How to justify $a=(a,0)$ in Theorem $\mathbf{1.29}$ in Baby Rudin? Rudin says in page fourtheen in theorem 1.29 : If $a$  and $b$ are real, then $(a,b)=a+bi$. Proof he gives: $$a+bi=(a,0)+(b,0)(0,1)\\=(a,0)+(0,b)=(a,b)$$
of course this is correct (if we accept $a=(a,0)$), however i'm interested in how to justify that $a=(a,0)$, it seems entirely unrigorous to me, on one hand we have a pair, so it will be equal to $\{\{a\},\{a,0\}\}$ and on the other we have only one number, so how to make that rigorous?
 A: Earlier he proves that that $\mathbb{R} \subseteq \mathbb{C}$ by the embedding $x \mapsto (x,0)$; he also defines $i$ to be $(0,1)$ and shows that $i^2 = -1$. Theorem 1.29 simply says that an arbitrary element $(a,b) \in \mathbb{C}$ (indeed we defined $\mathbb{C}$ to be pairs of reals with certain operations making it a field) can be expressed as $a+b i$, where '$a$' denotes $(a,0)$, the canonical embedding of the real $a$ into $\mathbb{C}$.
It's a completely trivial result that only serves to justify the common notion of writing complex numbers as $a+bi$, $(a,b \in \mathbb{R})$. I believe you are  really overthinking this as there's no reason to consider the set-theoretic definition of an ordered pair.
A: A definition: A complex number is a number of the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, satisfying $i^2 = −1$.
$a$ is the real part of the complex number $a + bi$; the real number $b$ is  the imaginary part of $a + bi$.
So $a = a+0\cdot i = (a,0)$.
