# Complex.Reciprocal(Complex) 方法

## 定义

``````public:
static System::Numerics::Complex Reciprocal(System::Numerics::Complex value);``````
``public static System.Numerics.Complex Reciprocal (System.Numerics.Complex value);``
``static member Reciprocal : System.Numerics.Complex -> System.Numerics.Complex``
``Public Shared Function Reciprocal (value As Complex) As Complex``

value
Complex

#### 返回

`value` 的倒数。The reciprocal of `value`.

## 示例

``````using System;
using System.Numerics;

public class Example
{
public static void Main()
{
Complex[] values = { new Complex(1, 1),
new Complex(-1, 1),
new Complex(10, -1),
new Complex(3, 5) };
foreach (Complex value in values)
{
Complex r1 = Complex.Reciprocal(value);
Console.WriteLine("{0:N0} x {1:N2} = {2:N2}",
value, r1, value * r1);
}
}
}
// The example displays the following output:
//       (1, 1) x (0.50, -0.50) = (1.00, 0.00)
//       (-1, 1) x (-0.50, -0.50) = (1.00, 0.00)
//       (10, -1) x (0.10, 0.01) = (1.00, 0.00)
//       (3, 5) x (0.09, -0.15) = (1.00, 0.00)
``````
``````Imports System.Numerics

Module Example
Public Sub Main()
Dim values() As Complex = { New Complex(1, 1),
New Complex(-1, 1),
New Complex(10, -1),
New Complex(3, 5) }
For Each value As Complex In values
Dim r1 As Complex = Complex.Reciprocal(value)
Console.WriteLine("{0:N0} x {1:N2} = {2:N2}",
value, r1, value * r1)
Next
End Sub
End Module
' The example displays the following output:
'       (1, 1) x (0.50, -0.50) = (1.00, 0.00)
'       (-1, 1) x (-0.50, -0.50) = (1.00, 0.00)
'       (10, -1) x (0.10, 0.01) = (1.00, 0.00)
'       (3, 5) x (0.09, -0.15) = (1.00, 0.00)
``````

## 注解

Number x的倒数或乘法倒数是数字y ，其中x乘以y将生成1。The reciprocal, or multiplicative inverse, of a number x is a number y where x multiplied by y yields 1. 复数的倒数是在两个数字相乘时产生 Complex.One 的复数。The reciprocal of a complex number is the complex number that produces Complex.One when the two numbers are multiplied. 如果复数由 + bi 表示，则其倒数由表达式 a/（a2+ b2） +-b/（a2 + b2）表示。If a complex number is represented by a +bi, its reciprocal is represented by the expression a/(a2+b2) + -b/(a2 + b2).