CODEWITHSUNDEEP
pythonmathsympysymbolic-math
josteinb
josteinb

Reputation: 2172

How to simplify an expression for a complex constant using sympy?

I have done some calculations in sympy, and the result is in the end a set of constants. One of them is inserted directly into the snippet below:

from sympy import *
expr = (18**(Rational(1, 3))/(6*(3 + sqrt(3)*I)**(Rational(1, 3)))
        + 12**(Rational(1, 3))*(3 + sqrt(3)*I)**(Rational(1, 3))/12)
print(expr.evalf())
print(expr.simplify())

This returns

0.56857902130163 + 0.e-22*I
18**(1/3)/(6*(3 + sqrt(3)*I)**(1/3)) + (36 + 12*sqrt(3)*I)**(1/3)/12

so the expression appears to be a real number, yet sympy cannot simplify it further. With pen and paper, I have simplified this to

cos(pi/18) / sqrt(3)

which agrees with the numerical value returned by evalf().

I have tried many of the different simplification functions, but none seem to be able to reduce the expression any further. Using substitutions like

expr.subs(3 + sqrt(3)*I, sqrt(12) * exp(I*pi/6))

improves the expression, but still sympy is unable to conclude that it is real. Using Euler's formula for substitution,

expr.subs(3 + sqrt(3)*I, sqrt(12) * (cos(pi/6) + I*sin(pi/6)))

sympy is finally able to conclude that the expression is real, but the expression itself explodes in size when printed (even if I attempt simplify after the substitution).

Is there a better way to try to reduce this? I have many similar expressions for complex constants that I would like to know for sure are real (or not).

Upvotes: 9

Views: 9754

Answers (2)

user6655984
user6655984

Reputation:

The method as_real_imag often helps to simplify a complex number, even though it's not listed among the Simplification methods. In your example, 

expr.as_real_imag()

returns (sqrt(3)*cos(pi/18)/3, 0)

If a complex number is desired (rather than a tuple as above), one should not just call complex on this tuple, since this would create an object of Python complex class, involving numeric evaluation. Instead, I'd write

pair = expr.as_real_imag()
result = pair[0] + pair[1]*I

Upvotes: 5

user3717023
user3717023

Reputation:

For the expression you gave, the command

(expr.conjugate().conjugate() - expr.conjugate()).simplify()

returns 0, which signifies that expr is real. (Double application of conjugation returns to the original value, but it gets expanded along the way, which enables subsequent simplification.) In general, the above formula returns the imaginary part multiplied by 2i.

To find the real part of the expression, you can use a similar trick: add it to its conjugate and simplify (and divide by 2):

((expr.conjugate().conjugate()+expr.conjugate())/2).simplify()

returns sqrt(3)*cos(pi/18)/3.

Upvotes: 5

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