Consider :
sage: reset() sage: z=var("z") sage: Eq=(3+4*I)^2-2*(z.conjugate())==z ; Eq
-2*conjugate(z) + 24*I - 7 == z sage: Sol=solve(Eq, z) ; Sol [z == (8*I -
7/3)] sage: Eq.subs(Sol[0]) (40*I - 7/3) == (8*I - 7/3) sage:
bool(Eq.subs(Sol[0])) False
which annoys me *immensely*. It reeks of conjugate() error, which, AFAICT
is ginac territory…
However :
sage: E2=(Eq.lhs()-Eq.rhs()).subs(z=x+I*y) ; E2 -3*x + I*y + 24*I - 7 sage:
S2=solve([E2.real(), E2.imag()], [x, y]) ; S2 [[x == (-7/3), y == -24]]
sage: E2.subs(S2[0]) 0 sage: Eq.subs(z==x+I*y).subs(S2[0]) (-24*I - 7/3) ==
(-24*I - 7/3)
This failure to solve a *simple* equation on complexes is troubling.
Re-reading myself did not led me to “obvious” problems.
Does this deserve a critical ticket ?
BTW, there is currently no *free* alternative :
sage: solve(Eq, z, algorithm="sympy") ConditionSet(z, Eq(-z -
2*conjugate(z) - 7 + 24*I, 0), Complexes) sage: solve(Eq, z,
algorithm="giac") Warning, the test a==b is performed by checking that the
internal representation of regroup(a-b) is not 0. Therefore a==b may return
false even if a and b are mathematically equal, if they have different
internal representations. You can explicitly call a simplification function
like simplify(a-b)==0 to avoid this. Warning, argument is not an equation,
solving false=0 [z] sage: solve(Eq, z, algorithm="fricas") [z == (8*I -
7/3)] sage: solve(Eq, z, algorithm="maxima") [z == (8*I - 7/3)] sage:
[[u[1].sage()==u[2].sage() for u in s] for s in mathematica.Solve(Eq, z)]
[[z == (-24*I - 7/3)]]
HTH,
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