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I have run across some confusing behaviour with square roots of complex numbers in python. Running this code: from cmath import sqrt a = 0.2 b = 0.2 + 0j print(sqrt(a / (a - 1))) print(sqrt(b / (b - 1)))
One way is to calculate integer square root of the number multiplied by required power of 10. For example, using the Newton's method: def isqrt(num): x = 1 << -(-num.bit_length() >> 1) # ceil. y = (x + num//x) >> 1. while x > y: x = y.
1 mar 2024 · This code snippet shows the use of Python’s built-in cmath.sqrt() function, which provides a straightforward approach for calculating the square root of individual complex numbers, although it’s less suited for array operations.
It's simple to calculate the square root of a value in Python using the exponentiation operator ** or math.sqrt(). It's worth mentioning that math.sqrt() is usually the faster of the two and that by using cmath.sqrt() you can get the square root of a complex number.
The cmath.sqrt() method returns the square root of a complex number. Note: The number must be greater than or equal to 0.
math.sqrt(x) is faster than math.pow(x, 0.5) or x ** 0.5 but the precision of the results is the same. The cmath module is extremely similar to the math module, except for the fact it can compute complex numbers and all of its results are in the form of a + bi. It can also use .sqrt(): import cmath. cmath.sqrt(4) # 2+0j.
Python can calculate the square root of a number in three ways: math.sqrt(), pow(), or **. This article explains each with easy-to-follow code.
