Python 对数据类型进行编码，即用于复数算术

Question

In Python -

x3 = 23/2j
print(x3)
#output = -11.5j

and when

x3 = 23j/2
print(x3)
#output = 11.5j

whats the reason

Answer 1

see, value of j is sq-root of -1, and so value of j-square j^2 = -1

Multiple both numerator and denominator by j in first equation

x3 = (23 * j)/ (2j * j)

=> x3 = 23j/(-2) because j-square = -1

=> x3 = -23j/2

Answer 2

Suppose you have 23j/2 ie a complex number and if you want j in denominator, you can shift it to denominator but the condition is you have to place a negative sign as it is an imaginary number( For ex. 2+3j, here 2 is real and 3 is imaginary part of the equation. You can also write it as 2-3/j) so your number becomes -23/2j. I hope this will be helpful for you

Answer 3

It's to do with syntax/representation of the complex part. (ie the #j is one single unit)

1. 23/2j = 23/(2j) = -11.5j and NOT (23/2)*j .

2. 23j/2 = (23j)/2 = 11.5j