Question 1208325
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True or False: If S, P, and A are the cube roots of a complex number, then Arg(S)+Arg(P)=2×Arg(A).
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<pre>
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  [{{{0}}},{{{2pi}}}),  while arg can differ from Arg by any integer multiple of {{{2pi}}}.


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, {{{2pi/3}}}  and {{{4pi/3}}}.

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

    s = 0,  p = {{{2pi/3}}} , a = {{{4pi/3}}},


where s = Arg(S), p = Arg(P), a = arg(A), then the equality s + p = 2a  becomes  0 + {{{2pi/3}}} = {{{2*(4pi/3)}}},
which does not work (is FALSE).  It works ONLY if we consider the equality by the modulo {{{2pi}}}.


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 {{{2pi}}}.


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 {{{2pi}}}).


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 + {{{2pi/3}}},  a = s + {{{4pi/3}}}.


So , from one side,        Arg(S) + Arg(P) = s + p = s + (s + {{{2pi/3)}}} = 2s + {{{2pi/3}}};

     from the other side,  2*Arg(A) = 2*(s + {{{4pi/3)}}} = 2s + {{{8pi/3}}}.


But  {{{8pi/3}}}  is the same as  {{{2pi/3}}}  modulo {{{2pi}}}.  Thus, we proved that

     Arg(S) + Arg(P) = 2*Arg(A)  (mod {{{2pi}}})     (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.
</pre>

Solved.


But please note that it is valid universally ONLY for the modified formulation,
where the equality is by modulo  {{{2pi}}}.


Otherwise, it is EITHER invalid, OR is invalid as a universal equality for any/arbitrary order of the cube roots
(and, respectively, for any/arbitrary order of their principal arguments).