Is it possible to use set-builder notation to define the reals using the complex numbers?

The set of complex numbers can be defined using the reals:

$$\mathbb{C}=\{a+bi\,|\,a,b\in\mathbb{R}\}.$$

Could I do the opposite and define the reals using the complex numbers?

$$\mathbb{R}=\{z\,|\,z\in\mathbb{C},\,\operatorname{Im}(z)=0\}.$$

Yes, you could, in principle. The reason we usually go the other way is because we already know what the reals are (namely "Cauchy sequences of rationals", or "Dedekind cuts"), so going $$\mathbb{R}$$ to $$\mathbb{C}$$ allows us to build a more complicated object from something we already know about. There's no reason to go $$\mathbb{C}$$ to $$\mathbb{R}$$, because in order to construct $$\mathbb{C}$$ you must have constructed $$\mathbb{R}$$ already.
So if you have somehow already defined $$\mathbb{C}$$ then you could use your definition as a definition of $$\mathbb{R}$$; but otherwise you can't, because your "definition" would be circular.
• You could try to get around this by instead defining $\Bbb C$ as the only algebraically complete uncountable field, or something like that. (I'm not even sure if that description is accurate, or provable without encountering $\Bbb R$ along the way, e.g. as the only complete ordered field.)
Yes, you can represent the real numbers like this. If you are pedantic, you may want to distinguish between the real number $$x$$ and the complex number $$x + 0i$$. If this is a concern, you can write $$\mathbb{R} = \bigl\{ \mathrm{Re}(z) \bigm| z \in \mathbb{C} \bigr\}.$$
You could, see the caveat by Patrick, though. Another way would be to say $$\Bbb R=\{z \in \Bbb C: \bar{z}=z\}$$ or $$\Bbb R=\{z \in \Bbb C: |z|=z \lor |z|=-z \}$$ and a few others.