k \sum_{j=0}^{k-1}a_j\omega_i^j\\ &= \sum_{j=0}^{k-1} a_j\sum_{i=1}^k\omega_i^j. \end{aligned} \] For \(j=0\), the inner sum is \(k\). For every \(1\le j<k\), step 1 shows that the inner sum is \(0\). Thus only the constant term remains: \[ \sum_{i=1}^kP(\omega_i)=ka_0. \]
3. Therefore\[ \boxed{ \frac1k\sum_{i=1}^kP(\omega_i)=P(0) }. \]
By step 2, \[ \frac1k\sum_{i=1}^kP(\omega_i)=a_0, \] and by definition \(a_0=P(0)\).
4. Q.E.D.
Step 3 is the required identity.