复合共轭物

复数是以(x + i y)形式表示的数字；其中x＆y是实数，而i =√(-1)称为iota(虚数单位)。它也被称为虚数或数量。复数可以是纯实数，也可以是纯虚数，具体取决于x和y的值。

例子：

(5 + 3 i), (0 + 7 i), (-4 + 9 i), (1 + 0 i), (0 – i), etc.

现在，我们将讨论复数的两个非常重要的概念，即共轭和模数。

复数的共轭

复数的共轭数是另一个复数，其实部Re(z)相等，而虚部Im(z)大小相等但符号相反。有时，复数的共轭也称为复共轭。复数z的共轭表示为${\overline{z}}$和(z＆ ${\overline{z}}$ )一起称为复共轭对，因为z和${\overline{z}}$是彼此共轭的。如果z =(x + iy)是复数，则z的共轭定义为${\overline{z}}$ =(x – iy)。从z与${\overline{z}}$ ，我们可以说，复数的共轭是通过用(-i)代替i来获得的。复数共轭的几何意义${\overline{z}}$是复数z在复平面或argand平面中围绕实轴(X轴)的反射或镜像。

在Argand Plane中复数的共轭

共轭物的性质

如果z，z ₁和＆z ₂是复数，则它们将具有以下共轭性质。

1. Conjugate of a purely real complex number is the number itself (z = ${\overline{z}}$) i.e. conjugate of (7 + 0 i) = (7 – 0 i) = 7
2. Conjugate of a purely imaginary complex number is negative of that number (z + ${\overline{z}}$ = 0) i.e. conjugate of (0 -7 i) = (0 + 7 i) = 7 i
3. ${\overline{({\overline{z}})}}$ = z
4. z + ${\overline{z}}$ = 2 Re(z)
5. z – ${\overline{z}}$ = 2 i . Im(z)
6. z ${\overline{z}}$ = {Re(z)}² + {Im(z)}²
7. ${\overline{(Z₁ + Z₂)}}$ = ${\overline{Z₁} + {\overline{Z₂}}}$
8. ${\overline{(Z₁ - Z₂)}}$ = ${\overline{Z₁} - {\overline{Z₂}}}$
9. ${\overline{Z₁.Z₂}}$ =  ${\overline{Z₁} × {\overline{Z₂}}}$
10. z = (z₁ / z₂) then ${\overline{Z}}$ = ${\overline{Z₁}}$ / ${\overline{Z₂}}$ ; z₂ ≠ 0

Note: In order to find out the conjugate of a complex number that complex number must be in its standard form that is Z = (x + i y). If the complex number is not in its standard form then it has to be converted into its standard form before finding its complex conjugate.

例子

例1：如果z =(5 + 7 i)是复数，则其共轭由下式给出：

解决方案：

${\overline{z}}$ = (5 – 7 i).

例2：如果z = 1 /(4 + 3 i)是复数，则其共轭为

解决方案：

First z has to be converted into its standard form by multiplying the numerator and

denominator with the conjugate of (4 + 3 i)

z = (1 / (4 + 3 i)) × ((4 – 3 i) / (4 – 3 i))

z = (4 – 3 i) / (16 + 9)

z = (4 / 25) – (3 / 25) i

The conjugate of z is ${\overline{z}}$ = (4 / 25) + (3 / 25) i

例3：如果(a + ib)是一个复数，它是(8 – 3 i)的复共轭，则a和b的值由下式给出：

解决方案：

Let z = a + i b

${\overline{(8 - 3 i)}}$ = (8 + 3 i)

z = ${\overline{(8 - 3 i)}}$

z = (8 + 3 i)

Two complex numbers are equal only when their corresponding real & imaginary parts are equal

Equating the real and imaginary parts of z & (8 + 3 i)

Re(z) = a = 8 

Im(z) = b = 3

Hence 8 & 3 are the respective values of a and b.

复数的模数

复数z =(x + iy)的模数定义为| z |。 =√(x ² + y ² );其中x = Re(z)和y = Im(z)。 | z |对于所有复数，≥0。我们无法比较两个复数z ₁和z _2，即(z ₁ > z ₂ )或(z ₁ 2 )没有意义，但是(| z ₁ |> | z ₂ |)或(| z ₁ | <| z ₂ |)具有确定的含义，因为| z ₁ | ＆| z ₂ |是实数。复数(x + iy)模量的几何含义是点P(x，y)到复平面或argand平面中的点O(0，0)的距离，即| z |。 = ${\overline{OP}}$ =√((x – 0) ² +(y – 0) ² )=√(x ² + y ² )。

Argand平面上复数的模

模量的性质

如果z，z ₁ ，z ₂和z _n是复数，则它们将具有以下模量特性。

1. If a complex number z has equal real and imaginary parts then its modulus is zero i.e. Re(z) = Im(z) ⇒ |z| = 0
2. Real and imaginary part of a complex number z always lies between -|z| and  |z|  i.e. -|z| ≤ Re(z) ≤ |z| and -|z| ≤ Im(z) ≤ |z|
3. |z| = |${\overline{z}}$| = |-z|
4. ${z.\overline{z}}$ = |z|²
5. |Z₁ .  Z₂ …….. Z_n| = |Z₁| . |Z₂| …….. |Z_n|
6. |Z₁ / Z₂| = |Z₁| / |Z₂| ; Z₂ ≠ 0
7. |Z₁ + Z₂|² = |Z₁|² + |Z₂|² + 2 Re(${z.\overline{z}}$)
8. |Z₁ – Z₂|² = |Z₁|² + |Z₂|² – 2 Re(${z.\overline{z}}$)
9. |Z₁ + Z₂|² + |Z₁ – Z₂|² = 2( |Z₁|² + |Z₂|²)
10. |b. Z₁ + a. Z₂|² + |a. Z₁ – b. Z₂|² = (a² + b²) ( |Z₁|² + |Z₂|²) ; here a & b are real numbers.

Note: To find out the modulus of a complex number that complex number must be in its standard form that is Z = (x + i y). If the complex number is not in its standard form then it has to be converted into its standard form before finding its modulus.

例子

示例1：如果(8 + 3 i)是复数，则其模数为

解决方案：

|8 + 3 i| = √(8² + 3²) 

|8 + 3 i| = √(64 + 9)

|8 + 3 i| = √(73)

示例2：如果(3 + 4 i) 和(√3+√6i) 如果是两个复数，则其乘积的模数由其各个模数的乘积给出。

解决方案：

|3 + 4 i| = √(3² + 4²)

|3 + 4 i| = √(9 + 16)

|3 + 4 i| = √(25) = 5

|√3 + √6 i| = √((√3)² + (√6)²)

|√3 + √6 i| = √(3 + 6) 

|√3 + √6 i| = √9 =3

|(3 + 4 i).(√3 + √6 i)| = 5 × 3

|(3 + 4 i).(√3 + √6 i)| = 15

例3：如果z =(2。i)/(3 – 4 i)是复数，则其模量为

解决方案：

First z has to be converted into its standard form by multiplying the numerator and

denominator with the conjugate of (3 – 4 i)

z = ((2.i) / (3 – 4 i)) × ((3 + 4 i) / (3 + 4 i))

z = (8 + 6 i) / (9 + 16)

z = (8 + 6 i) / 25

z = (8 / 25) + (6 / 25) i

The modulus of z is |z| = √((8 / 25)² + (6 / 25)²)

|z| = √((64 + 36) /625)

|z| = √(100 /625)

|z| = 10 / 25

|z| = 2 / 5