复数
ren_gao_zu
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学习·文化课
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})}
考虑模长为1的复数:
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)
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|x^3|=1\to|x|^3=1\to|x|=1
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3arg(x)=arg(x^3)=0°或360°或720°
arg(x)=0°或120°或240°
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
虚根成对原理
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