My take is that one uses complex numbers for the proof, but not for the theorem itself. The reason is that i2 can be thought of as the additive "destructor" of positive integers
(e.g., a), since a2 + (ia)2 = a2 - a2 = 0 .
Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2 , so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid