复数 - 洛谷仓库

复数集C=\{a+bi|a,b∈R\}这里i=\sqrt{-1}

复平面：两条垂直且原点重合的数轴，交点为原点。一般水平方向是实轴，竖直方向是虚轴。坐标为（a，b）的点，表示

a+bi∈C

定义

\sqrt{a^2+b^2}

为

z=a+bi∈C

的模(长)，记作

|z|

定义复数在平面上对应的点与原点连线和宾轴正半轴夹角为z的幅角，记作

arg(z)

两个复数相乘，模长如何变化？

z_{1}=a_{1}+b_{1}i,z_{2}=a_{2}+b_{2}i(a_{1},a_{2},b_{1},b_{2}∈R)

|z_{1}|=\sqrt{a_{1}^2+b_{1}^2},|z_{2}|=\sqrt{a_{2}^2+b_{2}^2}

z_{1}z_{2}=(a_{1}+b_{1}i)(a_{2}+b_{2}i)=a_{1}a_{2}+a_{1}b_{2}i+a_{2}b_{1}i+b_{1}b_{2}i^2

|z_{1}z_{2}|=\sqrt{(a_{1}a_{2}-b_{1}b_{2})^2+(a_{1}b_{2}+a_{2}b_{1})^2}=\sqrt{a_{1}^2a_{2}^2-2a_{1}a_{2}b_{1}b_{2}+b_{1}^2b_{2}^2+2a_{1}a_{2}b_{1}b_{2}+a_{2}^2b_{1}^2}=\sqrt{(a_{1}^2+b_{1}^2)·(a_{2}^2+b_{2}^2)}=|z_{1}|·|z_{2}|

arg(z_{1}z_{2})=arg(z_{1})+arg(z_{2})

我们知道

e^0=1

，这里

e=\lim_{n \to \infty}(1+\frac{1}{n})≈2.71828

我们令

e^{\theta i}

为模长为1，幅角为

\theta

的复数

则

e^{2\pi i}=(e^{\pi i})^2=(-1)^2=1

e^{\theta i}=-1,e^{\frac{\pi}{2} i}=i,e^{\theta_{1} i}·e^{\theta_{2} i}=e^{(\theta_{1}+\theta_{2})i}

模长为

r

，幅角为

\theta·re^{i \theta}

z_{1}=r_{1}e^{i \theta_{1}},z_{2}=r_{2}e^{i \theta_{2}}

z_{1}·z_{2}=r_{1}e^{i \theta_{1}}·r_{2}e^{i \theta_{2}}=r_{1}r_{2}·e^{i(\theta_{1}+\theta_{2})}

哪些复数模长为1？

1,i,-1,-i,e^{i \theta}

x^2+1=0,x=\pm i

x^2=1,x=\pm 1

x^3=1:

在R上：x=1

在C上：x=1,\omega,\omega^2(三次单位根)

x^3-1=(x-1)(x^2+x+1)=(x-1)(x-\omega)(x-\omega^2)

x^2+x+1=(x-\omega)(x-\omega^2)=0

x^2+x+1=0\to x^2+x+ \frac{1}{4}=-\frac{3}{4}\to (x+\frac{1}{2})^2=-\frac{3}{4}\to x+\frac{1}{2}=\pm\frac{\sqrt{3}}{2}i\to x=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i

(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)^2=(-\frac{1}{2})^2+2·(-\frac{1}{2})·\frac{\sqrt{3}}{2}i+(\frac{\sqrt{3}}{2}i)^2=\frac{1}{4}-\frac{\sqrt{3}}{2}i-\frac{3}{4}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i

(\omega^2)^2=\omega^4=\omega^3·\omega=1·\omega=\omega

设P(x)\in R[x]

若z=a+bi(b≠0)是P(x)的根，则\bar{z}=a-bi也是P(x)的根

\overline{x}\pm\overline{y}=\overline{x \pm y}

\overline{xy}=\overline{x}·\overline{y},\overline{x}^n=\overline{x^n}

P(x)=\overline{P(x)}

例：

a\overline{x}^2+b\overline{x}+c=\overline{a} ·\overline{x}^2+\overline{b}·\overline{x}+\overline{c}

=\overline{ax^2}+\overline{bx}+\overline{c}=\overline{ax^2+bx+c}

若P(z)=0,则P(\overline{z})=\overline{P(x)}=\overline{0}=0

复数