立方根
}
ω = exp  ( i ⋅ ( 2 π 3 + 2 k π ) ) , ω 2 = exp  ( i ⋅ ( 4 π 3 + 2 k π ) ) , ω ̄ = exp  ( i ⋅ ( − 2 π 3 + 2 k π ) ) . {\displaystyle \omega =\exp \left(i\cdot \left({\frac {2\pi }{3}}+2k\pi \right)\right),\quad \omega ^{2}=\exp \left(i\cdot \left({\frac {4\pi }{3}}+2k\pi \right)\right),\quad {\overline {\omega }}=\exp \left(i\cdot \left({\frac {-2\pi }{3}}+2k\pi \right)\right).}
ω + 1 = exp  ( i ⋅ ( π 3 + 2 k π ) ) = − ω 2 , ω ̄ + 1 = exp  ( i ⋅ ( − π 3 + 2 k π ) ) = − ω .