complex2

|a| = |b| = |c| = 1
aa' = bb' = cc' = 1 (a' = conjugate of a)
So,
a + b + c = 0
(a + b + c)' = 0
a' + b' + c' = 0
1/a + 1/b + 1/c = 0
(bc + ca + ab)/abc = 0
ab+bc+ca=0

Sorry, I am not sure if my answer is correct.
You can treat it as a reference.

In order to let you see clearly, I change them to capital alphabets.
AB + BC + CA = 0
'A' in AB is a double figure.
That means if A=5, B=4, AB=54
So The value of A in AB is actually 10 times.
Same as 'B' in BC and 'C' in CA, they are also double figures.

AB + BC + CA
= (10A + B) + (10B + C) + (10C + A)
= 10A + B + 10B + C + 10C + A
= 10A + A + 10B + B + 10C + C
= 11A + 11B + 11C
= 11(A + B + C)......(1)

Since A + B + C = 0.
Therefore Substitute A + B + C = 0 into (1),
AB + BC + CA = 11(0)
AB + BC + CA = 0 (proved)