As Robin as pointed out, for all primes $p$, $\mathbb{Q}_p$ is rigid, i.e., has no nontrivial automorphisms. It is sort of a coincidence that you ask, since I spent much of the last $12$ hours writing up some material on multiply complete fields which has applications here:
Theorem (Schmidt): Let $K$ be a field which is complete with respect to two inequivalent nontrivial norms (i.e., the two norms induce distinct nondiscrete topologies). Then $K$ is algebraically closed.
Corollary: Let $K$ be a field which is complete with respect to a nontrivial norm and not algebraically closed. Then every automorphism of $K$ is continuous with respect to the norm topology. (Proof: To say that $\sigma$ is a discontinuous automorphism is to say that the pulled back norm $\sigma^*|| \ ||: x \mapsto ||\sigma(x)||$ is inequivalent to $|| \ ||$. Thus Schmidt's theorem applies.
In particular this applies to show that $\mathbb{Q}_p$ and $\mathbb{R}$ are rigid, since every continuous automorphism is determined by its values on the dense subspace $\mathbb{Q}$, hence the identity is the only possibility. (It is possible to give a much more elementary proof of these facts, e.g. using the Ostrowski classification of absolute values on $\mathbb{Q}$.)
At the other extreme, each algebraically closed field $K$ has the largest conceivable automorphism group: $\# \operatorname{Aut}(K) = 2^{\# K}$: e.g. Theorem 80 of
http://alpha.math.uga.edu/~pete/FieldTheory.pdf.
There is a very nice theorem of Bjorn Poonen which is reminiscent, though does not directly answer, your other question. For any field $K$ whatsoever, and any $g \geq 3$, there exists a genus $g$ function field $K(C)$ over $K$ such that $\operatorname{Aut}(K(C)/K)$ is trivial. However there may be other automorphisms which do not fix $K$ pointwise.
There is also a sense in which for each $d \geq 3$, if you pick a degree $d$ polynomial $P$ with $\mathbb{Q}$-coefficients at random, then with probability $1$ it is irreducible and $\mathbb{Q}[t]/(P)$ is rigid. By Galois theory this happens whenever $P$ is irreducible with Galois group $S_d$, and by Hilbert Irreducibility the complement of this set is small: e.g. it is "thin" in the sense of Serre.
Addendum: Recall also Cassels' embedding theorem (J.W.S. Cassels, An embedding theorem for fields, Bull. Austral. Math. Soc. 14 (1976), 193-198): every finitely generated field of characteristic $0$ can be embedded in $\mathbb{Q}_p$ for infinitely many primes $p$. It would be nice to know some positive characteristic analogue that would allow us to deduce that a finitely generated field of positive characteristic can be embedded in a rigid field (so far as I know it is conceivable that every finitely generated field of positive characteristic can be embedded in some Laurent series field
$\mathbb{F}_q((t))$, but even if this is true it does not have the same consequence, since Laurent series fields certainly have nontrivial automorphisms).
The following answer was communicated to me by Keith Conrad:
See:
M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25
(1970), 638--641.
http://alpha.math.uga.edu/~pete/Isaacs70.pdf
Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are
relatively prime, then
$K(a,b)$ = $K(a+b)$, which answers the students question in the
affirmative. His proof shows the same conclusion holds under the
weaker assumption that
$[K(a,b):K] = [K(a):K][K(b):K]$.
since Isaacs uses the relative primality assumption on the degrees
only to get that degree formula above, which can occur even in cases
where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.
Best Answer
As soon as you get to general $$ z^3 = a + b i $$ this may be impossible. In solving a cubic $z^3 + p z + q = 0$ with, say, rational coefficients, one may use Cardano's formula. If there is only one real (irrational) root and two complex conjugate roots, then this works in the sense of being able to separately calculate real and imaginary parts using real square and cube roots. However, if there are three real irrational roots, a situation called CASUS IRREDUCIBILIS, then Cardano's formula is just a sum of cube roots of complex numbers $a+bi,$ with no way to separate real and imaginary parts. The terms are summed in a way that guarantees real answers, but that is not satisfying.
Now, if you begin with general $ z^3 = a + b i, $ say with $a,b \in \mathbb Q,$ and carefully write out equations for the real and imaginary parts of this $z,$ you get cubics with three real roots, where you cannot separate parts for those...it is all pretty circular, and hopeless.