A Complex Problem

I believe the answer may be wrong, or my solution may have a problem. Here it goes:

The magnitude of $a, b, c$ is $1$ so, $\overline{a}a = \overline{b}b = \overline{c}c = 1 \Rightarrow$

We also have $|a+b+c|^2=(a+b+c)\overline{(a+b+c)} = ... = 3 + a\overline{b} +a\overline{c} + b\overline{a} + b\overline{c} + c\overline{a} + c\overline{b} = X ^2$

Using the initial equalities we get:

$X^2= 3 + \frac {a}{b} +\frac {a}{c} +\frac {b}{a} +\frac {b}{c} + \frac {c}{a} +\frac {c}{b} = 3 + \frac{3}{3} \cdot \frac{a^2c +a^2b +b^2c +b^a +c^2b +c^2a}{abc} =$

$3 +\frac{(a+b+c)^3 - (a^3+b^3+c^3+6abc)}{3abc} =$

$3 + \frac{(a+b+c)^3-5abc}{3abc} =$

$\frac {4}{3}+ \frac{(a+b+c)^3}{3abc}$

But, $\frac{(a+b+c)^3}{3abc}$ should be a real number so,

$\frac{(a+b+c)^3}{3abc} = \frac{|(a+b+c)|^3}{3|abc|} =\frac{|(a+b+c)|^3}{3} = \frac{X^3}{3}$

We have:

$X^2 = \frac {4}{3} + \frac{X^3}{3}$

Which gives $X= 2,-1$

The grader accepted 3 as an answer but we cannot accept the $-1$ value as magnitude.

If you find an error in my solution please, tell me to edit it.

:)

For magnitudes of 1, only angles are considered.

Satisfied p {0, -60, 60} and q {0, 180, 180} makes | a + b + c | of | 2 | and | -1 | respectively for 2 + 1 = 3.

Note that -180 $^\circ$ and 180 $^\circ$ are of equal place and therefore only 180 $^\circ$ can be taken as sole definition.

1
2
3
4
5
6
7

0   1
-60 0.5-0.866025403784439i
60  0.5+0.866025403784439i

0   1
180 -1
180 -1

Where magnitudes of 2 and 1 can be there as p and q, we don't take -60 $^\circ$ and 60 $^\circ$ as ordered in their set. Just try your own ways to evaluate the two sets to your favorites.

Answer: $\boxed{3}$