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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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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.
Solved.
But please note that it is valid universally ONLY for the modified formulation,
where the equality is by modulo .
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).