math@常见的复数运算
文章目录
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- 复数一元运算
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- 共轭运算
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- 共轭运算的基本性质
- 复数二元运算
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- 符号说明
- 复数加减法
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- 复数相反数
- 复数乘法
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- 复数乘方
- 复数的倒数
- 复数除法
- 复数共轭运算下基础运算
- 复数的运算性质小结
- 求和式共轭与累积式共轭
- 建立了复数的概念后,接着就是建立复数集里的各种运算
- 由于实数是复数的一部分,所以建立复数运算时,应当遵循的一个原则是,作为复数的实数在集合里的运算和在实数集里的运算是一致的
复数一元运算
共轭运算
- 根据共轭复数的定义,复数 z = a + b i z=a+bi z=a+bi的共轭复数为 z ‾ = a − b i \overline{z}=a-bi z=a−bi,
共轭运算的基本性质
- 由共轭复数的定义
- 实数a的共轭复数是a本身
- 共轭运算满足自反律,即 z ‾ ‾ = z \overline{\overline{z}}=z z=z
- 在复平面中,表示共轭复数的两个点关于 x x x轴(实轴)对称
- 共轭复数具有相同的模: ∣ z ‾ ∣ = ∣ z ∣ = a 2 + b 2 |\overline{z}|=|z|=\sqrt{a^2+b^2} ∣z∣=∣z∣=a2+b2
- k z ‾ = k z ‾ \overline{kz}=k\overline{z} kz=kz
- 设 z = a + b i z=a+bi z=a+bi
- k z = k ( a + b i ) = k a + k b i kz=k(a+bi)=ka+kbi kz=k(a+bi)=ka+kbi
- k z ‾ = k a − k b i \overline{kz}=ka-kbi kz=ka−kbi
- k z ‾ = k ( a − b i ) = k a − k b i k\overline{z}=k(a-bi)=ka-kbi kz=k(a−bi)=ka−kbi
- 可见 k z ‾ = k z ‾ \overline{kz}=k\overline{z} kz=kz
复数二元运算
- 以下讨论复数的加减乘除二元运算
符号说明
- 设任意两个复数: z 1 = a + b i z_1=a+b_i z1=a+bi, z 2 = a 2 + b 2 i z_2=a_2+b_2i z2=a2+b2i, a i , b i ∈ R , i = 1 , 2 a_i,b_i\in{\mathbb{R}},i=1,2 ai,bi∈R,i=1,2
复数加减法
- 复数加法 z 1 + z 2 = ( a 1 + a 2 ) + ( b 1 + b 2 ) i z_1+z_2=(a_1+a_2)+(b_1+b_2)i z1+z2=(a1+a2)+(b1+b2)i该结果仍然是复数
- 容易验证加法满足交换律和结合律
复数相反数
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设 z = a + b i z=a+bi z=a+bi,由复数加法定义有 a + b i + ( − a − b i ) = 0 a+bi+(-a-bi)=0 a+bi+(−a−bi)=0,称 − a − b i -a-bi −a−bi(即 − ( a + b i ) -(a+bi) −(a+bi))为 z z z的相反数,记为 − z -z −z
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根据相反数的概念,规定复数减法: z 1 − z 2 = ( a 1 − a 2 ) + ( b 1 − b 2 ) i z_1-z_2=(a_1-a_2)+(b_1-b_2)i z1−z2=(a1−a2)+(b1−b2)i
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两个复数之差还是复数
复数乘法
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z 1 z 2 = ( a 1 a 2 − b 1 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i z_1z_2=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i z1z2=(a1a2−b1b2)+(a1b2+a2b1)i
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可见两个复数的乘积仍为复数
- 复数乘法可以按照多项式乘法的方式计算:
- z 1 z 2 = ( a 1 + b 1 i ) ( a 2 + b 2 i ) z_1z_2=(a_1+b_1i)(a_2+b_2i) z1z2=(a1+b1i)(a2+b2i)
- a 1 a 2 + a 1 b 2 i + a 2 b 1 i + b 1 b 2 i 2 a_1a_2+a_1b_2i+a_2b_1i+b_1b_2i^{2} a1a2+a1b2i+a2b1i+b1b2i2,其中 i 2 = − 1 i^2=-1 i2=−1
- ( a 1 a 2 − b 1 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i (a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i (a1a2−b1b2)+(a1b2+a2b1)i
- z 1 z 2 = ( a 1 + b 1 i ) ( a 2 + b 2 i ) z_1z_2=(a_1+b_1i)(a_2+b_2i) z1z2=(a1+b1i)(a2+b2i)
- 复数乘法可以按照多项式乘法的方式计算:
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容易验证,复数乘法满足交换律和结合律,以及"乘法对加法"的分配律
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z ⋅ z ‾ = ( a + b i ) ( a − b i ) = ( a 2 + b 2 ) + ( a b − a b ) i = a 2 + b 2 z\cdot{\overline{z}}=(a+bi)(a-bi)=(a^2+b^2)+(ab-ab)i=a^2+b^2 z⋅z=(a+bi)(a−bi)=(a2+b2)+(ab−ab)i=a2+b2
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z 2 = ( a + b i ) 2 = ( a 2 − b 2 ) + 2 a b i z^2=(a+bi)^2=(a^2-b^2)+2abi z2=(a+bi)2=(a2−b2)+2abi, z 2 ‾ = ( a 2 − b 2 ) − 2 a b i \overline{z^2}=(a^2-b^2)-2abi z2=(a2−b2)−2abi
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z ‾ 2 = ( a − b i ) 2 = ( a 2 − b 2 ) − 2 a b i \overline{z}^2=(a-bi)^2=(a^2-b^2)-2abi z2=(a−bi)2=(a2−b2)−2abi
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z z ‾ = ∣ z ∣ 2 = ∣ z ‾ ∣ 2 = a 2 + b 2 z\overline{z}=|z|^2=|\overline{z}|^2=a^2+b^2 zz=∣z∣2=∣z∣2=a2+b2
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z 1 z 2 ‾ = ( a 1 a 2 − b 1 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i ‾ = ( a 1 a 2 − b 1 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i \overline{z_1z_2}=\overline{(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i}=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i z1z2=(a1a2−b1b2)+(a1b2+a2b1)i=(a1a2−b1b2)−(a1b2+a2b1)i
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z 1 ‾ z 2 ‾ = ( a 1 − b 1 i ) ( a 2 − b 2 i ) = ( a 1 a 2 − b 1 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i \overline{z_1}\;\overline{z_2}=(a_1-b_1i)(a_2-b_2i)=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i z1z2=(a1−b1i)(a2−b2i)=(a1a2−b1b2)−(a1b2+a2b1)i
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综上:
- z 2 ‾ = z ‾ 2 \overline{z^2}=\overline{z}^2 z2=z2
- z 1 z 2 ‾ = z 1 ‾ z 2 ‾ \overline{z_1z_2}=\overline{z_1}\;\overline{z_2} z1z2=z1z2
复数乘方
- z m z n = z m + n z^mz^n=z^{m+n} zmzn=zm+n
- ( z m ) n = z n m (z^m)^{n}=z^{nm} (zm)n=znm
- ( z 1 z 2 ) n = z 1 2 z 2 2 (z_1z_2)^{n}=z_1^2z_2^2 (z1z2)n=z12z22
- 虚数单位i的方幂
- i 1 = i i^1=i i1=i
- i 2 = − 1 i^2=-1 i2=−1
- i 3 = − i i^3=-i i3=−i
- i 4 = 1 i^4=1 i4=1
- i 5 = i 4 i 1 = i i^5=i^4i^1=i i5=i4i1=i
- 不难看出,第5个式子和第一个式子的结果相等,而 i n + 1 = i n ⋅ i i^{n+1}=i^{n}\cdot{i} in+1=in⋅i,由此可见,数列 a n = i n a_n=i^n an=in中,若存在 r ∈ N r\in\mathbb{N} r∈N使得 a r = a 1 a_r=a_1 ar=a1,则数列是某一个子序列的重复,不妨将子序列的长度记为 T T T,将最短的序列长度记为 T 0 T_0 T0
- 使 a r = a 1 a_r=a_1 ar=a1成立的最小整数 r r r记为 r 0 r_0 r0,则 T 0 = r 0 − 1 T_0=r_0-1 T0=r0−1
- 本例中, r 0 = 5 r_0=5 r0=5, T 0 = r 0 − 1 = 5 − 1 = 4 T_0=r_0-1=5-1=4 T0=r0−1=5−1=4
- 因此数列 a n = i n a_n=i^n an=in是 i , − 1 , − i , 1 i,-1,-i,1 i,−1,−i,1的重复
- 并且,可以进一步确定:
- i 4 n + 1 = i i^{4n+1}=i i4n+1=i
- i 4 n + 2 = − 1 i^{4n+2}=-1 i4n+2=−1
- i 4 n + 3 = − i i^{4n+3}=-i i4n+3=−i
- i 4 n = 1 i^{4n}=1 i4n=1
- 其中 n = 1 , 2 , ⋯ n=1,2,\cdots n=1,2,⋯
- 例如, 99 = 4 × 24 ⋯ 3 99=4\times{24}\cdots{3} 99=4×24⋯3, i 99 = i 3 = i 2 ⋅ i = − i i^{99}=i^{3}=i^2\cdot{i}=-i i99=i3=i2⋅i=−i
复数的倒数
- 设 z = a + b i z=a+bi z=a+bi,若存在 z ′ z' z′使得 z ⋅ z ′ = 1 z\cdot{z'}=1 z⋅z′=1则 z ′ z' z′称为 z z z的倒数,记为 z − 1 z^{-1} z−1或 1 z \frac{1}{z} z1
- 用待定系数法计算 z ′ z' z′:
- 设 z ′ = x + y i z'=x+yi z′=x+yi则 z ⋅ z ′ = ( a + b i ) ( x + y i ) = ( a x − b y ) + ( a y + b x ) i = 1 z\cdot{z'}=(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1 z⋅z′=(a+bi)(x+yi)=(ax−by)+(ay+bx)i=1
- a x − b y = 1 ax-by=1 ax−by=1
- a y + b x = 0 ay+bx=0 ay+bx=0,即 y = − b a x y=-\frac{b}{a}x y=−abx,代入第一个方程
- 解得 x = a a 2 + b 2 x=\frac{a}{a^2+b^2} x=a2+b2a; y = − b a 2 + b 2 y=-\frac{b}{a^2+b^2} y=−a2+b2b
- 即 z − 1 = a a 2 + b 2 − b a 2 + b 2 i z^{-1}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i z−1=a2+b2a−a2+b2bi
- z − 1 = 1 a 2 + b 2 ( a − b i ) z^{-1}=\frac{1}{a^2+b^2}(a-bi) z−1=a2+b21(a−bi)= ∣ z ∣ − 2 z ‾ |z|^{-2}\overline{z} ∣z∣−2z= ( ∣ z ∣ 2 ) − 1 z ‾ (|z|^2)^{-1}\overline{z} (∣z∣2)−1z= 1 ∣ z ∣ 2 z ‾ \frac{1}{|z|^2}\overline{z} ∣z∣21z
- z − 1 ‾ = 1 ∣ z ∣ 2 z ‾ ‾ \overline{z^{-1}}=\overline{\frac{1}{|z|^2}\overline{z}} z−1=∣z∣21z= 1 ∣ z ∣ 2 z ‾ ‾ {\frac{1}{|z|^2}}\overline{\overline{z}} ∣z∣21z= 1 ∣ z ∣ 2 z {\frac{1}{|z|^2}}z ∣z∣21z
- z ‾ − 1 = 1 z ‾ = z z ‾ z \overline{z}^{-1}=\frac{1}{\overline{z}}=\frac{z}{\overline{z}z} z−1=z1=zzz
- = z ∣ z ∣ 2 =\frac{z}{|z|^2} =∣z∣2z
- 可见 z − 1 ‾ = ( z ‾ ) − 1 \overline{z^{-1}}=(\overline{z})^{-1} z−1=(z)−1
- 设 z ′ = x + y i z'=x+yi z′=x+yi则 z ⋅ z ′ = ( a + b i ) ( x + y i ) = ( a x − b y ) + ( a y + b x ) i = 1 z\cdot{z'}=(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1 z⋅z′=(a+bi)(x+yi)=(ax−by)+(ay+bx)i=1
复数除法
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有了倒数概念,可以定义两个复数除法的运算法则
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( a 1 + b 1 i ) ÷ ( a 2 + b 2 i ) = a 1 + b 1 i a 2 + b 2 i = ( a 1 + b 1 i ) ( a 2 + b 2 i ) − 1 (a_1+b_1i)\div{(a_2+b_2i)}=\frac{a_1+b_1i}{a_2+b_2i}=(a_1+b_1i)(a_2+b_2i)^{-1} (a1+b1i)÷(a2+b2i)=a2+b2ia1+b1i=(a1+b1i)(a2+b2i)−1
- = ( a 1 + b 1 i ) ( a 2 2 + b 2 2 ) − 1 ( a 2 − b 2 i ) =(a_1+b_1i)(a_2^2+b_2^2)^{-1}(a_2-b_2i) =(a1+b1i)(a22+b22)−1(a2−b2i)
- = [ ( a 1 a 2 + b 1 b 2 ) + ( a 2 b 1 − a 1 b 2 ) i ] ( a 2 2 + b 2 2 ) − 1 =[(a_1a_2+b_1b_2)+(a_2b_1-a_1b_2)i](a_{2}^2+b_2^2)^{-1} =[(a1a2+b1b2)+(a2b1−a1b2)i](a22+b22)−1
- a 1 a 2 + b 1 b 2 a 2 2 + b 2 2 + a 2 b 1 − a 1 b 2 a 2 2 + b 2 2 i \frac{a_1a_2+b_1b_2}{a_{2}^2+b_2^2}+\frac{a_2b_1-a_1b_2}{a_{2}^2+b_2^2}i a22+b22a1a2+b1b2+a22+b22a2b1−a1b2i
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上述结论不必可以将 z 1 z 2 \frac{z_1}{z_2} z2z1视为分式,分子分母同时乘以分母的共轭复数 z 2 ‾ \overline{z_2} z2
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z 1 z 2 = z 1 z 2 ‾ z 2 z 2 ‾ = z 1 z 2 ‾ ∣ z 2 ∣ 2 \frac{z_1}{z_2}=\frac{z_1\overline{z_2}}{z_2\overline{z_2}} =\frac{z_1\overline{z_2}}{|z_2|^2} z2z1=z2z2z1z2=∣z2∣2z1z2
- 其中 z 1 z 2 ‾ z_1\overline{z_2} z1z2和 ∣ z ∣ 2 |z|^2 ∣z∣2分别使用乘法公式和模平方公式计算即可,分别为 ( a 1 a 2 + b 1 b 2 ) + ( a 2 b 1 − a 1 b 2 ) i (a_1a_2+b_1b_2)+(a_2b_1-a_1b_2)i (a1a2+b1b2)+(a2b1−a1b2)i, a 2 2 + b 2 2 a_2^2+b_2^2 a22+b22
复数共轭运算下基础运算
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z 1 + z 2 ‾ = ( a 1 + b 1 ) − ( b 1 + b 2 ) i \overline{z_1+z_2}=(a_1+b_1)-(b_1+b_2)i z1+z2=(a1+b1)−(b1+b2)i
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z 1 ‾ + z 2 ‾ = ( a 1 + a 2 ) − ( b 1 + b 2 ) i \overline{z_1}+\overline{z_2}=(a_1+a_2)-(b_1+b_2)i z1+z2=(a1+a2)−(b1+b2)i
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可见共轭运算对加法运算满足分配律: z 1 + z 2 ‾ = z 1 ‾ + z 2 ‾ \overline{z_1+z_2}=\overline{z_1}+\overline{z_2} z1+z2=z1+z2
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z 1 z 2 = ( a 1 a 2 − b 1 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i z_1z_2=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i z1z2=(a1a2−b1b2)+(a1b2+a2b1)i
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z 1 z 2 ‾ = ( a 1 a 2 − b 1 b 2 ) + ( a 1 b 2 + a 2 b 1 ) i ‾ = ( a 1 a 2 − b 1 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i \overline{z_1z_2}=\overline{(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i}=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i z1z2=(a1a2−b1b2)+(a1b2+a2b1)i=(a1a2−b1b2)−(a1b2+a2b1)i
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z 1 ‾ z 2 ‾ = ( a 1 − b 1 i ) ( a 2 − b 2 i ) = ( a 1 a 2 − b 1 b 2 ) − ( a 1 b 2 + a 2 b 1 ) i \overline{z_1}\;\overline{z_2}=(a_1-b_1i)(a_2-b_2i)=(a_1a_2-b_1b_2)-(a_1b_2+a_2b_1)i z1z2=(a1−b1i)(a2−b2i)=(a1a2−b1b2)−(a1b2+a2b1)i
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可见,共轭运算对乘法运算满足分配律: z 1 z 2 ‾ = z 1 ‾ z 2 ‾ \overline{z_1z_2}=\overline{z_1}\;\overline{z_2} z1z2=z1z2
- 特别的, − z ‾ = − z ‾ \overline{-z}=-\overline{z} −z=−z
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( z 1 z 2 ) ‾ = z 1 ⋅ z 2 − 1 ‾ = z 1 ‾ ⋅ z 2 − 1 ‾ = z 1 ‾ ⋅ z 2 ‾ − 1 = z 1 ‾ z 2 ‾ \overline{(\frac{z_1}{z_2})} =\overline{z_1\cdot{z_2^{-1}}} =\overline{z_1}\cdot\overline{z_2^{-1}} =\overline{z_1}\cdot{\overline{z_2}^{-1}} =\frac{\overline{z_1}}{\overline{z_2}} (z2z1)=z1⋅z2−1=z1⋅z2−1=z1⋅z2−1=z2z1
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可见共轭运算对除法运算满足分配律
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复数共轭运算对方幂的性质: z n ‾ = z ‾ n \overline{z^n}=\overline{z}^n zn=zn
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证法1:
- 对等式两边同时乘以 z ‾ \overline{z} z
- 例如: z ‾ = z ‾ \overline{z}=\overline{z} z=z
- L H S = z ‾ ⋅ z ‾ = z ⋅ z ‾ = z 2 ‾ LHS=\overline{z}\cdot\overline{z}=\overline{z\cdot{z}}=\overline{z^2} LHS=z⋅z=z⋅z=z2,逆用共轭运算对乘法运算的分配律
- R H S = z ‾ ⋅ z ‾ RHS=\overline{z}\cdot{\overline{z}} RHS=z⋅z= z ‾ 2 \overline{z}^2 z2,直接进行乘方运算
- 对 z ‾ = z ‾ \overline{z}=\overline{z} z=z执行上述操作 n − 1 n-1 n−1次,可以得到: z n ‾ = z ‾ n \overline{z^n}=\overline{z}^n zn=zn
- 对等式两边同时乘以 z ‾ \overline{z} z
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证法2:
- n = 2 n=2 n=2
- z 2 ‾ = z ⋅ z ‾ = z ˉ z ˉ = z ˉ 2 \overline{z^2}=\overline{z\cdot{z}}=\bar{z}\bar{z}=\bar{z}^2 z2=z⋅z=zˉzˉ=zˉ2,显然结论成立
- 设 n = k n=k n=k时,结论成立,即 z k ‾ = z ‾ k \overline{z^k}=\overline{z}^k zk=zk
- 当 n = k + 1 n=k+1 n=k+1时, z 3 ‾ = z 2 ⋅ z 1 ‾ = z 2 ‾ ⋅ z ˉ \overline{z^3}=\overline{z^2\cdot{z^1}}=\overline{z^2}\cdot\bar{z} z3=z2⋅z1=z2⋅zˉ= z ‾ 2 ⋅ z ‾ = z ‾ 3 \overline{z}^2\cdot{\overline{z}}=\overline{z}^3 z2⋅z=z3
- L H S = z k + 1 ‾ = z k ⋅ z ‾ LHS=\overline{z^{k+1}}=\overline{z^k\cdot{z}} LHS=zk+1=zk⋅z= z k ‾ ⋅ z ‾ \overline{z^k}\cdot{\overline{z}} zk⋅z
- 代入归纳假设,得 L H S = z ‾ k ⋅ z ‾ LHS=\overline{z}^k\cdot{\overline{z}} LHS=zk⋅z= z ‾ k + 1 \overline{z}^{k+1} zk+1
- R H S = z ‾ k + 1 RHS=\overline{z}^{k+1} RHS=zk+1
- 可见 n = k + 1 n=k+1 n=k+1时,结论依然成立
- 综上,n取任意正整数结论都成立
- n = 2 n=2 n=2
复数的运算性质小结
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复数 (wikipedia.org)
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加法： ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i 减法： ( a + b i ) − ( c + d i ) = ( a − c ) + ( b − d ) i 乘法： ( a + b i ) ( c + d i ) = a c + b c i + a d i + b d i 2 = ( a c − b d ) + ( b c + a d ) i 除法： ( a + b i ) ( c + d i ) = ( a + b i ) ( c − d i ) ( c + d i ) ( c − d i ) = ( a c + b d c 2 + d 2 ) + ( b c − a d c 2 + d 2 ) i 加法：(a+bi)+(c+di)=(a+c)+(b+d)i\\ 减法：(a+bi)-(c+di)=(a-c)+(b-d)i\\ 乘法：(a+bi)(c+di)=ac+bci+adi+bdi^{2}=(ac-bd)+(bc+ad)i\\ 除法：{\frac {(a+bi)}{(c+di)}} ={\frac {(a+bi)(c-di)}{(c+di)(c-di)}} =\left({ac+bd \over c^{2}+d^{2}}\right)+\left({bc-ad \over c^{2}+d^{2}}\right)i 加法：(a+bi)+(c+di)=(a+c)+(b+d)i减法：(a+bi)−(c+di)=(a−c)+(b−d)i乘法：(a+bi)(c+di)=ac+bci+adi+bdi2=(ac−bd)+(bc+ad)i除法：(c+di)(a+bi)=(c+di)(c−di)(a+bi)(c−di)=(c2+d2ac+bd)+(c2+d2bc−ad)i
求和式共轭与累积式共轭
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这两条性质是共轭运算对加法(乘法)运算分配律的一般形式
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∑ a i ‾ = ∑ a i ‾ ∏ i a i ‾ = ∏ i a i ‾ ∑ i ∏ j a i , j i ‾ = ∑ i ∏ j a i , j i ‾ ∑ i k i ∏ j a i , j i ‾ = ∑ i k i ∏ j a i , j i ‾ \sum{\overline{a_i}}=\overline{\sum{a_i}} \\ \prod_{i}\overline{a_i}=\overline{\prod_{i}a_i} \\ \sum_{i}{\prod_{j}\overline{a_{i,j_i}}}= \overline{\sum_{i}{\prod_{j}a_{i,j_i}}} \\ \sum_{i}k_i{\prod_{j}\overline{a_{i,j_i}}}= \overline{\sum_{i}k_i{\prod_{j}a_{i,j_i}}} ∑ai=∑aii∏ai=i∏aii∑j∏ai,ji=i∑j∏ai,jii∑kij∏ai,ji=i∑kij∏ai,ji