determine if a subfield of complex numbers I'm almost certain that this is a field but I'm not sure if I've missed anything not so obvious.

Let $F\subset C$ be the set of complex numbers of the form
$$z=a+ib$$,   $$a,b\in Q$$
is $F$ a subfied of $C$

 A: It is not a subfield: for instance $\pi+\dfrac i\pi\in F$, but 
$$\Bigl(\pi+\dfrac i\pi\Bigr)^{\!2}=\pi^2-\frac1{\pi^2}+ 2i\notin F,$$
as it would imply $\pi$ is an algebraic number.
Edit:
After the update of the question, if the condition is ‘$a$ and $b\in\mathbf Q$’, $F$ is indeed  a subfield: since $\mathbf Q$ is itself a subfield, we have
$$a, b, a',b'\in\mathbf Q\Rightarrow \begin{cases}a+a',\, b+b'\in\mathbf Q,\\
aa'-bb',\, ab'+ba'\in\mathbf Q\\\dfrac{a}{a^2+b^2},\,\dfrac{-b}{a^2+b^2}\in\mathbf Q.\end{cases}$$
This subfield is denoted $\mathbf Q(i)$. It is the field  of fractions of the ring of Gaussian integers
$$\mathbf Z[i]=\bigl\{a+ib\mid a,\,b\in\mathbf Z\bigr\}.$$
A: $G \subset H$ and $H$ is a field, then $G$ "inherits" associativity, commutativity, distribution, identities and inverses for $H$.  So to show $G$ is a field it only remains to show $G$ is closed under multiplication and addition, that the identities and inverses are in $G$
$\mathbb C$ is a field and $1 + 0i$ is multiplicative identity and $0+ 0i$ is the additive identity.
Just need to show that is is closed under addition and multiplication, and that the inverses are in $F$.
So if $a+bi,c+di \in F$ then $a,b,c,d\in \mathbb Q$ and ....
$(a + bi) + (c + di)= (a+c) +(b+d)i\in F$ 
$(a+bi)(c+di) = (ac - bd) + (bc + ad)i \in F$. 
$-(a+bi) = -a - bi \in F$.
And $(a+b)^{-1} = \frac {a}{a^2 + b^2} - \frac {b}{a^2 + b^2} i \in F$.
So is a field.
A: Observe
\begin{align*}
\frac{1}{a + b i} & = \frac{a - bi}{(a + bi)(a - bi)} \\
& = \frac{a - bi}{a^2 + b^2} \\
& = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2} i.
\end{align*}
If $a, b$ are rational, then so will be the coefficients above. Observe that $\mathbb{Q}$ could be replaced with any subfield of $\mathbb{C}$.
