SymPy和复数的平方根 [英] SymPy and square roots of complex numbers

问题描述

当使用solve计算二次方程式的根时，SymPy返回可以简化的表达式，但我无法简化它们.一个最小的示例如下所示:

When using solve to compute the roots of a quadratic equation, SymPy returns expressions which could be simplified but I can't get it to simplify them. A minimal example looks like so:

from sympy import *
sqrt(-24-70*I)

在这里，SymPy仅返回sqrt(-24-70*I)，而Mathematica或Maple会以5-7*I等效.

Here, SymPy just returns sqrt(-24-70*I) while Mathematica or Maple will answer with the equivalent of 5-7*I.

我知道有两个平方根，但是这种行为需要例如SymPy从中返回相当复杂的解决方案

I'm aware that there are two square roots, but this behavior entails that SymPy will, for example, return pretty complicated solutions from

z = symbols("z")
solve(z ** 2 + (1 + I) * z + (6 + 18 * I), z)

同时，Maple和Mathematica都将愉快地给我提供两个可解决该方程的高斯整数.

while, again, Maple and Mathematica will both happily give me the two Gaussian integers that solve this equation.

是否有选项或我缺少的东西?

Is there an option or something that I'm missing?

推荐答案

在逻辑上，找到z的平方根与求解方程(x + I * y)** 2 = z相同.因此，您可以这样做:

Finding the square root of z is logically the same as solving the equation (x+I*y)**2 = z. So you can do just that:

from sympy import *
z = -24-70*I
x, y = symbols('x y', real=True)
result = solve((x+I*y)**2 - z, (x, y))

结果为[(-5, 7), (5, -7)]

为方便起见，可以将其包装为一个函数:

For convenience, this can be wrapped as a function:

def my_sqrt(z):
    x, y = symbols('x y', real=True)
    sol = solve((x+I*y)**2 - z, (x, y))
    return sol[0][0] + sol[0][1]*I

现在您可以使用my_sqrt(-24-70*I)并获取-5 + 7*I

Now you can use my_sqrt(-24-70*I) and get -5 + 7*I

相同的策略在二次方程式中对您的示例有帮助:

The same strategy helps in your example with a quadratic equation:

x, y = symbols('x y', real=True)
z = x + I*y
solve(z ** 2 + (1 + I) * z + (6 + 18 * I), (x, y))

输出:[(-3, 3), (2, -4)]

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