Solution: Solution: A. Take z 1= 1, z2=i, z3=0, which satisfies z 12+z22+z32=0, but z 1=z2=z3=0 is not valid;

B let z 1=2-i, z2=3-i, z3=5+i, which satisfies z 12+z22+z32 > 0, but

z

2

1

+

z

2

2

= 1 1- 10i-

z

2

three

=-24- 10i, both of which are complex numbers, and the sizes cannot be compared, so z 12+z22 >-z32 does not hold;

C.∫ satisfies z 12+z22 >-z32, ∴

z

2

1

+

z

2

2

and

z

2

three

Are all real numbers, ∴ z 12+z22+z32 > 0;

D let z 1=a+bi(a, b∈R), √.

.

z 1

=-z 1, ∴a-bi=-a-bi, ∴a=0, ∴z 1 are imaginary numbers, not necessarily pure imaginary numbers, so they are incorrect.

To sum up, only C is correct.

So choose: C.