不使用内置类型和运算符的复数Python划分 [英] Python Division Of Complex Numbers Without Using Built In Types and Operators

问题描述

我必须实现一个名为ComplexNumbers的类，该类表示一个复数，并且不允许为此使用内置类型. 我已经覆盖了允许执行基本操作的运算符(__add__，__sub__，__mul__，__abs__，__str_). 但是现在我不得不覆盖__div__运算符.

I have to implement a class called ComplexNumbers which is representing a complex number and I'm not allowed to use the built in types for that. I already have overwritten the operators (__add__, __sub__, __mul__, __abs__, __str_ which allows to perform basic operations. But now I'm stuck with overwriting the __div__ operator.

允许使用:

我用float表示数字的虚部，用float表示相关部分.

I'm using float to represent the imaginary part of the number and float to represent the rel part.

我已经尝试过的:

• 我查看了如何对复数进行除法(手写)
• 我已经完成了示例计算
• 考虑如何以编程方式实现它而没有任何良好结果

关于如何对复数进行除法的说明:

Explanation of how to divide complex numbers:

http://www .mathwarehouse.com/algebra/complex-number/divide/how-to-divide-complex-numbers.php

我的乘法实现:

 def __mul__(self, other):
        real = (self.re * other.re - self.im * other.im)
        imag = (self.re * other.im + other.re * self.im)
        return ComplexNumber(real, imag)

推荐答案

我认为这足够了:

def conjugate(self):
    # return a - ib

def __truediv__(self, other):
    other_into_conjugate = other * other.conjugate()
    new_numerator = self * other.conjugate()
    # other_into_conjugate will be a real number
    # say, x. If a and b are the new real and imaginary
    # parts of the new_numerator, return (a/x) + i(b/x)

__floordiv__ = __truediv__

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