Solving complex coefficient polynomials using sympy

I am very new to python, so forgive me if this has a simple fix. I am trying to solve polynomials with complex coefficients using sympy. I find that I get an empty output if k is "too complicated" ... I'm not quite sure how to determine what this means. As a first example, consider this fourth-order polynomial with complex coefficients,

In [424]: solve(k**4+ 2*I,k)
Out[424]: 
[-2**(1/4)*sqrt(-sqrt(2)/4 + 1/2) - 2**(1/4)*I*sqrt(sqrt(2)/4 + 1/2),
 2**(1/4)*sqrt(-sqrt(2)/4 + 1/2) + 2**(1/4)*I*sqrt(sqrt(2)/4 + 1/2),
 -2**(1/4)*sqrt(sqrt(2)/4 + 1/2) + 2**(1/4)*I*sqrt(-sqrt(2)/4 + 1/2),
 2**(1/4)*sqrt(sqrt(2)/4 + 1/2) - 2**(1/4)*I*sqrt(-sqrt(2)/4 + 1/2)]

no problem getting the result. I am interested, however, in solving something like

In [427]: solve(k**6 + 3*I*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1,k)
Out[427]: []

which is much more complicated and returns an empty list. I can, however, solve this using maple, for example. Also note that there are no problems removing complex odds,

In [434]: solve(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1,k)
Out[434]: 
[CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 0),
 CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 1),
 CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 2),
 CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 3),
 CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 4),
 CRootOf(k**6 + 3*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1, 5)]

Elements of the resulting array can be calculated numerically.

, , ? , [427]?

nsolve() , .

+4
1

comment Stelios sympy.polys.polytools.nroots:

>>> from sympy import solve, nroots, I
>>> from sympy.abc import k
>>> solve(k**6 + 3*I*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1,k)
[]
>>> nroots(k**6 + 3*I*k**5 - 2*k**4 + 9*k**3 - 4*k**2 + k - 1)
[-2.05972684672 - 0.930178254620881*I, -0.0901851681681614 + 0.433818575087712*I, -0.0734840785305346 - 0.434217215694685*I, 0.60726931721974 - 0.0485101438937812*I, 0.745127208196241 + 0.945593905069312*I, 0.870999568002712 - 2.96650686594768*I]
+1

Source: https://habr.com/ru/post/1653190/


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