复数最早是在16世纪中期发展起来的，作为解决某些三次方程的一种方法。它们由虚部（以 i 或 − 1 {\displaystyle {\sqrt {-1}}} 67 {\displaystyle {\sqrt {67}}}
以前，人们无法解决任何包含负数根的方程。例如，二次方程 y = x 2 + x + 2 {\displaystyle y=x^{2}+x+2}
It cannot be factorised
It cannot be solved with the quadratic equation (as the discriminate would be negative, and one could not square root a negative number).
但是，使用“虚数”允许对负数开平方根
−
1
=
i
{\displaystyle {\sqrt {-1}}=i}
应用这一点和对根式的了解，我们可以将其用于任何负数。因此
−
9
=
9
×
−
1
=
±
3
i
{\displaystyle {\sqrt {-9}}={\sqrt {9}}\times {\sqrt {-1}}=\pm 3i}
−
10
=
10
×
−
1
=
±
10
i
{\displaystyle {\sqrt {-10}}={\sqrt {10}}\times {\sqrt {-1}}=\pm {\sqrt {10}}i}
−
8
=
8
×
−
1
=
±
2
2
i
{\displaystyle {\sqrt {-8}}={\sqrt {8}}\times {\sqrt {-1}}=\pm 2{\sqrt {2}}i}
复数是一个包含虚部的数，换句话说，它包含 i {\displaystyle i}
z
=
1
+
3
i
{\displaystyle {\mathit {z}}=1+3{\mathit {i}}\!}
z
=
3
2
+
3
i
{\displaystyle z={\frac {\sqrt {3}}{2}}+3i}
z
=
−
2
+
2
i
{\displaystyle z=-2+{\sqrt {2}}i}
z
=
3
+
5
i
2
{\displaystyle z={\frac {3+{\sqrt {5}}i}{2}}}
即使在更复杂的例子中，我们也应该始终能够将实部与虚部分开。考虑上面的最后一个例子
z
=
3
+
5
i
2
{\displaystyle z={\frac {3+{\sqrt {5}}i}{2}}}
在这个数字中，实部是 3 2 {\displaystyle 3 \over 2} 5 2 i {\displaystyle {\frac {\sqrt {5}}{2}}i}
类似地，如果我们要考虑
z
=
3
(
2
+
i
)
{\displaystyle z={\sqrt {3}}({\sqrt {2}}+i)}
实部将是 3 × 2 = 6 {\displaystyle {\sqrt {3}}\times {\sqrt {2}}={\sqrt {6}}} 3 i {\displaystyle {\sqrt {3}}i}
我们也应该注意这里：在 2 i {\displaystyle {\sqrt {2}}i} i {\displaystyle {\mathit {i}}\!} 2 × i {\displaystyle {\sqrt {2}}\times i}
上面的每个例子（以及所有复数）都可以用以下方式表示
z
=
a
+
b
i
{\displaystyle {\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!}
a
∈
R
{\displaystyle a\in \mathbb {R} \!}
b
∈
R
{\displaystyle b\in \mathbb {R} \!}
值得熟悉一下像 a ∈ R {\displaystyle a\in \mathbb {R} \!} R {\displaystyle \mathbb {R} \!} R {\displaystyle \mathbb {R} \!}
每个复数都有一个“共轭复数”。尽管名字很复杂，但这是一个非常简单的概念。
The Complex Number
z
=
a
+
b
i
{\displaystyle {\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!}
z
∗
=
a
−
b
i
{\displaystyle {\mathit {z}}*={\mathit {a}}-{\mathit {b}}{\mathit {i}}\!}
共轭复数有很多用处。例如，如果一个方程有一个复根 (1+3i)，那么它的共轭复数 (1-3i) 也是一个根。同样地，为了除复数，我们必须使用共轭复数来“有理化”分母，这将在后面讲解。我们通过用 z ∗ z ∗ {\displaystyle {\frac {z*}{z*}}} a+ib 的形式。
对于任何复数，其共轭复数用 * 表示。例如，复数 'a' 的共轭复数为 'a*'。复数 'y' 的共轭复数为 'y*'。考试中通常会使用复数 z 和 w。
假设我们要定义
z
=
3
+
2
i
{\displaystyle z=3+{\sqrt {2}}i}
w
=
−
2
+
3
2
i
{\displaystyle w=-2+3{\sqrt {2}}i}
复数的加法是分别对实部和虚部进行的 - 也就是说，将实部加在一起，将虚部加在一起。使用我们之前定义的例子
z
+
w
=
(
3
+
2
i
)
+
(
−
2
+
3
2
i
)
=
(
3
−
2
)
+
(
2
+
3
2
)
i
=
1
+
4
2
i
{\displaystyle {\begin{aligned}z+w&=(3+{\sqrt {2}}i)+(-2+3{\sqrt {2}}i)\\&=(3-2)+({\sqrt {2}}+3{\sqrt {2}})i\\&=1+4{\sqrt {2}}i\\\end{aligned}}}
复数的加法与实数的加法一样，与加法的顺序无关 - 换句话说， z + w ≡ w + z {\displaystyle z+w\equiv w+z}
同样地，如果我们考虑之前定义的那对复数，它们可以用相同的方式进行减法。
z
−
w
=
(
3
+
2
i
)
−
(
−
2
+
3
2
i
)
=
(
3
−
(
−
2
)
)
+
(
2
−
3
2
)
i
=
5
−
2
2
i
{\displaystyle {\begin{aligned}z-w&=(3+{\sqrt {2}}i)-(-2+3{\sqrt {2}}i)\\&=(3-(-2))+({\sqrt {2}}-3{\sqrt {2}})i\\&=5-2{\sqrt {2}}i\\\end{aligned}}}
与实数的减法一样，复数的减法也与减法的顺序无关 - 换句话说
z
−
w
≠
w
−
z
{\displaystyle z-w\neq w-z}
因此
w
−
z
=
(
−
2
+
3
2
i
)
−
(
3
+
2
i
)
=
(
−
2
−
3
)
+
(
3
2
−
2
)
i
=
−
5
+
2
2
i
{\displaystyle {\begin{aligned}w-z&=(-2+3{\sqrt {2}}i)-(3+{\sqrt {2}}i)\\&=(-2-3)+(3{\sqrt {2}}-{\sqrt {2}})i\\&=-5+2{\sqrt {2}}i\\\end{aligned}}}
用常数乘以复数非常简单 - 对于常数 'k'，只需将实部乘以 k，将虚部乘以 k。因此，使用我们之前定义的 z
k
z
=
k
(
3
+
2
i
)
=
3
k
+
k
2
i
{\displaystyle kz=k(3+{\sqrt {2}}i)=3k+k{\sqrt {2}}i}
如果我们要用一个复数乘以另一个复数，我们需要遵循与展开像 ( 2 x + 2 ) ( x + 3 ) {\displaystyle ({\sqrt {2}}x+2)(x+3)}
z
2
=
z
×
z
=
(
3
+
2
i
)
(
3
+
2
i
)
=
3
×
3
+
3
×
2
i
+
3
×
2
i
+
2
i
×
2
i
=
9
+
6
2
i
+
2
i
2
{\displaystyle {\begin{aligned}z^{2}&=z\times z\\&=(3+{\sqrt {2}}i)(3+{\sqrt {2}}i)\\&=3\times 3+3\times {\sqrt {2}}i+3\times {\sqrt {2}}i+{\sqrt {2}}i\times {\sqrt {2}}i\\&=9+6{\sqrt {2}}i+2i^{2}\\\end{aligned}}}
我们可能会想当然地将解写成与乘未知数时一样的形式。但是， i 2 {\displaystyle i^{2}\!}
z
2
=
9
+
6
2
i
+
2
i
2
=
9
+
6
2
+
2
×
(
−
1
)
=
9
−
2
+
6
2
=
7
+
6
2
{\displaystyle {\begin{aligned}z^{2}&=9+6{\sqrt {2}}i+2i^{2}\\&=9+6{\sqrt {2}}+2\times (-1)\\&=9-2+6{\sqrt {2}}=7+6{\sqrt {2}}\\\end{aligned}}}
同样地，我们可以用不同的复数相乘
z
×
w
=
(
3
+
2
i
)
(
−
2
+
3
2
i
)
=
3
×
(
−
2
)
+
3
×
3
2
i
+
(
−
2
)
×
2
i
+
2
i
×
3
2
i
)
=
−
6
+
7
2
i
+
6
×
i
2
=
(
−
6
−
6
)
+
7
2
i
=
−
12
+
7
2
i
{\displaystyle {\begin{aligned}z\times w&=(3+{\sqrt {2}}i)(-2+3{\sqrt {2}}i)\\&=3\times (-2)+3\times 3{\sqrt {2}}i+(-2)\times {\sqrt {2}}i+{\sqrt {2}}i\times 3{\sqrt {2}}i)\\&=-6+7{\sqrt {2}}i+6\times i^{2}\\&=(-6-6)+7{\sqrt {2}}i=-12+7{\sqrt {2}}i\\\end{aligned}}}
这是一个稍微复杂的过程，类似于在涉及根式的运算中将分母有理化。
例如： 2 + 3 i 1 2 + 5 i {\displaystyle {\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+{\sqrt {5}}i}}} 1 2 + 5 i {\displaystyle {\frac {1}{2}}+5i} 1 2 − 5 i {\displaystyle {\frac {1}{2}}-5i} 1 2 − 5 i 1 2 − 5 i {\displaystyle {\frac {{\frac {1}{2}}-5i}{{\frac {1}{2}}-5i}}}
这个过程可以通过数学方法解决。
z
=
2
+
3
i
{\displaystyle z=2+{\sqrt {3}}i}
w
=
1
2
+
5
i
{\displaystyle w={\frac {1}{2}}+5i}
w
∗
=
1
2
−
5
i
{\displaystyle w*={\frac {1}{2}}-5i}
z
w
=
2
+
3
i
1
2
+
5
i
=
2
+
3
i
1
2
+
5
i
×
1
2
−
5
i
1
2
−
5
i
=
(
2
+
3
i
)
(
1
2
−
5
i
)
(
1
2
+
5
i
)
(
1
2
−
5
i
)
=
2
×
1
2
+
2
×
(
−
5
i
)
+
3
i
×
1
2
+
3
i
×
(
−
5
i
)
1
2
×
1
2
+
1
2
×
(
−
5
i
)
+
5
i
×
1
2
+
5
i
×
(
−
5
i
)
=
1
+
(
−
10
i
)
+
3
2
+
(
−
5
3
i
)
1
4
+
(
−
5
2
i
)
+
5
2
i
+
(
−
25
i
2
)
=
(
1
+
3
2
)
+
(
−
10
−
5
3
)
i
1
4
+
(
−
25
×
(
−
1
)
)
=
2
+
3
2
+
(
−
10
−
5
3
)
i
101
4
=
4
101
(
2
+
3
2
+
(
−
10
−
5
3
)
i
)
{\displaystyle {\begin{aligned}{\frac {z}{w}}&={\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+5i}}\\&={\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+5i}}\times {\frac {{\frac {1}{2}}-5i}{{\frac {1}{2}}-5i}}\\&={\frac {(2+{\sqrt {3}}i)({\frac {1}{2}}-5i)}{({\frac {1}{2}}+5i)({\frac {1}{2}}-5i)}}\\&={\frac {2\times {\frac {1}{2}}+2\times (-5i)+{\sqrt {3}}i\times {\frac {1}{2}}+{\sqrt {3}}i\times (-5i)}{{\frac {1}{2}}\times {\frac {1}{2}}+{\frac {1}{2}}\times (-5i)+5i\times {\frac {1}{2}}+5i\times (-5i)}}\\&={\frac {1+(-10i)+{\frac {\sqrt {3}}{2}}+(-5{\sqrt {3}}i)}{{\frac {1}{4}}+({\frac {-5}{2}}i)+{\frac {5}{2}}i+(-25i^{2})}}\\&={\frac {(1+{\frac {\sqrt {3}}{2}})+(-10-5{\sqrt {3}})i}{{\frac {1}{4}}+(-25\times (-1))}}\\&={\frac {{\frac {2+{\sqrt {3}}}{2}}+(-10-5{\sqrt {3}})i}{\frac {101}{4}}}\\&={\frac {4}{101}}({\frac {2+{\sqrt {3}}}{2}}+(-10-5{\sqrt {3}})i)\\\end{aligned}}}
虽然这个过程为了清晰起见进行了扩展，但它确实强调了在除复数时需要谨慎和准确。同样，虽然数字可能有点复杂，但过程相当简单。它也非常类似于“有理化分母”，即消去分母中的“i”项，使分母成为整体表达式的因子。
z 2 {\displaystyle {\mathit {z}}^{2}} [ 编辑 | 编辑源代码 ] 想象一下复数
z 2 = − 8 + 6 i {\displaystyle {\mathit {z}}^{2}=-8+6{\mathit {i}}\!}
那么如何找到 z {\displaystyle {\mathit {z}}\!} a + b i {\displaystyle {\mathit {a}}+{\mathit {b}}{\mathit {i}}\!}
Let us initially define
z
{\displaystyle {\mathit {z}}\!}
z
=
a
+
b
i
{\displaystyle {\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!}
z
2
=
(
a
+
b
i
)
2
=
a
2
+
2
a
b
i
+
i
2
b
=
(
a
2
−
b
2
)
+
2
a
b
i
{\displaystyle {\begin{aligned}{\mathit {z}}^{2}&=({\mathit {a}}+{\mathit {b}}{\mathit {i}})^{2}\\&={\mathit {a}}^{2}+2{\mathit {a}}{\mathit {b}}{\mathit {i}}+{\mathit {i}}^{2}{\mathit {b}}\!\\&=({\mathit {a}}^{2}-{\mathit {b}}^{2})+2{\mathit {a}}{\mathit {b}}{\mathit {i}}\!\\\end{aligned}}}
这种 z 2 {\displaystyle {\mathit {z}}^{2}\!} ( a 2 − b 2 ) {\displaystyle ({\mathit {a}}^{2}-{\mathit {b}}^{2}\!)} ( 2 a b i ) {\displaystyle (2{\mathit {a}}{\mathit {b}}{\mathit {i}})\!}
We can relate the given value of
z
2
{\displaystyle {\mathit {z}}^{2}\!}
−
8
+
6
i
=
(
a
2
−
b
2
)
+
2
a
b
i
{\displaystyle -8+6{\mathit {i}}=({\mathit {a}}^{2}-{\mathit {b}}^{2})+2{\mathit {a}}{\mathit {b}}{\mathit {i}}\!}
(
a
2
−
b
2
)
=
−
8
{\displaystyle ({\mathit {a}}^{2}-{\mathit {b}}^{2})=-8\!}
2
a
b
=
6
{\displaystyle 2{\mathit {a}}{\mathit {b}}=6\!}
现在，很明显， a {\displaystyle {\mathit {a}}\!} b {\displaystyle {\mathit {b}}\!}
2
a
b
=
6
{\displaystyle 2{\mathit {a}}{\mathit {b}}=6\!}
∴
a
b
=
3
{\displaystyle \therefore {\mathit {a}}{\mathit {b}}=3\!}
∴
a
=
3
b
{\displaystyle \therefore {\mathit {a}}={\frac {3}{b}}}
(
3
b
)
2
−
b
2
=
−
8
{\displaystyle ({\frac {3}{b}})^{2}-{b}^{2}=-8\!}
∴
9
b
2
−
b
2
=
−
8
{\displaystyle \therefore {\frac {9}{b^{2}}}-b^{2}=-8}
∴
9
−
b
4
=
−
8
b
2
{\displaystyle \therefore 9-{\mathit {b}}^{4}=-8{\mathit {b}}^{2}\!}
∴
b
4
−
8
b
2
−
9
=
0
{\displaystyle \therefore {\mathit {b}}^{4}-8{\mathit {b}}^{2}-9=0}
现在我们得到了一个关于 b 2 {\displaystyle {\mathit {b}}^{2}\!}
b
4
−
8
b
2
−
9
=
0
{\displaystyle {\mathit {b}}^{4}-8{\mathit {b}}^{2}-9=0\!}
(
b
2
−
9
)
(
b
2
+
1
)
=
0
{\displaystyle ({\mathit {b}}^{2}-9)({\mathit {b}}^{2}+1)=0\!}
⇒
b
2
=
9
,
−
1
{\displaystyle \Rightarrow {\mathit {b}}^{2}=9,-1\!}
b
∈
ℜ
{\displaystyle {\mathit {b}}\in \Re \!}
⇒
b
2
=
9
{\displaystyle \Rightarrow {\mathit {b}}^{2}=9\!}
∴
b
=
±
3
{\displaystyle \therefore {\mathit {b}}=\pm 3\!}
然后我们可以将这些值代回我们最初的任何一个方程。
2
a
b
=
6
{\displaystyle 2{\mathit {a}}{\mathit {b}}=6\!}
b
=
3
⇒
3
×
2
a
=
6
∴
a
=
1
{\displaystyle {\mathit {b}}=3\Rightarrow 3\times 2{\mathit {a}}=6\therefore {\mathit {a}}=1\!}
b
=
−
3
⇒
−
3
×
2
b
=
6
∴
a
=
−
1
{\displaystyle {\mathit {b}}=-3\Rightarrow -3\times 2{\mathit {b}}=6\therefore {\mathit {a}}=-1\!}
最后，我们得到了 a {\displaystyle {\mathit {a}}\!} b {\displaystyle {\mathit {b}}\!}
When
z
2
=
−
8
+
6
i
{\displaystyle {\mathit {z}}^{2}=-8+6{\mathit {i}}\!}
z
=
±
(
1
+
3
i
)
{\displaystyle z=\pm (1+3i)}
二次方程（即 x {\displaystyle x\!} x 2 {\displaystyle x^{2}\!} ( x + 2 ) 2 = 0 {\displaystyle (x+2)^{2}=0\!}
现在考虑二次方程 z 2 + 1 = 0 {\displaystyle {\mathit {z}}^{2}+1=0\!} ± − 1 {\displaystyle \pm {\mathit {-1}}\!} i {\displaystyle {\mathit {i}}\!}
Using the quadratic formula gives:
z
=
−
b
±
b
2
−
4
a
c
2
a
=
−
0
±
0
2
−
4
×
1
×
1
2
=
−
0
±
−
4
2
=
±
−
4
2
=
±
4
−
1
2
=
±
2
i
2
=
±
i
{\displaystyle {\begin{aligned}z&={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}\\&={-0\pm {\sqrt {0^{2}-4\times 1\times 1}} \over 2}\\&={-0\pm {\sqrt {-4}} \over 2}\\&={\pm {\sqrt {-4}} \over 2}\\&={\pm {\sqrt {4}}{\sqrt {-1}} \over 2}\\&={\pm 2i \over 2}\\&={\pm i}\\\end{aligned}}}
然后我们可以使用因式定理证明，要么 z = i {\displaystyle z=i\!} z = − i {\displaystyle z=-i\!} z 2 + 1 = 0 {\displaystyle z^{2}+1=0\!}
Let
P
(
z
)
=
z
2
+
1
{\displaystyle P(z)=z^{2}+1\!}
z
=
i
⇒
P
(
z
)
=
i
2
+
1
=
(
−
1
)
+
1
=
0
{\displaystyle {\begin{aligned}z&=i\Rightarrow P(z)=i^{2}+1\!\\&=(-1)+1\!\\&=0\!\\\end{aligned}}}
P
(
i
)
=
0
⇒
i
{\displaystyle P(i)=0\Rightarrow i\!}
z
2
+
1
{\displaystyle z^{2}+1\!}
对于 − i {\displaystyle -i\!} − 1 = ± i {\displaystyle {\sqrt {-1}}=\pm i\!}
⇒ P ( z ) = ( z + i ) ( z − i ) {\displaystyle \Rightarrow P(z)=(z+i)(z-i)}
对于复数根也可以做同样的事情
Consider the quadratic equation:
z
2
−
14
z
+
53
{\displaystyle z^{2}-14z+53\!}
z
=
−
b
±
b
2
−
4
a
c
2
a
=
−
(
−
14
)
±
(
−
14
)
2
−
4
×
1
×
53
2
=
14
±
196
−
212
2
=
14
±
−
16
2
=
14
±
4
i
2
=
7
±
2
i
{\displaystyle {\begin{aligned}z&={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}\\&={-(-14)\pm {\sqrt {(-14)^{2}-4\times 1\times 53}} \over 2}\\&={14\pm {\sqrt {196-212}} \over 2}\\&={14\pm {\sqrt {-16}} \over 2}\\&={14\pm 4i \over 2}\\&=7\pm 2i\!\\\end{aligned}}}
z
2
−
14
z
+
53
=
(
z
−
(
7
+
2
i
)
)
(
z
−
(
7
−
2
i
)
)
{\displaystyle z^{2}-14z+53=(z-(7+2i))(z-(7-2i))\!}
再次注意， 7 ± 2 i {\displaystyle 7\pm 2i\!} 7 + 2 i {\displaystyle 7+2i\!} 7 − 2 i {\displaystyle 7-2i\!}
这是通用的阿根图，显示了复数 z = a + bi 当我们考虑到复数的一部分是完全虚数时，这是一个有趣的概念。如何用数学方式表示完全虚数的东西呢？“阿根图”就是用来图形化表示复数的工具。它将复数的实部作为 x 坐标，虚部作为 y 坐标进行处理。
阿根图在确定复数的“幅角”方面非常有用。它还可以更直观地表示复数的“模”的含义。
这与在图上找到任何给定直线的长度非常相似。假设你有两个点 - 原点 (0,0) 和点 A (3,4)。画出连接这两个点的线段 OA。为了找到这条线的长度，可以使用在 C1 和 C2 中学习过的勾股定理。
L
e
n
g
t
h
O
A
=
(
3
−
0
)
2
+
(
4
−
0
)
2
=
9
+
16
=
25
=
5
{\displaystyle {\begin{aligned}LengthOA&={\sqrt {(3-0)^{2}+(4-0)^{2}}}\\&={\sqrt {9+16}}\\&={\sqrt {25}}\\&=5\end{aligned}}}
