Equivalent of j in NumPy

Question

What is the equivalent of Octave's j in NumPy? How can I use j in Python?

In Octave:

octave:1> j
ans =  0 + 1i
octave:1> j*pi/4
ans =  0.00000 + 0.78540i

But in Python:

>>> import numpy as np
>>> np.imag
<function imag at 0x2368140>
>>> np.imag(3)
array(0)
>>> np.imag(3,2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: imag() takes exactly 1 argument (2 given)
>>> np.imag(32)
array(0)
>>> 
>>> 0+np.imag(1)
1

Answer 1

In Python, 1j or 0+1j is a literal of complex type. You can broadcast that into an array using expressions, for example

In [17]: 1j * np.arange(5)
Out[17]: array([ 0.+0.j,  0.+1.j,  0.+2.j,  0.+3.j,  0.+4.j])

Create an array from literals:

In [18]: np.array([1j])
Out[18]: array([ 0.+1.j])

Note that what Michael9 posted creates a complex, not a complex array:

In [21]: np.complex(0,1)
Out[21]: 1j
In [22]: type(_)
Out[22]: complex

Answer 2

You can create one if needed or use 1j which instance of complex class

 >>> 1j #complex object
 1j
 >>> type(1j)
 <class 'complex'>
 >>> j = np.complex(0,1) #create complex number
 >>> j
 1j

Answer 3

You must use Python built-in 1j in order to distinguish this constant from variable j:

from numpy import pi, arange, exp, imag, real
from matplotlib import pyplot as plt

# Some sine wave parameters and sampling times
A = 1; f = 1; af = 2*pi*f; p0 = pi/2
Ts = 5e-3
tn = arange(-1, 1, Ts)

# Build complex array, plot real and imag parts
zn = A * exp(1j * (af*tn + p0))
plt.plot(tn, real(zn), tn, imag(zn))
plt.legend(('real', 'imag'))

zn array content:

array([
-1.83697020e-16+1.00000000e+00j,
-3.14107591e-02+9.99506560e-01j,
-6.27905195e-02+9.98026728e-01j, ...