1 | initial version |
solns = [x == -I, x == I, x == -sqrt(3), x == sqrt(3)]
# We want to eliminate the I imaginaries from solns
realsols=[]
for sol in solns:
if(type(sol.rhs().n())==sage.rings.complex_number.ComplexNumber):
pass
else:
realsols.append(sol.rhs().n())
print(realsols)
# Or if you want symbolic values, drop the .n() and you will get -sqrt(3) and sqrt(3)
Answer:
[-1.73205080756888, 1.73205080756888] or
[-sqrt(3), sqrt(3)]
2 | No.2 Revision |
This strikes me as the easiest hack:
solns x, y, z = [x var('x, y, z')
P = solve([x^2 * y * z == -I, 18, x * y^3 * z == I, 24, x * y * z^4 == -sqrt(3), x == sqrt(3)]
# We want to eliminate the I imaginaries from solns
realsols=[]
6], x, y, z)
for sol solutions in solns:
P:
if(type(sol.rhs().n())==sage.rings.complex_number.ComplexNumber):
if 'i' not in str(solutions).lower():
pass
else:
realsols.append(sol.rhs().n())
print(realsols)
# Or if you want symbolic values, drop the .n() and you will get -sqrt(3) and sqrt(3)
Answer:
[-1.73205080756888, 1.73205080756888] or
[-sqrt(3), sqrt(3)]
print(solutions)