单位根
[egin{aligned}
&omega_n^k=e^{frac{2pi i}{n}}=cosfrac{2πk}{n}+isinfrac{2πk}{n} \
&omega_{n}^{n}=1 \
&omega_{2n}^{2k}=omega_n^k \
&omega_{n}^{k+frac{n}{2}}=-omega_n^k \
&sum_{i=1}^{n-1} omega_n^i = 0
end{aligned}
]
单位根反演
[[nmid k]=frac{1}{n}sum_{i=0}^{n-1}omega_n^{ik}
]
若 (nmid k) 时，得
[frac{1}{n}sum_{i=0}^{n-1}omega_n^{ik}=frac{1}{n}sum_{i=0}^{n-1}omega_n^n=1
]
若 (n mid k) 时，得
[frac{1}{n}sum_{i=0}^{n-1}omega_{n}^{ik}=frac{1}{n}omega_n^{0}frac{1-omega_n^{nk}}{1-omega_n^{k}}=0
]
可以用来求多项式某个倍数的系数和
[egin{aligned}
&sum_{i=0}^{lfloor frac{n}{k}
floor} [x^{ik}] f(x) \
=&sum_{i=0}^n [x^{i}] f(x)[k mid i] \
=&sum_{i=0}^n [x^{i}] f(x)frac{1}{k}sum_{j=0}^{k-1}omega_k^{ij} \
=&frac{1}{k} sum_{j=0}^{k-1} sum_{i=0}^n [x^{i}] f(x)(omega_k^j)^i \
=&frac{1}{k} sum_{j=0}^{k-1} f(omega_k^j) \
end{aligned}
]