Eigenvalues of a circulant matrix

Computation. Let $v_k=(1,\omega^k,\omega^{2k})^T$. Then
$$(Cv_k)_1 = a\cdot1+b\cdot\omega^k+c\cdot\omega^{2k} = \lambda_k.$$
$$(Cv_k)_2 = c\cdot1+a\cdot\omega^k+b\cdot\omega^{2k} = \omega^k(a+b\omega^k+c\omega^{2k}) = \omega^k\lambda_k.$$
$$(Cv_k)_3 = b\cdot1+c\cdot\omega^k+a\cdot\omega^{2k} = \omega^{2k}(a+b\omega^k+c\omega^{2k}) = \omega^{2k}\lambda_k.$$
So $Cv_k=\lambda_k v_k$, and $\lambda_k=a+b\omega^k+c\omega^{2k}$.

Diagonalizability. For $k=0,1,2$, the vectors $v_k$ are linearly independent (Vandermonde-type argument with distinct $\omega^k$). So $C$ is diagonalizable over $\mathbb{C}$, with eigenvalues $\lambda_0,\lambda_1,\lambda_2$.

Conclusion. The eigenvalues of the circulant matrix $C$ are given by the discrete Fourier transform of $(a,b,c)$.