# How to find arbitrary complex constants in symbolic expressions?

How do I go about finding terms in a symbolic expression containing arbitrary complex constants? I had hoped that the code fragment below would yield something like [2Ix], [Ix], [I], and [2I]. However, instead it yields [], [I*x], [], and []. Is there another way to isolate the terms containing an arbitrary complex constant?

w0 = SR.wild(0)

f = x^2 + 2*I*x + 1
g = x^2 + I*x + 1
h = x^2 + 2*x + I
k = x^2 + x + 2*I

print(f.find(I*w0))
print(g.find(I*w0))
print(h.find(I*w0))
print(k.find(I*w0))

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Edited. This was the original answer

It seems that, in your expressions, $x$ denotes a real number. If this were the case, you can get the coefficients of I by taking the imaginary part of the expression:

sage: f = x^2 + 2*I*x + 1
sage: g = x^2 + I*x + 1
sage: h = x^2 + 2*x + I
sage: k = x^2 + x + 2*I
sage: assume(x,'real')
sage: for expression in [f, g, h, k]:
....:     print(expression.imag())
2*x
x
1
2


New answer. I have improved your approach, based on wildcards. It seems that I is not detected because it is not a symbolic variable. So, we can temporarily transform I into such a variable, find the coefficients of I in the expressions and then restore I:

w0 = SR.wild(0)
var("I")
f = x^2 + 2*I*x + 1
g = x^2 + I*x + 1
h = x^2 + 2*x + I
k = x^2 + x + 2*I
m = sqrt(1-x^2) + I*(1-x^3)^(1/3)
n = sqrt(1-x^2) + 2*I*(1-x^3)^(1/3)
p = x^2 + 2*x - I
q = x^2 + 2*x

expressions = [f, g, h, k, m, n, p, q]
for expression in expressions:
print("\nExpression: ",expression)
if expression.has(I):
if expression.has(w0*I):
aux = expression.find(w0*I)
coef = aux[0].match(w0*I)[w0]
else:
coef = 1
print(f"The coefficient of I is {coef}")
else:
print("The expression does not contain I")
restore("I")


This is the output:

Expression:  2*I*x + x^2 + 1
The coefficient of I is 2*x

Expression:  I*x + x^2 + 1
The coefficient of I is x

Expression:  x^2 + I + 2*x
The coefficient of I is 1

Expression:  x^2 + 2*I + x
The coefficient of I is 2

Expression:  (-x^3 + 1)^(1/3)*I + sqrt(-x^2 + 1)
The coefficient of I is (-x^3 + 1)^(1/3)

Expression:  2*(-x^3 + 1)^(1/3)*I + sqrt(-x^2 + 1)
The coefficient of I is 2*(-x^3 + 1)^(1/3)

Expression:  x^2 - I + 2*x
The coefficient of I is -1

Expression:  x^2 + 2*x
The expression does not contain I

more

Thanks for the answer. You are correct that I meant x to be a real number. I thought of this approach too, but unfortunately, it breaks down when the expression becomes a more complex. Here is an example.

w0 = SR.wild(0)

f = sqrt(1-x^2) + I*(1-x^3)^(1/3)
g = sqrt(1-x^2) + 2*I*(1-x^3)^(1/3)

assume(x, 'real')

print(f.find(I*w0))
print(g.find(I*w0))

print(f.imag())
print(g.imag())


In this example, .find(I*w0) yields the term containing the cube root from f, but not from g. However, .imag() yields complicated expressions for both f and g because it takes into account that abs(x) may be greater than 1.

( 2020-06-27 22:53:35 +0200 )edit

Thanks again. The updated approach works when we know the expressions up front. However, one step more difficult is when they result from a computation. Here is another example. Interestingly enough, when solving the equation for x, the .find(I*w0) approach works. However, when solving for y, it doesn't.

w0 = SR.wild(0)

var('y')
assume(x, 'real')
assume(y, 'real')

eqn = (x^3 + y^3 / 27 == 1)

sols = solve(eqn, x)

f, g, h = [sol.rhs() for sol in sols]

expressions = [f, g, h]

for expr in expressions:
print(expr.find(I*w0))

sols = solve(eqn, y)

f, g, h = [sol.rhs() for sol in sols]

expressions = [f, g, h]

for expr in expressions:
print(expr.find(I*w0)

( 2020-06-28 12:59:16 +0200 )edit

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Last updated: Jun 28 '20