FFT（临时）

快速傅里叶变换（FFT）

单位根及其性质

FFT 中需要用到单位根这一工具，笔者在这里简要介绍一下。

对于复数 $z$，我们用辐角和模长来表示它，有

$$\begin{array}{}\text{(1)}& z=|z|\cdot (\cos \theta +i\sin \theta )\end{array}$$

对于复数 $z_1$ 和 $z_2$，分别设 ${\theta }_1$ 和 ${\theta }_2$ 为它们的辐角。考虑 $z_1\cdot z_2$，根据三角函数的和角公式，有

$$\begin{array}{}\text{(2)}& \begin{array}{rl}{z}_1\cdot z_2& =|z_1|\cdot (\cos {\theta }_1+i\sin {\theta }_1)\cdot |z_2|\cdot (\cos {\theta }_2+i\sin {\theta }_2)\\ & =|z_1||z_2|\cdot (\cos {\theta }_1+i\sin {\theta }_1)(\cos {\theta }_2+i\sin {\theta }_2)\\ & =|z_1||z_2|\cdot (\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2+i\sin {\theta }_1\cos {\theta }_2+i\cos {\theta }_1\sin {\theta }_2)\\ & =|z_1||z_2|\cdot ((\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)+i(\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2))\\ & =|z_1||z_2|\cdot (\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2))\end{array}\end{array}$$

$$$$

考虑方程

$$\begin{array}{}\text{(3)}& x^n=1\end{array}$$

我们早