# 复数

## 虚数单位

### 为何需要虚数单位

• 解方程：${\displaystyle 0.5x^{2}-6x+55.5=0}$

### 运算

${\displaystyle {\sqrt {-9}}=3i}$
${\displaystyle {\sqrt {-2}}={\sqrt {2}}i}$
${\displaystyle {\sqrt {-x}}={\sqrt {x}}i}$，其中${\displaystyle x\geq 0}$
${\displaystyle {\sqrt {-9}}\times {\sqrt {-2}}=3i\times {\sqrt {2}}i=-{\sqrt {18}}}$

${\displaystyle {\sqrt {-9}}\times {\sqrt {-2}}={\sqrt {(-9)(-2)}}={\sqrt {18}}}$

${\displaystyle {\sqrt {x}}\times {\sqrt {y}}={\sqrt {xy}}}$只能应用于${\displaystyle x,y\geq 0}$时，因为负数的开方是不连续的。

${\displaystyle i}$ 的高次方会不断作以下的循环：

${\displaystyle i^{0}=1\,}$
${\displaystyle i^{1}=i\,}$
${\displaystyle i^{2}=-1\,}$
${\displaystyle i^{3}=-i\,}$

${\displaystyle i^{4}=1\,}$
${\displaystyle i^{5}=i\,}$
${\displaystyle i^{6}=-1\,}$
${\displaystyle i^{7}=-i\,}$

...

### 练习

${\displaystyle n}$是整数，试计算以下的值：

1. ${\displaystyle i^{4n}}$
2. ${\displaystyle i^{4n+1}}$
3. ${\displaystyle i^{4n+2}}$
4. ${\displaystyle i^{4n+3}}$

## 运算

### 四则运算

• 加、减法：实部加实部，虚部加虚部：${\displaystyle (a+bi)+(c+di)=(a+c)+(b+d)i}$
• 乘法：${\displaystyle (a+bi)(c+di)=ac+adi+bci+bidi=ac+bdi^{2}+(ad+bc)i=(ac-bd)+(ad+bc)i}$
• 除法：可将分母“实数化”，方法是分子、分母乘以分母的轭作扩分：${\displaystyle {\frac {a+bi}{c+di}}={\frac {(a+bi)(c-di)}{(c+di)(c-di)}}={\frac {(ac+bd)+(bc-ad)i}{c^{2}+d^{2}}}}$

### 开方

${\displaystyle (a+bi)^{2}=a^{2}+2abi+(bi)^{2}=(a^{2}-b^{2})+(2ab)i}$
${\displaystyle i=0+1i}$
${\displaystyle a^{2}-b^{2}=0;2ab=1}$

## 练习解答

### 练习一

1. 1
2. ${\displaystyle i}$
3. -1
4. ${\displaystyle -i}$