SOLUTION: True or False: If S, P, and A are the cube roots of a complex number, then Arg(S)+Arg(P)=2×Arg(A).

Mathematical symbol Arg (written with capital "A") is the principal value of the argument 
of a complex number. In distinction from "arg" (written with lowercase  "a"), Arg is always
in the interval  [,),  while arg can differ from Arg by any integer multiple of .


Let's consider  first a simplest case of cube roots S, P and A of the number 1 (real one).


They have principal arguments  0,   and .

If to take these cube roots and their principal arguments in this order 

    s = 0,  p =  , a = ,


where s = Arg(S), p = Arg(P), a = arg(A), then the equality s + p = 2a  becomes  0 +  = ,
which does not work (is FALSE).  It works ONLY if we consider the equality by the modulo .


If we take the roots and the sum of their principal arguments in other order, like s + a = 2p, we see that it works.


So, in order for the equality  s + p = 2a  be universal and work for any order of addends,
the formulation of the problem must be changed. The equality  s + p = 2a  should be considered by modulo .


Then this modified statement will be true for any permutations (and for any order) of cube roots.

So, the correct formulation in this problem MUST BE

    Arg(S) + Arg(P) = 2*Arg(A)  (mod ).


In this formulation, the statement of the problem is true independently of the order of addends.


For the proof in the general case, let S be the cube root of the complex number with the smallest principal argument s;
let P be the cube root of the complex number with the intermediate principal argument p;
and let A be the cube root of the complex number with the greatest principal argument "a".


Then, according to the de Moivre formula

    p = s + ,  a = s + .


So , from one side,        Arg(S) + Arg(P) = s + p = s + (s +  = 2s + ;

     from the other side,  2*Arg(A) = 2*(s +  = 2s + .


But    is the same as    modulo .  Thus, we proved that

     Arg(S) + Arg(P) = 2*Arg(A)  (mod )     (1)


in this order.


Furthermore, it is obvious, that equality (1) is valid for any permutation of the roots;
so, it is valid UNIVERSALLY for any order of the principal arguments.


At this point the proof and the solution are completed.