复数的代数运算11年级数学

复数是以a + bi的形式表示的数字，其中a和b是实数，i代表虚数单位，满足等式i²= -1 。例如，5 + 6i是复数，其中5是实数，6i是虚数。因此，实数和虚数的组合是复数。复数可以有四种类型的代数运算，如下所述。复数上的四个运算包括：

• 添加
• 减法
• 乘法
• 分配

复数的加法

要添加两个复数，只需添加相应的实部和虚部。

(a + bi) + (c + di) = (a + c) + (b + d)i 

例子：

• (7 + 8i) + (6 + 3i)  = (7 + 6) + (8 + 3)i = 13 + 11i
• (2 + 5i) + (13 + 7i) = (2 + 13) + (7 + 5)i = 15 + 12i
• (-3 – 6i) + (-4 + 14i) = (-3 – 4) + (-6 + 14)i = -7 + 8i
• (4 – 3i ) + ( 6 + 3i) = (4+6) + (-3+3)i = 10
• (6 + 11i) + (4 + 3i) = (4 + 6) + (11 + 3)i = 10 + 14i

减复数

要减去两个复数，只需减去相应的实部和虚部。

(a + bi) − (c + di) = (a − c) + (b − d)i 

例子：

• (6 + 8i)  –  (3 + 4i) = (6 – 3) + (8 – 4)i = 3 + 4i
• (7 + 15i) – (2 + 5i) = (7 – 2) + (15 – 5)i = 5 + 10i
• (-3 + 5i) – (6 + 9i) = (-3 – 6) + (5 – 9)i = -9 – 4i
• (14 – 3i) – (-7 + 2i) = (14 – (-7)) + (-3 – 2)i = 21 – 5i
• (-2 + 6i) – (4 + 13i) = (-2 – 4) + (6 – 13)i = -6 – 7i

两个复数的乘法

两个复数的乘法与两个二项式的乘法相同。让我们假设我们必须将a + bi和c + di相乘。我们将逐项相乘。

(a + bi) ∗ (c + di) = (a + bi) ∗ c + (a + bi) ∗ di

                              = (a ∗ c + (b ∗ c)i)+((a ∗ d)i + b ∗ d ∗ −1)

                              = (a ∗ c − b ∗ d + i(b ∗ c + a ∗ d))   

示例1 ：分别乘以(1 + 4i)和(3 + 5i)。

(1 + 4i) ∗ (3 + 5i) = (3 + 12i) + (5i + 20i²)

                            = 3 + 17i − 20

                            = −17 + 17i

Note: Multiplication of complex numbers with real numbers or purely imaginary can be done in the same manner. 

示例2：分别乘以5和(4 + 7i)。

5 ∗ (4+7i) can be viewed as (5 + 0i) ∗ (4 + 7i)

                                            = 5 ∗ (4 + 7i)

                                            = 20 + 35i   

示例3：分别乘以3i和(2 + 6i)。

3i ∗ (2 + 6i) can be viewed as (0 + 3i) ∗ (2 + 6i)

                                               = 3i ∗ (2 + 6i)

                                               = 6i + 18i²

                                               = 6i − 18

                                               = −18 + 6i   

示例4：分别乘以(5 + 3i)和(3 + 4i)。

(5+3i) ∗ (3+4i) = (5 + 3i) ∗ 3 + (5 + 3i) ∗ 4i

                        = (15 + 9i) + (20i + 12i²)

                        = (15 − 12) + (20 + 9)i

                        = 3 + 29i  

复数加法，减法和乘法复习

1. (a + bi) + (c + di) = (a + c) + (b + d)i
2. (a + bi) − (c + di) = (a − c) + (b − d)i
3. (a + bi) ∗ (c + di) = ((a ∗ c − b ∗ d) + (b ∗ c + a ∗ d)i)

复数的共轭

在任何两个复数中，如果仅虚部的符号不同，则它们被称为彼此的复共轭。因此，复数a + bi的共轭将是a-bi。

$The\ complex\ conjugate\ of\ z\ is \ denoted\ by\ \bar z .$

复数共轭有什么用?

$\frac{1 + 2i}{4-5i} \ this\ can\ be\ written\ as\: \frac{1 + 2i}{4-5i}*\frac{4 + 5i}{4+5i} \\ \ \\ = \frac{(1 + 2i)*(4 + 5i)}{(4-5i)*(4 + 5i)} \\ \ \\ = \frac{(4-10)+(8+5)i}{(16+25)+(20 - 20)i} \\ \ \\ = \frac{-6+13i}{16+25} \\ \ \\ = \frac{-6}{41} + \frac{13}{41}i \\$

Thus we can observe that multiplying a complex number with its conjugate gives us a real number. Thus the division of complex numbers is possible by multiplying both numerator and denominator with the complex conjugate of the denominator.

复杂共轭物的例子

$1. \ \overline {4 + 7i} = 4 - 7i \\ 2. \ \overline {-6 + 12i} = -6 - 12i \\ 3. \ \overline {34 - 7i} = 34 + 7i \\ 4. \ \overline {-15 - 7i} = - 15 + 7i$

复杂共轭物的性质

属性1：

${ \bullet } \ \bar {\bar z} = z \\ Proof: \\ Let\ z = a + ib .\ Then\ by\ definition, (conjugate of z) = a - ib. \\ Therefore, (conjugate\ of\ \bar z ) = a + ib = z$

属性2：

${ \bullet } \ \overline {z1+z2} = \overline {z1} + \overline {z2} \\ Proof: \\ If\ z1 = a + ib\ and\ z2 = c + id\ then\ \overline {z1} = a - ib \ and\ \overline {z2} = c - id \\ Now, z1 + z2 = a + ib + c + id = (a + c) + i(b + d) \\ Therefore, \overline {z1+z2} = a + c - i(b + d) \\ = a - ib + c - id \\ = \overline {z1} + \overline {z2}$

属性3：

${ \bullet } \ \overline {z1-z2} = \overline {z1} - \overline {z2} \\ Proof: \\ If\ z1 = a + ib\ and\ z2 = c + id\ then\ \overline {z1} = a - ib \ and\ \overline {z2} = c - id \\ Now, z1 + z2 = a + ib - (c + id) = (a- c) + i(b - d) \\ Therefore, \overline {z1-z2} = a - c - i(b - d) \\ = (a - ib) - (c - id) \\ = \overline {z1} - \overline {z2}$

物业4：

${ \bullet } \ \overline {z1*z2} = \overline {z1}*\overline {z2} \\ Proof: \\ If\ z1 = a + ib\ and\ z2 = c + id\ then\ \overline {z1} = a - ib \ and\ \overline {z2} = c - id \\ Now, z1*z2 = a + ib - (c + id) = (ac-bd) + i(bc + ad) \\ Therefore, \overline {z1*z2} = (ac-bd) - i(bc+ ad) \\ Also,\ \overline {z1}*\overline {z2} = (a - ib)(c - id) = (ac - bd) - i(bc + ad)$

属性5：

${ \bullet } \ \overline {(\frac{z1}{z2})} = \frac{\overline {z1}}{\overline {z2}} ,\ provided\ z2\neq 0 \\ Proof: \\ According\ to\ the\ problem \\ z2 ≠ 0 ⇒ \overline {z2} ≠ 0 \\ Let, \frac{z1}{z2} = z3 \\ z1 = z2*z3 \\ ⇒ \overline {z1} = \overline {z2*z3} \\ ⇒ \overline {z1} = \overline {z2}*\overline {z3} \\ ⇒ \frac{\overline {z1}}{\overline {z2}} = \overline {z3} \\ \ \\ ⇒ \overline {(\frac{z1}{z2})} = \frac{\overline {z1}}{\overline {z2}}, [Since\ z3 = \frac{z1}{z2}]$

两个复数的除法

通过将分子和分母都乘以分母的复共轭来完成复数的除法。

$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} \\ \ \\ = \frac{(ac+bd)+ (bc-ad)i}{c^2+d^2} \\ \ \\ = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i \\$

范例1：

${ \bullet } \ \frac{2 + 5i}{3+4i} \\ \ \\ = \frac{2 + 5i}{3+4i}*\frac{3- 4i}{3-4i} \\ \ \\ = \frac{(2 + 5i)*(3- 4i)}{(3+4i)*(3 -4i)} \\ \ \\ = \frac{(6+20)+(15-8)i}{(9+16)+(12- 12)i} \\ \ \\ = \frac{26+7i}{25} \\ \ \\ = \frac{26}{25} + \frac{7}{25}i \\$

范例2：

${ \bullet } \ \frac{4 + 2i}{-1+i} \\ \ \\ = \frac{4 + 2i}{-1+i}*\frac{-1-i}{-1-i} \\ \ \\ = \frac{(4 + 2i)*(-1- i)}{(-1+i)*(-1 -i)} \\ \ \\ = \frac{(-4+2)+(-2-4)i}{(1+1)+(1- 1)i} \\ \ \\ = \frac{-2-6i}{2} \\ \ \\ = \frac{-2}{2} - \frac{6}{2}i \\ = -1 -3i$

范例3：

${ \bullet } \ \frac{-2}{1+i} \\ \ \\ = \frac{-2}{1+i}*\frac{1-i}{1-i} \\ \ \\ = \frac{-2*(1- i)}{(1+i)*(1 -i)} \\ \ \\ = \frac{-2+2i}{(1+1)+(1- 1)i} \\ \ \\ = \frac{-2+2i}{2} \\ \ \\ = \frac{-2}{2} + \frac{2}{2}i \\ = -1 + i$

范例4：

${ \bullet } \ \frac{4 + 5i}{2i} \\ \ \\ = \frac{4 + 5i}{2i}*\frac{-2i}{-2i} \\ \ \\ = \frac{(4 + 5i)*-2i}{-4i^2} \\ \ \\ = \frac{10-8i}{4} \\ \ \\ = \frac{10-8i}{4} \\ \ \\ = \frac{10}{4} - \frac{8}{4}i \\ \ \\ = \frac{5}{2} -2i$