Symbolic simplification of algebraic expressions composed of complex numbers

I have a question concerning the symbolic simplification of algebraic expressions composed of complex numbers. I have executed the following Python script:

from sympy import *

expr1 = 3*(2 - 11*I)**Rational(1, 3)*(2 + 11*I)**Rational(2, 3)
expr2 = 3*((2 - 11*I)*(2 + 11*I))**Rational(1, 3)*(2 + 11*I)**Rational(1, 3)

print("expr1 = {0}".format(expr1))
print("expr2 = {0}\n".format(expr2))
print("simplify(expr1) = {0}".format(simplify(expr1)))
print("simplify(expr2) = {0}\n".format(simplify(expr2)))
print("expand(expr1) = {0}".format(expand(expr1)))
print("expand(expr2) = {0}\n".format(expand(expr2)))
print("expr1.equals(expr2) = {0}".format(expr1.equals(expr2)))

The output is:

expr1 = 3*(2 - 11*I)**(1/3)*(2 + 11*I)**(2/3)
expr2 = 3*((2 - 11*I)*(2 + 11*I))**(1/3)*(2 + 11*I)**(1/3)

simplify(expr1) = 3*(2 - 11*I)**(1/3)*(2 + 11*I)**(2/3)
simplify(expr2) = 15*(2 + 11*I)**(1/3)

expand(expr1) = 3*(2 - 11*I)**(1/3)*(2 + 11*I)**(2/3)
expand(expr2) = 15*(2 + 11*I)**(1/3)

expr1.equals(expr2) = True

My questions is why the simplifications does not work for expr1 but
works for expr2 thoug the expressions are algebraically equal.
What has to be done to get the same result from simplify for expr1 as for expr2?

Thanks in advance for your replys.

Kind regards

Klaus

You can use the minimal polynomial to place algebraic numbers into a canonical representation:

In [30]: x = symbols('x')

In [31]: p1 = minpoly(expr1, x, polys=True)

In [32]: p2 = minpoly(expr2, x, polys=True)

In [33]: p1
Out[33]: Poly(x**2 - 60*x + 1125, x, domain='QQ')

In [34]: p2
Out[34]: Poly(x**2 - 60*x + 1125, x, domain='QQ')

In [35]: [r for r in p1.all_roots() if p1.same_root(r, expr1)]
Out[35]: [30 + 15⋅ⅈ]

In [36]: [r for r in p2.all_roots() if p2.same_root(r, expr2)]
Out[36]: [30 + 15⋅ⅈ]

This method should work for any two expressions representing algebraic numbers through algebraic operations: either they give the precise same result or they are distinct numbers.

It works (but nominally) for expr1 because when the product in the radical is expanded you get the cube root of 125 which is reported as 5. But SymPy tries to be careful about putting radicals together under a common exponent, an operation that is not generally valid (e.g. root(-1, 3)*root(-1,3) != root(1, 3) because the principle values are used for the roots. But if you want the bases to combine under a common exponent, you can force it to happen with powsimp:

>>> from sympy.abc import x, y
>>> from sympy import powsimp, root, solve, numer, together
>>> powsimp(root(x,3)*root(y,3), force=True)
(x*y)**(1/3)

But that only works if the exponents are the same:

>>> powsimp(root(x,3)*root(y,3)**2, force=True)
x**(1/3)*y**(2/3)

As you saw, equals was able to show that the two expressions were the same. One way this could be done is to solve for root(2 + 11*I, 3) and see if any of the resulting expression are the same:

>>> solve(expr1 - expr2, root(2 + 11*I,3))
[0, 5/(2 - 11*I)**(1/3)]

We can check the non-zero candidate:

>>> numer(together(_[1]-root(2+11*I,3)))
-(2 - 11*I)**(1/3)*(2 + 11*I)**(1/3) + 5
>>> powsimp(_, force=True)
5 - ((2 - 11*I)*(2 + 11*I))**(1/3)
>>> expand(_)
0

So we have shown (with force) that the expression was the same as that for which we solved. (And, as Oscar showed while I was writing this, minpoly is a nice candidate when it works: e.g. minpoly(expr1-expr2) -> x which means expr1 == expr2.)