$a+bi$ is algebraic over $\mathbb{Q}$ iff $a$ and $b$ are algebraic over $\mathbb{Q}$. A complex number $a + bi$($a$,$b\in \mathbb{R}$,$i^2=-1$) is algebraic over $\mathbb{Q}$ if and only if $a$ and $b$ are algebraic over $\mathbb{Q}$.

Actually,I want to know how to get $a$ and $b$ are algebraic over $\mathbb{Q}$ from $a+bi$ is algebraic over $\mathbb{Q}$.
Consider $F=\mathbb{Q}(a,b,i)$, this is an finite extension of $\mathbb{Q}$ (prove it, it's not difficult, use that for finite extensions $L/H$ and $H/K$ you have that $[L:K]=[L:H][H:K]$, and a finite extension is in particular an Algebraic extension), so $F$ is algebraic over $\mathbb{Q}$. Therefore, every element of $F$ is an algebraic element over $\mathbb{Q}$, so $a+ib\in F$ is.

On the other hand, if $a+ib$ is algebraic over $\mathbb{Q}$, we have that if $b=0$ the result follows, so we suppose that $b\neq 0$. Let $f(x)\in\mathbb{Q}[x]$ the minimal monic polynomial associated with $a+ib$ (over $\mathbb{Q}$), so $a-ib$ is a root of $f(x)$, so $a-ib$ is algebraic over $\mathbb{Q}$, then $((a+ib)+(a-ib))/2=a$ is algebraic over $\mathbb{Q}$ (the set of algebraic elements over a field is a field). Now, in the same way $((a+ib)-a)/i=b$ is algebraic over $\mathbb{Q}$.
Hint: Investigate sums, products, and complex conjugates of algebraic numbers.