Symbolic definition for $i$?

It sounds like you have picked up on the fact that conjugation $$f: a+bi \mapsto a-bi$$ is a ring isomorphism where $i$ and $-i$ correspond to each other.

• Addition preserving: $f((a+bi)+(c+di))=f((a+c)+(b+d)i)=(a+c)-(b+d)i=(a-bi)+(c-di)=f(a+bi)+f(c+di)$
• Multiplication preserving: $f((a+bi)(c+di))=f((ac-bd)+(ad+bc)i)=(ac-bd)-(ad+bc)i=(a-bi)(c-di)=f(a+bi)f(c+di)$
• $f(1)=1$

Conjugation is clearly bijective too.

An idea: using the "identificacion" (in fact isomorphism, but making it a simple identification) of $\;\Bbb C\;$ with $\;\Bbb R^2\;$, we have that $\;z=x+iy \stackrel{\text{Ident.}}\sim (x,y)\;$, and thus

$$i\sim(0,1)\;,\;\;\text{whereas}\;\;-i\sim (0,-1)$$

and that's one way (a rather simple and algebraic one) to distinguish completely $\;i\;$ from $\;-i\;$ .

As @DanielFischer commented, $\mathbb C$ has the complex-conjugation automorphism that interchanges the two square roots of $-1$ (however we decide to label them). Thus, any algebraic relation (with real coefficients) that holds for one holds for the other. In different words, $\mathbb R(i)$ is constructed algebraically as $\mathbb R[x]/\langle x^2+1\rangle$, and the image of $x$ is a canonical square root of $-1$ in that model. (But/and, also, the image of $-x$ is another.) But/and I think this is not the type of distinction you want.

It gets worse in the Hamiltonian quaternions: there are infinitely-many square roots of $-1$, and they are all conjugate to each other in the quaternions.

Yes, if we choose to represent complex numbers as the real plane, we can label/name the square root of $-1$ that is in the upper half-plane "$i$". But, as in my first remark, flipping/interchanging upper and lower half-planes is an isomorphism of $\mathbb C$ to itself, and it is essentially impossible to distinguish... so this hasn't really accomplished anything.

For contrast, the case of $\sqrt{2}$ is somewhat different. It is the same, in the sense that the field $\mathbb Q[x]/\langle x^2-2\rangle$ is abstractly made by adjoining a square root of $2$ to $\mathbb Q$, because the image of $x$ in the quotient is a sort of canonical $\sqrt{2}$ in that model. BUT the distinction that matters in practice is that $\mathbb Q(\sqrt{2})$ admits two different imbeddings into $\mathbb R$, and in one the abstract square root of $2$ goes to $1.414...$ while in the other it is the negative of that. The "standard/canonical" square root of $2$ we usually refer to is actually the real number $1.1414...$, rather than an abstract one.