I am using Jupyter in particular. e.g from the equation e2+j, how can I seperate it in it's real part (e2) and into it's imaginary (ej)?
I have tried:
exp(complex(2,1)).real
However the error that follows it is: 'Mul' object has no attribute 'real'.
An other solution might be to implement the eulers formula to seperate it as cos(2)+j·sin(1) but no success so far. Generally the problem is when complex numbers appear in power position rather than in usual format (2+j). If someone has any idea upon this matter would be greatly appreciated!
Important edit that fits more in my problem:
In my case the complex equation that i have, came through the dsolve() for a 2nd order differential equation. For the exp() element that exist in nympy this is a valid solution, it's not the same as an arbitary equation. However my equation is just a the above one regarding it's complexity
I include my code:
import scipy as sp
from sympy import*
import sympy as syp
from scipy.integrate import odeint
t, z, w, C2=symbols('t, z, w, C2')
x=Function('x')
eq=x(t).diff(t,2)+2*z*w*x(t).diff(t,1)+w**2*x(t)
sol=dsolve(eq,x(t),ics={x(0):0,x(t).diff(t,1).subs(t,0):2*C2*w*sqrt(z**2-1)})
next I want to substitute the z,w parameters in order to fit my data and then with a loop to make an array that takes the numerical solutions in order to plot them. I've tried the following:
for i in range(1000):
step.append(i)
numdata=[]
for i in range(1000):
numdata.append(N(sol.rhs.subs(t,i).subs(w,10).subs(z,0.001)))
However this can't work as the sol is an complex function. After this long journey I try to find (meaning of my life) ways to seperate the real and imaginary parts of this kind of function.
Thank you for being with me so far, you are a hero reagrdless of the result.
sympy has re and im functions:
In [113]: exp(complex(2,1))
Out[113]:
1.0⋅ⅈ
7.38905609893065⋅ℯ
In [114]: re(exp(complex(2,1)))
Out[114]: 3.99232404844127
In [115]: im(exp(complex(2,1)))
Out[115]: 6.21767631236797
In [116]: exp(complex(2,1)).evalf()
Out[116]: 3.99232404844127 + 6.21767631236797⋅ⅈ
.real and .imag are attributes (possibly implemented as properties) of numpy arrays (and complex python numbers).
Exploring sympy a bit more:
In [152]: expand(exp(y),complex=True)
Out[152]:
re(y) re(y)
ⅈ⋅ℯ ⋅sin(im(y)) + ℯ ⋅cos(im(y))
In [153]: expand(exp(complex(2,1)),complex=True)
Out[153]: 3.99232404844127 + 6.21767631236797⋅ⅈ
Your sol:
In [157]: sol
Out[157]:
⎛ ________⎞ ⎛ ________⎞
⎜ ╱ 2 ⎟ ⎜ ╱ 2 ⎟
t⋅w⋅⎝-z - ╲╱ z - 1 ⎠ t⋅w⋅⎝-z + ╲╱ z - 1 ⎠
x(t) = - C₂⋅ℯ + C₂⋅ℯ
In [181]: f1 = sol.rhs.subs({w:10, z:0.001,C2:1})
In [182]: f1
Out[182]:
10⋅t⋅(-0.001 - 0.999999499999875⋅ⅈ) 10⋅t⋅(-0.001 + 0.999999499999875⋅ⅈ)
- ℯ + ℯ
Making a numpy compatible function:
In [187]: f = lambdify(t, f1)
In [188]: print(f.__doc__)
Created with lambdify. Signature:
func(t)
Expression:
-exp(10*t*(-0.001 - 0.999999499999875*I)) + exp(10*t*(-0.001 +...
Source code:
def _lambdifygenerated(t):
return (-exp(10*t*(-0.001 - 0.999999499999875*1j)) + exp(10*t*(-0.001 + 0.999999499999875*1j)))
Imported modules:
evaluate it at a range of values:
In [189]: f(np.arange(10))
Out[189]:
array([0.+0.j , 0.-1.07720771j, 0.+1.78972745j, 0.-1.91766624j,
0.+1.43181934j, 0.-0.49920326j, 0.-0.57406585j, 0.+1.44310044j,
0.-1.83494157j, 0.+1.63413971j])
same values with just sympy:
In [199]: [im(f1.evalf(subs={t:i})) for i in range(10)]
Out[199]:
[0, -1.0772077135423, 1.78972744700845, -1.9176662437755, 1.43181934232583, -0.499203257243971, -0.
574065847629935, 1.44310044143674, -1.83494157235822, 1.63413971490123]