I guess that since a real vector is an element of vector space R^n over R.

We can say that a complex vector is an element of vector space C^n over C.

Thus, if,

v=(a1+ib1,a2+ib2,...,an+ibn)

It conjugate vector is,

u=(a1-ib1,a2-ib2,...,an-ibn)

Thus,

vu

Is the product of all ye components.

Since, for all k from 1 to n

(ak+ibk)(ak-ibk)=(ak)^2+(bk)^2 element of R

We see that each component becomes real.