let a = sqrt(7+4*sqrt(3)) prove that a is algebraic over Q
now let a = sqrt(7) + i show that a is now algebraic over R
You just gotta find a polynomial over the respective field such that a is a root. Just takes some playing around with the numbers.
1)
so has a as a root. Dunno if it is irreducible or not, but you didn't ask for the degree of the extension. Clearly either 2 or 4 depending on whether p(x) factors into square terms. (Rational root test shows no linear terms).
2)We are in now, so in particular, .
Consider so a is a root to This pretty much has to be minimal as i is in complex and is a degree 2 extension over the reals, so it should be a degree 2 extension over the reals.
Hope this helped