复数的加法恒等式和倒数

实数和虚数的组合称为复数。这些是可以写成a + ib形式的数字。其中 a 和 b 都是实数。它用 z 表示。在复数形式中，值“a”称为实部，用 Re(z) 表示，“b”称为虚部 Im(z)。它也被称为虚数。在复数形式中，a + bi 'i' 是一个称为“iota”的虚数。 i 的值为 (√-1) 或者我们可以写成 i ² = -1。

例如，

• 3 + 4i 是复数，其中 3 是实数 (Re)，4i 是虚数 (Im)。
• 2 + 5i 是复数，其中 2 是实数 (Re)，5i 是虚数 (im)

Complex number	Real number	Imaginary number
-2 + 5i	-2	5i
2 – 5i	2	-5i
-5i	0	-5i (Pure Imaginary)
5	5	0i (Pure real)

虚数规则

1. i = √-1
2. i² = -1
3. i³ = -i
4. i⁴ = 1
5. i⁴ⁿ = 1
6. i^{4n – 1} = -1

复数的加法恒等式和倒数

解决方案：

Additive identity 

Additive identity property states that when a number is added to zero will give the same number in result . 

Here , “Zero” is called the identity element, (also known as additive identity) If we add any complex number with zero, the resulting number will be the same complex number. This is true for any real numbers, complex numbers and even for imaginary numbers.

Suppose, ‘a’ is any real number, then

x + 0 = x =  0 + x

For Example: Let assume z = a+ib , then as per the property z + 0= z

Therefore, (a + ib ) + (0 + 0i)

= a + bi + 0 + 0i

=  a + bi

= z, for all z ∈ C.

Hence, The additive identity of complex numbers is written as (x + yi) + (0 + 0i) = x + yi. 

So, the additive identity is 0 + 0i.

Additive inverse 

An additive inverse of a complex  number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.  

Suppose, x is the original number, then the additive inverse of x will be minus of x i.e., -x , such that;

x + (-x) = x – x = 0

It is also called the opposite of the number, negation of number or changed sign of original number. Suppose the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero. 

Therefore, the additive inverse of A + iB is  -(A + iB)

Example: Additive inverse of 3 + 3i is -(3 + 3i)

3 + 3i + [-(3 + 3i)]

= 3 + 3i – 3 – 3i

=  0 

Hence proved the property .

示例问题

问题 1：求复数 5 + 5i 的加法逆？

解决方案：

An additive inverse of a complex number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.  

So here the additive inverse of complex number 5 + 5i is -(5 + 5i) 

= -5 – 5i

问题2：求8+3i的加法逆并证明恒等式？

解决方案：

An additive inverse of a complex  number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.  

So here the additive inverse of complex number 8 + 3i is -(8 + 3i)

= -8 – 3i

Now to prove identity the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero .

Therefore, 8 + 3i + (-8 – 3i)

= 8 + 3i – 8 – 3i

= 0 

Hence proved 

问题 3：证明加法恒等复数 7 + 4i？

解决方案：

Additive identity property states that when a number is added to zero will give the same number in result. The additive identity of complex numbers is written as (x + yi) + (0 + 0i) = x + yi. 

So now, to prove additive identity for complex number 7 + 4i 

= (7 + 4i) + (0 + 0i)

 = 7 + 4i

Hence proved

问题 4：简化 {(-3 – 5i) / (2 +2i)}？

解决方案：

Given {(-3 – 5i) / (2 + 2i)}

Conjugate of denominator 2 + 2i is 2 – 2i

Multiply the numerator and denominator with the conjugate,

Therefore, {(-3 – 5i) / (2 + 2i) } × {(2 – 2i) / (2 – 2i)}

= {-6 – 6i – 10i +10i²} / {2² – (2i)²} {Difference of squares formula. i.e; (a+b)(a-b) = a² – b²}

= {-6 – 6i – 10i + 10(-1)} / {4 – 4(-1)} {i² = -1}

= {-6 – 6i – 10i – 10} / {4 + 4}

= (-16 – 16i) / 8

= -16 /8 – 16i /8

= -2  – 2i

问题 5：求复数 3 + 5i 的加法逆并证明？

解决方案：

An additive inverse of a complex number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.  

So here the additive inverse of complex number 3 + 5i is -(3 + 5i)

= -3 -5i

To prove: the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero .

Therefore: 3 + 5i + (-3 – 5i)

= 3 + 5i – 3 + 3i

= 0

Hence proved 

问题 6：化简 (3 + 4i) / (3 + 2i) 并求其加法逆？

解决方案：

Multiplying the numerator and denominator with the conjugate of denominators.

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

= (9 – 6i + 12i – 8i²) / {9 – (2i)²}

= (17 + 6i) / (13)

= (17 + 6i) / 13

= 17/13 + 6i/13

Now Additive inverse of complex number,

17/13 + 6i/13 is – (17/13 + 6i/13)

= -17/13 – 6i/13

问题7：进行如下操作，求a+ib形式的结果，求其逆？

• (2 – √-25) / (1 – √-16)

解决方案：

Given: (2 – √-25 ) / (1 – √-16)                                                  

(2 – √-25) / (1 – √-16) = {2 – (i)(5)} / {1 – (i)(4)}, {i = √-1}

= (2 – 5i ) / (1 – 4i)

= {(2 – 5i) / (1 – 4i)} × {(1 + 4i) / (1 + 4i)}

= {(2 – 5i) (1 + 4i)} / {(1 – 4i) (1 + 4i)}

= {2 + 8i – 5i – (20i²)} / {(1 – 16i²)} {i² = -1}

= {2 + 3i + 20} / {1 – 16(-1)}

= (22 + 3i) / (1 + 16)

= (22 + 3i)/17

= {(22/17) + (3i/17)}

= 22/17 + 3i/17

Now Additive inverse of complex number of 22/17 + 3i/17

= -(22/17 + 3i/17)

= -22/17 – 3i/17